ANT-intro

8.7. EXERCISES 351

8.7 Exercises

Analysis

Exercise 8.7.1† (Vandermonde’s Identity). Let x and y be p-adic integers. Prove that

x+y = x y
j
ki
i+j=k,i,j≥0

for any k.

Solution

When x and y are natural integers, this follows from considering the coefficient of Xk in (X +
1)x+y = (X + 1)x(X + 1)y. For arbitrary p-adic integers, this follows from the density of N in
Zp.

Exercise 8.7.2† (Mahler’s Theorem). Prove that a function f : Zp → Qp is continuous if and only if

there exist ai → 0 such that ∞x
f (x) = ai i

i=0

for all x ∈ Zp. These ai are called the Mahler coefficients of f . Moreover, show that max(|f (x)|p) =
max(|ai|p).

Solution

It is clear that any such function is continuous on Zp, hence we need to prove that the reverse

holds as well. Let ∆f = x → f (x + 1) − f (x) denote the discrete derivative operator from

Exercise A.3.6†. The coefficients ak are then ∆kf (0): indeed, a straightforward shows that
n n ∞ x
f (n) = k=0 ak k for any n ∈ N, so if these ak go to 0, f must be equal to x → x=0 ak k

by density and continuity.

Thus, it only remains to show that ∆kf (0) → 0. To prove this, we will show that they eventually

all become divisible by p. We can then subtract p ∆kf(0) ∆kf (0) x from f (x) and divide
k

everything by p to conclude that p2 | ∆kf (0) for large k. Iterating this process yields that

vp(∆kf (0)) → +∞ as desired.

To show this, let N be such that p | f (x + pN ) − f (x) for any x. There exists such an N since f
is continuous by assumption. Then, by Exercise A.3.7†,

N pN f (x + k)

∆pN f (x) = (−1)pN −k k

k=0

for any x ∈ Zp. Now, by Frobenius, (1 + X)pN ≡ 1 + XpN (mod p) which means that p | pN
k
for any 1 ≤ k ≤ pN − 1. Hence,

∆pN f (x) ≡ f (x + pN ) + (−1)pN f (x) (mod p).

When p is odd this is f (x + pN ) − f (x) which is divisible by p by construction, and when p is even
the same holds since −1 ≡ 1. Hence, p | ∆pN f (x) for all x ∈ Zp which implies that p | ∆nf (x)
for all n ≥ pN as well by applying ∆ multiple times to ∆pN f (x). In particular, p | ∆nf (0) for

sufficiently large n as wanted.

352 CHAPTER 8. P -ADIC ANALYSIS

Exercise 8.7.4†. Prove that the following power series converge if and only if for |x|p < 1 and
|x|p < p−1/(p−1) respctively:

logp(1 + x) = ∞ (−1)k−1xk , expp(x) = ∞ xk
.
k=1 k k=0 k!

In addition, prove that

1. expp(x + y) = expp(x) expp(y) for |x|p, |y|p < p−1/(p−1).

2. logp(xy) = logp(x) + logp(y) for |x|p, |y|p < 1

3. expp(log(1 + x)) = 1 + x for |x|p < p−1/(p−1).

4. logp(exp(x)) = x for |x|p < p−1/(p−1).

Solution

We shall only prove the convergence, the claimed equalities follow from the general theory of
power series: if g(x), (f ◦ g)(x) and f (g(x)) all converge, we have (f ◦ g)(x) = f (g(x)) (this is
even easier over Qp because we have the strong triangle inequality). The convergence for logp
follows from the fact that that |xk/k|p = |x|pk/|k|p goes to 0 when |x|p < 1 since |k|p > 1/k, but
does not go to 0 when |x|p = 1 since |k|p ≤ 1 for all k.

The convergence for expp is very similar: by Legendre’s formula,

vp (xk /k!) = kvp(x) − p k 1 + sp(k) = k vp(x) − p 1 1 + o(k)
− p−1 −

where o(k)/k → 0. This forces vp(x) ≥ 1 , i.e. |x|p ≤ p−1/(p−1). Finally, we need to see that
p−1

we can’t have equality. This is easy: when vp(x) = 1 , vp (xk /k!) is sp (k) which is bounded
p−1 p−1

when k is a power of p, so does not go to infinity.

Exercise 8.7.5†. Prove that

n 2k → ∞.
v2 k

k=1

Solution

The problem is equivalent to showing that ∞ 2k = 0 in Q2. Note that this sum is exactly
k=1 k
log2(−1), which is 1/2 log2(1) = log2(1) = 0 by Exercise 8.7.4†.

Exercise 8.7.6† (Mean Value Theorem). Let f (x) = ∞ ai xi be a p-adic power series converging
i=0

for |x|p ≤ 1, i.e. ai → 0. Prove that

|f (t + h) − f (t)|p ≤ |h|p max(|ai|p)

i

for any |t|p ≤ 1 and |h|p ≤ p−1/(p−1).

8.7. EXERCISES 353

Solution

We shall prove that |(t + h)n − tn|p ≤ |h|p for any |t|p ≤ 1 and |h|p ≤ p−1/(p−1). The strong
triangle inequality then implies that

∞ ∞∞

ai(t + h)i − aiti = ai((t + h)i − ti)

i=0 i=0 i=0

≤ max(|ai((t + h)i − ti)|p)

i

≤ |h|p max(|ai|p)

i

as wanted. Our claim is however very easy to prove: since |h|p ≤ p−1/(p−1), we have |hk/k!|p ≤ 1
by Legendre’s formula so that

n

(t + h)n − tn = tn−kn(n − 1) · . . . · (n − (k − 1))hk/k!

k=0

has absolute value at most |h|p by the strong triangle inequality.

Absolute Values

Exercise 8.7.7†. We say an absolute value | · | over a field K, i.e. a function | · | → R≥0 such that

• |x| = 0 ⇐⇒ x = 0

• |x + y| ≤ |x| + |y|

• |xy| = |x| · |y|

is non-Archimedean if the sequence |m| ≤ 1 for all m ∈ Z and Archimedean otherwise. Prove that m
is non-Archimedean if and only if it satisfies the strong triangular inequality |x + y| ≤ max(|x|, |y|)
for all x, y ∈ K. In addition, prove that, if | · | is non-Archimedean, we have |x + y| = max(|x|, |y|)
whenever |x| = |y|.

Solution

It is clear that | · | is non-Archimedean if it satisfies the strong triangle inequality. Thus, suppose
that |m| ≤ 1 for all m ∈ Z. Now, notice that, for any positive integer n,

|x + y|n = |(x + y)n|

n n xkyn−k
k
=

k=0

n n |x|k|y|n−k
k
=

k=0

≤ n max(|x|, |y|)n.

Taking the limit as n goes to ∞, we get

|x + y| ≤ n1/n max(|x|, |y|) → max(|x|, |y|)

as wanted. For the equality, if |x| > |y|, note that, by the same inequality, we also have |x| ≤
max(| − y|, |x + y|). Since | − y| = |y| < |x|, we must have max(|x + y|, | − y|) = |x + y| so
|x + y| ≥ |x| ≥ |x + y| as wanted.

354 CHAPTER 8. P -ADIC ANALYSIS

Exercise 8.7.8†. Let K be a field and let | · | : K → R≥0 be a multiplicative function which is an
absolute value on Q. Suppose that | · | satisfies the modified triangular inequality |x + y| ≤ c(|x| + |y|)
for all x, y ∈ K, where c > 0 is some constant. Prove that it satisfies the triangular inequality.

Solution

The argument is very similar to our proof of Exercise 8.7.7†. Let x, y be elements of K. For any
positive integer n,

|x + y|n = |(x + y)n|

n n xn−kyk
k
≤c

k=0

n n |x|n−k|y|k
k
=c

k=0

n n |x|n−k|y|k
k
≤c

k=0

= c(|x| + |y|)n.

Indeed, a straightforward induction shows that |m| ≤ m for m ∈ N since | · | is an absolute value
on Q so |m + 1| ≤ |m| + |1| = |m| + 1 for any m ∈ N since |1|2 = |1| and |1| = 0. Taking the nth
root and letting n tend to infinity, we get

|x + y| ≤ c1/n(|x| + |y|) → |x| + |y|

as wanted.

Exercise 8.7.9† (Ostrowski’s Theorem). Let | · | be an absolute value of Q. Prove that | · | is equal to

| · |pr for some prime p and some r ≥ 1, or to | · |r∞ for some 0 < r ≤ 1 or is the trivial absolute value
| · |0 which is 0 at 0 and 1 everywhere else.

Solution

First, note that f (1)2 = f (1) so f (1) = 1 since f (x) = 0. For the same reason, f (−1) = 1. Now,

suppose that there is some a ∈ N such that |a| > 1 and let b ∈ N be any integer. By the previous

remark, we have a > 1 so let am = n logb(a) aibi. We get
i=0

n logb(a)

|a|m ≤ |ai||b|i

i=0

which implies that |b| > 1 as well. But then,

m logb(a)

|a|m ≤ |ai||b|i ≤ C|b| n logb(a)

i=0

for some constant C = max(|1|, |2|, . . . , |b − 1|) > 0 which implies that |a| ≤ |b|logb(a) when we
take m → ∞, i.e. |a|1/ log a ≤ |b|1/ log b. Since the reverse inequality is true as well by symmetry,
we get that |a|1/ log a = c is constant. This gives us |a| = alog c := ar. It is then easy to see that
this extends to |a|∞r on all of Q using the multiplicativity of | · |. Finally, it is easy to check that
this satisfies the triangular inequality only for 0 < r ≤ 1.

Now suppose that |n| ≤ 1 for all n ∈ Z. By Exercise 8.7.7†, | · | satisfies the strong triangle
inequality. Without loss of generality, assume that | · | is non-trivial and let p ∈ N be the smallest

8.7. EXERCISES 355

positive integer such that |p| < 1. Since | · | is multiplicative, p must be prime as it has no
non-trivial divisor and is distinct from 1. By assumption, |a| = 1 for any 1 ≤ a ≤ p − 1. We shall
prove that |n| = 1 for any p n to conclude that, in general,

|n| = |p|vp(n)|n/p| = |p|vp(v) = |n|−p logp |p|

Consider any p n now and express it in base p as i aipi. Since p n, we have a0 < p, so
1 = |a0| > maxi≥1 |aipi|. By the previous inequality, we are in the equality case of

aipi ≤ max |aipi| = 1

i
i

so |n| = 1 as wanted. To conclude, it is this time easy to see that |x + y| ≤ max(|x|, |y|) only
when r ≥ 1.

Exercise 8.7.10† (Bolzano-Weierstrass Theorem). Prove that a set S ⊆ Rn is sequentially compact if
and only if it closed, meaning that any sequence of elements of S converging in Rn (for the Euclidean

distance) converges in S, and bounded.

Solution

Clearly, if S is unbounded or not closed, one can extract a sequence which diverges to infinity

or converges to an element not in S, and thus has no convergent subsequence. Now, suppose

that S is closed and bounded and let s = (sm)m≥0 be a sequence of elements of S. Without
loss of generality, by translating S, suppose that all its elements have coordinates in [0, M ]. We
shall proceed by dichotomy to extract a convergent in Rn subsequence of s, it will thus also
be convergent in S since S is closed. By the (infinite) pigeonhole principle, there must some
I1(1), . . . , I1(n) ∈ {[0, M/2], [M/2, M ]} such that

S ∩ I1(1) × . . . × I1(n)

is infinite. Pick an element r1 in this product of intervalls and then repeat the operation: if

a(2i) +b2(i)
2
I1(i) = [a(1i), b1(i)], there must be some I2(i) ∈ a , , , b(i) a(2i)+b(2i)(i) such that
2
22

S ∩ I2(1) × . . . × I2(n)

is infinite. Pick an element in this product of intervalls r2, and proceed inductively that way to
get chains of intervalls Im(i) = [am(i), b(mi)] of length M/2m such that Im(i)+1 ⊆ In(m) and

S ∩ Im(1) × . . . × Im(n)

is infinite and in particular contains rm. Since the length of Im(i) is M/2n, the sequences (am(i))m≥0
and (bm(i))m≥0 are Cauchy, say they converge to ci. Then, the sequence (rm)m≥1 we produced
converges to (c1, . . . , cn) as desired.

Exercise 8.7.11† (Extremal Value Theorem). Let M be a metric space, i.e. a set with a distance
d : M → R≥0 such that d(x, y) = 0 iff x = y, d(x, y) = d(y, x) (commutativity) and d(x, y) ≤
d(x, z) + d(z, y) (triangle inequality) for any x, y, z ∈ M and let S be a sequentially compact subset of
M . Suppose f : S → R is a continuous function. Prove that f has a maximum and a minimum.

356 CHAPTER 8. P -ADIC ANALYSIS

Solution

Suppose otherwise. There is a sequence (sn)n≥0 of elements of S such that

f (sn) → s ∈ im f

(s can be ±∞). Let (rn)n be subsequence of (sn)n converging to r ∈ S. Then, we get

f (r) = lim f (rn) = s

n→∞

which is a contradiction.

Exercise 8.7.12† (Equivalence of Norms). Let (K, | · |) be a complete valued field in characteristic
0, i.e. a field with an absolute value | · | which is complete1 for the distance induced by this absolute
value. A norm on a vector space V over K is a function · : V → R≥0 such that

• x = 0 ⇐⇒ x = 0

• x+y ≤ x + y

• ax = |a| x

for all x, y ∈ V and a ∈ K. We say two norms · 2 and · 2 are equivalence of norms if there are
two positive real numbers c1 and c2 such that x 1 ≤ c1 x 2 and x 2 ≤ c2 x 1 for all x ∈ V .2 Prove
that any two norms are equivalent over a finite-dimensional K-vector space V . In addition, prove that
V is complete under the induced distance of any norm · .

Solution

Since we wish to show that all norms are equivalent, it suffices to prove that any norm is equivalent
to a fixed norm we choose. A particularly simple one is the maximum norm

x ∞ = max |ai|

i

where e1, . . . , en is a basis of V and x = n aiei for some ai ∈ K. In other, words this is simply
i=1

the maximum of the coefficients of x in the basis (e1, . . . , en). Clearly, V is complete under this
n
norm, since if xk = i=1 ak,iei is a Cauchy sequence, then so is every (ak,i)k≥0 for the distance

induced by | · | which means that ak,i −→ ai for some ai and

k→+∞

n

xk → aiei.

i=1

Since two equivalent norms induce the same topology (a sequence is Cauchy for one norm if and

only if it is Cauchy for the other), we are done if we prove that any norm · is equivalent to

| · ∞. One inequality is very easy: if x = n aiei, we have
i=1

n

x= aiei

i=1

n

≤ |ai| ei

i=1

n

≤ x ∞· ei .

i=1

1Recall that completeness means that all Cauchy sequences converge. A Cauchy sequence (un)n≥0 is a sequence such

that, for any ε > 0, there is an N such that |um − un| ≤ ε for all m, n ≥ N .
2This means that they induce the same topology on V .

8.7. EXERCISES 357

For the other inequality, suppose for the sake of a contradiction that there doesn’t exist a c > 0

such that x ≤ c x ∞ for all x ∈ V . In other words, for all ε, there is some x such that

x < ε x ∞. In particular, x = 0. Since we have infinitely many x, by the pigeonhole principle,
n
we can assume that x ∞ = |ak| for a fixed k, where x = i=1 ai ei . By dividing x by ak , we

may also assume that ak = 1. This gives us a sequence

xm = ym + ek

converging to 0, where ym is in the space W spanned by e1, . . . , ek−1, ek+1, . . . , en. In particular,

ym − y ≤ ym + ek + y + ek

also converges to 0 when min(m, ) → +∞. In other words, (ym)m≥0 is a Cauchy sequence. Now,
we use induction on n = dim V . When n = 1 the result is trivial since V = K and · = 1 | · |.
For the inductive step, notice that W has dimension n−1 so, by assumption, it is complete under

· . Hence, (ym)m≥0 converges to some y ∈ W . This means that

y + ek = lim ym + ek = 0,

m→+∞

which is impossible since y + ek = 0.

Exercise 8.7.13†. Let K = Qp be a local field3, where p be a prime number or ∞ and let L be a
finite extension of K. Prove that there is only one absolute value of L extending | · |p on K, and that
it’s given by | · |p = NL/K (·) p1/[L/K].456

Solution

For simplicity purposes, we write | · | for | · |p. We first prove the uniqueness. Suppose that | · |(1)
and | · |(2) are two absolute values extending | · |. Then, they are norms over the K vector space
L. By Exercise 8.7.12†, they must be equivalent:

a|x|(1) ≤ |x|(2) ≤ b|x|(1)

for some positive real numbers a, b. In particular, if we let x = yn, we get a|y|n(1) ≤ |y|(n2) ≤ b|y|n(1).
By taking nth roots and letting n tend to infinity, this gives us

|y|(1) ← a1/n|y|(1) ≤ |y|(2) ≤ b1/n|y|(1) → |y|(1)

so |y|(1) = |y|(2) as wanted. Note that we didn’t use the fact that K was a field of the form Qp
here.

Now, we prove the existence. Multiplicativity is obvious, and |x| = 0 iff x = 0 too. The tricky
part is to prove that it satisfies the triangular inequality |x + y| ≤ |x| + |y|. After dividing by |y|,
this is equivalent to |x+1| ≤ |x|+1. We will however not prove this directly, but rather that there
is a constant c > 0 such that |x + 1| ≤ c(|x| + 1). Assuming we have proven this, Exercise 8.7.8†
tells us that we can in fact pick c = 1, i.e. that | · | satisfies the triangular inequality (and is thus
an absolute value).

3This result is true for any complete valued field (K, | · |), but it is harder to prove.
4In particular, this absolute value is still non-Archimedean if it initially was. For instance, by Exercise 8.7.7†, if p is
prime, the extension of | · |p still satisfies the strong triangle inequality. In fact, this is the only interesting case since it’s
too hard to treat the case K = R separately.
5Here is why this absolute value is intuitive: by symmetry between the conjugates, we should have |α|p = |β|p if α
and β are conjugates. Taking the norm yields |NK/Qp (α)|p = |α|[pK:Qp] as indicated.
6One might be tempted to also define a p-adic valuation for elements of K as vp(·) = − log(| · |p)/ log(p), and this is
also what we will do in some of the exercises. However, we warn the reader that, if α ∈ Z is an algebraic integer and αp
is a root of its minimal polynomial in Qp, vp(αp) ≥ 1 does not mean anymore that p divides α in Z, it only means that
p divides αp in Zp := {x ∈ Qp | |x|≤1}.

358 CHAPTER 8. P -ADIC ANALYSIS

It remains to prove that such a c exists. Let e1, . . . , en be a K-basis of L (for instance ei = αi
for some primitive element α). Define the maximum norm as

aiei = max |ai|.

i
i∞

The point is that this defines a distance d(x, y) = |x − y|(∞) and that the unit sphere S = {x |
x ∞ = 1} is sequentially compact for this distance, so that our extension of | · | will have a

(non-zero) minimum there by the extreme value theorem from Exercise 8.7.11†.

It is also not very hard to see that the unit sphere is indeed sequentially compact: this is the
Bolzano-Weierstrass theorem from Exercise 8.7.10† for p = ∞, i.e. K = R, and is very easy when

p is prime by an argument similar to the proof of ??.

To conclude, our extension of |·|, n |N (·)| is continuous for the distance induced by |·|(∞) because

N ( i aiei) is polynomial in the ai. Thus, there are positive a and b such that a ≤ |x| ≤ b for
|x|(∞) = 1 by the extremal value theorem from Exercise 8.7.11†. Note that a is positive as | · |

doesn’t vanish on S. From this we conclude that a x ∞ ≤ |x| ≤ b x ∞ for any x. But then, we

have b

|x + 1| ≤ b|x + 1|∞ ≤ b(|x|∞ + 1) ≤ (|x| + 1)
a

which is what we wanted to show.

Exercise 8.7.14†. Let (K, ·) be a complete valued field in characteristic 0 and let f ∈ K[X] be a
polynomial. Prove that f either has a root in K, or there is a real number c > 0 such that |f (x)| ≥ c
for all x ∈ K.

Solution

Suppose without loss of generality that f is irreducible and that there does not exist a c > 0 such
that |f (x)| ≥ c for all x ∈ K. In other words, |f (x)| takes arbitrarily small values for x ∈ K.
We will produce a Cauchy sequence (xn)n≥0 such that |f (xn)| → 0. The limit x of (xn)n≥0 will
then clearly be a root of f .

We use the Newton method to find such a sequence. Let x0 ∈ K be such that |f (x0)| < 1 is
small (we will specify this later). Note that |x0| is bounded since, by the triangular inequality, if
f = anXn + . . . + a0, we have

|f (x0)| ≥ |an||x|n − |an−1||x|n−1 − . . . − |a0|.

Define the sequence (xn)n≥0 by xn+1 = xn + εn, where εn will be chosen in the next sentences.

Given an element x ∈ K such that |x2 + 1| is small, we define the sequence (xn)n≥0 as follows.
Set x0 = x. Then, set xn+1 = xn + ε for some small ε. We have, by Taylor’s formula 5.3.1

n−1 εnk f (k)(xn)
k!
f (xn+1) = = f (xn) + εnf (xn) + O(εn2 ).

k=0

Hence, to kill the greatest term of this sum, we choose εn = − f (xn ) . Let’s justify a bit the
f (xn )

notation O(εn2 ): we have shown that, if f (x) is very small then x is bounded, so the derivatives
f (k)(x) are bounded as well. We also need to ensure that f (x) is not too small when f (x) is, so

that εn = − f (xn ) is very small. This follows from Bézout’s lemma: since f is irreducible, it is
f (xn )

coprime with its derivative f (we are in characteristic zero) so there are u, v ∈ K[X] such that

uf + vf = 1.

8.7. EXERCISES 359

When f (x) is very small, u(x) is bounded (since x is) so |v(x)f (x)| is very close to 1. Since v(x)
is also bounded, we get that |f (x)| is bounded below as wanted.

To conclude, we have

|f (xn+1)| = n−1 εknf (k)(xn) < c|εn|2
k!
k=0

when f (xn) is very small. Since

|εn|2 = |f (xn)|2 ,
|f (xn)|2

there is some θ < 1 such that

|f (xn+1)| ≤ θ|f (xn)|

when f (xn) is sufficiently small (in particular, it suffices to have f (x0) sufficiently small). Hence,
pick an x0 such that |f (x0)| is sufficiently small and this inequality is true. Then, |f (xn)| ≤ θn−1

by induction so that,

|xn+1 − xn| = |f (xn)| ≤ cθn
|f (xn)|

for some constant c > 0. It is not hard to see that this implies that (xn)n≥0 is Cauchy, so we are
done.

Exercise 8.7.15† (Ostrowski). Let (K, ·) be a complete valued Archimedean field in characteristic
07. Prove that it is isomorphic to to (R, | · |∞) or (C, | · |∞).

Solution

Without loss of generality, suppose that Q ⊆ K. By Exercise 8.7.9†, we may also assume that
| · | extends the usual absolute value | · |∞ of Q, by replacing | · | by | · |r for some suitable r ≥ 1.
This new absolute value might not satisfy the triangular inequality, but in fact it does. Indeed,

by the power mean inequality, we have

|x|r + |y|r ≥ |x| + |y| r |x + y|r
2 2 2r .
≥

Setting c = 2r−1, we get that this absolute value, which we will from now one abusively denote |·|
as well, satisfies the modified triangular inequality |x + y| ≤ c(|x| + |y|). Then, by Exercise 8.7.8†,

| · | satisfies the triangular inequality as desired.

Now, note that K contains (a field isomorphic to) R since it is complete and R is the set of limits

of Cauchy sequences of rational numbers. | · | is then the usual absolute of R, by construction of

R. Without loss of generality, suppose also that C ⊆ K, by extending | · | to K(i) if necessary.
By Exercise 8.7.13†, we know that we should extend | · | to K(i) by

|α + βi| = |α2 + β2|,

but we don’t know if it is indeed an absolute value. To show that it is, note that, if i ∈ K, by
Exercise 8.7.14† there is a constant c > 0 such that

|α2 + β2| ≥ c(|α|2 + |β|2)

for all α, β ∈ K. Indeed, if |x2 + 1| ≥ c/2 for all x ∈ K, we have, for any |β| ≥ |α|,

|α2 + β2| =|β|2|(α/β)2 + 1|
≥ |β|2c/2
≥ c(|α|2 + |β|2)

7In fact it is quite easy to show that char K = 0 follows from the assumption that | · | is Archimedean, but we add
this assumption for the convenience of the reader.

360 CHAPTER 8. P -ADIC ANALYSIS

Thus, for any α, β, γ, δ ∈ K,

|(α + βi) + (γ + δi)|2 = |(α + γ)2 + (β + δ)2|
≤ 2(|α|2 + |β|2 + |γ|2 + |δ|2)
≤ 2 (|α + βi|2 + |γ + δi|2)
c

where the third line follows from the triangular inequality and the inequality between the arith-
metic and geometric mean, so | · | satisfies the triangular inequality by Exercise 8.7.8† (and the
quadratic-geometric mean inequality).

We will now prove that any element of K is in fact in C, thus showing that K = C as wanted.

Let α be an element of K and let m be the minimum of |α − x| for x ∈ C. This minimum exists
by the Bolzano-Weierstrass theorem: we have |α − x| ≥ |x| − |α| so |α − x| → ∞. If we choose r
such that |α − x| > |α| for |x| > r, we get that the minimum of |α − x| over C is also its minimum
over the ball {x | |x| ≤ r}. However, this ball is compact by the Bolzano-Weierstrass theorem,
and the function x → |α − x| is continuous by the triangular inequality, so a minimum exists by

the extremal value theorem. We wish to prove that this minimum m is zero.

The idea is now to take an x such that |α − x| is large, and, at the same time, A − x divides a

polynomial f such that |f (α)| is small. If we let g = f , we get that |g(α)| is quite small so
A−x
that one of |α − z| where z is a root of g is small, and in particular smaller than m. Since the

remainder of a polynomial f modulo A − x is f (x), we can relax the condition to |f (α)| small

and |f (x)| as well. With these conditions, it is natural to pick f first and then x: an obvious

candidate for f is

f = (A − y)n

where y is such that |α − y| = m. Now, we need to estimate |f (α) − f (x)|. By the triangular

inequality, it is at most mn + |x − y|n. In particular, if ε = |x − y| < 1, it is at most mn plus

something very small. In addition, by definition, we know that |g(α)| ≥ mn−1, where g = f −f (x) .
A−x

Hence,

|α − x|mn−1 ≤ |g(α)||α − x| = |f (α) − f (x)| ≤ mn + εn.

This means that, if m is non-zero, by dividing by mn,

|α − x| ≤ m (1 + (ε/m)n) → m.

Thus, |α − x| = m for all |x − y| < 1. Iterating this process, we get |α − x| = m for all x ∈ C
which is obviously a contradiction since |α − x| goes to ∞ when |x| → ∞. Hence, |α − z| = 0 for
some z ∈ C, i.e. α = z ∈ C as wanted.

Diophantine Equations

Exercise 8.7.16† (Brazilian Mathematical Olympiad 2010). Find all positive rational integers n and
x such that 3n = 2x2 + 1.

Solution

√ √√
We proceed as in Proposition 8.6.1: working in Q( −2), we find 1 + −2x = (1 ± 2)n, i.e.

√√
(1 + −2)n + (1 − −2)n = ±2.

√√
To solve this, we shall work in Q11. We thus consider α = 1 ± −2 and β = 1 ∓ −2 as elements
of Q11; Hensel’s lemma gives us α ≡ 20 (mod 121) and β ≡ 103 (mod 121). We wish to find the
zeros of the linear recurrence αn − βn ± 2. Note that we have αn − βn ≡ ±2 modulo 11 only
when n ∈ {0, 1, 2}, so we restrict our attention to these n.

8.7. EXERCISES 361

Set a = α5 − 1 ≡ 0 (mod 11) and b = β5 − 1 ≡ 0 (mod 11). We shall compute the Strassmann
bounds of the analytic functions

fr(s) = αr(1 + a)s − βr(1 + b)s

for r ∈ {0, 1, 2}. Modulo 112, we have

fr(s) ≡ αr(1 + as) + βr(1 + bs) − 2.

The coefficient of s is αra + βrb. However, for r ∈ {1, 2}, this is respectively 44 and 88 modulo
112 so non-zero in both cases. Hence, the Strassmann bounds for f1 and f2 are 1. It remains to
compute the Strassmann for f0. This time, we have a + b ≡ 0 (mod 112) so we need to expand
one more term. We get

f0(s) = (1 + a)s + (1 + b)s − 2 ≡ 1 + as + a2 s + 1 + b2 s −2 (mod 113).
2 2

The coefficient of s2 is thus a2 +b2 modulo 113. However, we can check with Hensel’s lemma
2
that α ≡ 587 (mod 113) and β ≡ 746 (mod 113), which yields a ≡ 1012 (mod 113) and b ≡ 317

(mod 113). One can then verify that

a2 + b2 ≡ 847 ≡ 0 (mod 113).

Hence, the Strassmann bound for 0 is 2.

To finish, we need to find solutions: two solutions congruent to 0 modulo 5, one congruent to 1
modulo 5, and one congruent to 2 modulo 5. It is not hard to see that we indeed have

30 = 2 · 02 + 1
31 = 2 · 12 + 1
32 = 2 · 22 + 1
35 = 2 · 112 + 1.

Hence, we have found all solutions: (n, x) ∈ {(0, 0), (1, 1), (2, 2), (5, 11)}.

Exercise 8.7.19†. Solve the diophantine equation x2 − y3 = 1 over Z.

Solution

Write this equation as (x − 1)(x + 1) = y3. The gcd of the two factors divides 2, so we have
x − 1 = a3 and x + 1 = b3 or x ± 1 = 2a3 and x ∓ 1 = 4b3 for some a, b ∈ Z. The former is

impossible, so we must be in the latter case. The problem thus reduces to solving the equation
a3 − 2b3 = ±1 in rational integers. We know by Section 7.4 that

√
a − b 3 2 = ±θn

√
for some n, where θ is the fun√damental unit of Q( 3 2). In addition, by Exercis√e 7.5.18†, we√know
that we can choose θ = 1 − 3 2 = − 1+ √3 21+ √3 4 . Hence, we wish to have a − b 3 2 = ±(1 − 3 2)n.
As we saw in the proof of Theorem 7.4.2, for a given n, there are such a, b if and only if

√√ √
(1 − 3 2)n + j(1 − j 3 2)n + j2(1 − j2 3 2)n = 0.

We work in Q3(α, j), where α3 = 2 and j is now a tryadic root of unity of order 3. Note that

this has degree 6 over Q3 since j ∈ Q3 and α ∈ Q3(j) (for instance because Gal(Q3(j)/Q3) is
abelian but Gal(Q3(α)/Q3) isn’t). In particular, α0 = α, α1 = jα and α2 = j2α are conjugate.

362 CHAPTER 8. P -ADIC ANALYSIS

We wish to find when the linear recurrence (1 − α)n + j(1 − jα)n + j2(1 − j2α)n is zero. Here is
the magic: this is already almost a tryadic analytic function. Indeed, we can rewrite it as

2n((1 + π0)n + j(1 + π1)n + j2(1 + π2))

whe√re πk = −(1 + αjk)/2 has √norm −3/8 and thus tryadic absolute value 3−1/3 < 1. (In fact,
1 + 3 2 is prime in OQ( √3 2) = Z[ 3 2].) However, 3−1/3 is still too large: it’s greater than 3−1/(3−1).
Hence, we consider the function

fr(s) = (1 + π0)r(1 + π0)3s + j(1 + π1)r(1 + π1)3s + j2(1 + π2)r(1 + π2)3s

for r ∈ [3]. Indeed, these converge since (1 + πk)3 = 1 + 3(−3/8 − αk/8 + αk2/8) has absolute
value 3−1 < 3−1/(3−1). It is then straightforward to compute the Strassmann bounds: we claim
that it is 1 for r = 0 and r = 1, and 0 for r = 2. Let us start with r = 0. In that case,

2

f0(s) ≡ s j3(−3/8 − αjk/8 + α2j2k/8) (mod 27).

k=0

It is in fact very easy to compute a sum of the form 2 j k i aijki: this is a unity root filter
k=0
so is the sum 3 i≡−1 (mod 3) aiαi (see Exercise A.3.9†). This is actually normal: it’s why we

considered this sum in the first place. In particular, this also explains why this congruence holds

modulo 33 instead of simply 32: it’s because of the additional factor of 3 added by the unity root

filter. Hence, the coefficient of s of f0 is 9α2/8 which has tryadic valuation 2 < 3 and 1 is thus

the Strass√mann bound for f0. Conversely, it is clear that s = 0 is a solution (corresponding to
1 = (1 − 3 2)0).

We now consider f1. As before, we are done if the coefficient of s has tryadic valuation 2 since

all the following ones have valuation at least 3. We expand f1 modulo 27, and remember that
we only care about the coefficient of α3n+2:

2

f1(s) = jk(1 − αk)/2 · (1 + 3(−3/8 − αk/8 + αk2/8))s

k=0

22 (mod 27)

≡ jk(1 − αk)/2 + 3s jk(1 − αk)/2 · (−3/8 − αk/8 + αk2/8)

k=0 k=0

= −9sα2/8

which h√as absolute√value 3−2 as desired. It is again clear that s = 0 is a solution (corresponding
to 1 − 3 2 = (1 − 3 2)1).

Finally, we consider f2. Here is what changes: the coefficient of s0 is no longer zero because
(1 − α)2 now has a non-zero coefficient for some α3n+2. More specifically,

2

f2(s) = jk(1 − αk)2/4 · (1 + 3(−3/8 − αk/8 + αk2/8))s

k=0

2

≡ jk(1 − αk)2/4 (mod 9)

k=0

= −3α2/4

which has absolute value 3−1 as desired. √This shows t√hat the Strassmann bound is 0, and
concludes our study of the equation a − b 3 2 = ±(1 − 3 2)n: the only solutions are a = ±1,

b = 0 as well as a = b = ±1. If we go back to the original problem, these correspond to
x ± 1 = 2a3 = ±2, i.e. x ∈ {±1, ±3}. These then yield (x, y) ∈ {(±1, 0), (±3, 2)}, which are, in
conclusion, the only rational integer solutions to the equation x2 − y3 = 1.

8.7. EXERCISES 363

Exercise 8.7.20† (Lebesgue). Solve the equation x2 + 1 = yn over Z, where n ≥ 3 is an odd integer.

Solution

Suppose (x, y) is a solution. By the unique factorisation in Z[i], we have xi + 1 = ε(a + bi)n for
some a, b ∈ Z and ε ∈ Z[i] a unit. Note that y = a2 + b2 is odd, since x2 + 1 is never divisible by
4, so one of a, b is even and the other is odd. Since the units of Z[i] have the form ik for some k
by Exercise 2.2.3∗, they are all nth powers since n is odd, so we can assume ε = 1.

Hence, we wish to find the a, b ∈ Z such that (a + bi)n + (a − bi)n = 2. Since n is odd, this is
divisible by 2a so a is ±1. Since y = a2 + b2 is odd, b must be even. Expanding the real part of
(a + bi)n (which must be 1), we get

n−1

2 n an−2k(−1)kb2k = 1.
2k

k=0

Modulo b2 we get b2 | 1 − an ∈ {0, 2}, and since b2 is at least 4 since b is even, a must be 1. In
other words, our equation becomes

(1 + bi)n + (1 − bi)n = 2.

We wish to expand the LHS as a dyadic analytic function, but this is not possible because |b|2

might be equal to 2− 1 = 2−1, i.e. b might have dyadic valuation equal to 1. To remedy this
2−1

situation, we use the LTE lemma:

(1 + bi)2 = (1 + 2b(i − b/2)).

Since n is odd, we can set n = 2m + 1 and reduce the problem to finding the zeros of the now
dyadic analytic function

f (m) = (1 + bi)(1 + 2b(i − b/2))m + (1 − bi)(1 + 2b(−i − b/2))m − 2

where i is now a square root of −1 in Q2. Since this a root of unity filter (see Exercise A.3.9†),
as in Exercise 8.7.19†, f (m) is twice the "real dyadic" part of (1 + bi)(1 + 2b(i − b/2))m − 1, i.e.

the coefficient of 1 in this expression. Now expand this as

∞ m (2b(i − b/2))k .
k
−1 + (1 + bi)

k=0

Suppose that b = 0, otherwise we get x = 0 and y = 1. Since v2(k!) ≤ k − 1 by Legendre’s
formula, every term except the first two vanish modulo 2b2. Hence, modulo b2, this is simply

−1 + (1 + bi)(1 + 2bm(i − b/2)).

If we expand this while focusing only on the real dyadic part, we get
(1 − b2m) + bi · 2bim − 1 = −3b2m.

Since |b2|2 > |2b2|2, we conclude that the Strassmann bound is (at most) 1. Since m = 0 is a
trivial solution, we conclude that it is the only solution (corresponding to n = 1, which is not
the case). Thus, the only solution (x, y) = (0, 1).

Remark 8.7.1

364 CHAPTER 8. P -ADIC ANALYSIS

We can also finish directly with a slightly ad-hoc dyadic method once we reach

n−1

2 n (−1)kb2k−2 = 0.
2k

k=1

Let m be the dyadic valaution of n = n(n−1) . We will prove that 2m+1 divides n , which is
2 2 2
n
of course a contradiction. The denominator of 2k is (2k)!. By Legendre’s formula, we have

v2((2k)!) = 2k − s2(2k) ≤ 2k − 1. As a result, b2k−2 has dyadic valuation at least −1. Since
(2k)!

(n − 1)(n − 3) divides (2k)! n , we conclude that
2k

v2 n b2k ≥ v2((n − 1)(n − 3)) + v2(b2k/(2k)!) ≥ m + 2 − 1 = m + 1
2k

as wanted since m = v2 n−1 = v2(n − 1) − 1. Hence, 2m+1 divides every term of the sum
2

n−1

2 n (−1)kb2k = 0.
2k + 2

k=0

except the first one, which means that it also divides the first one.

Exercise 8.7.21†. Solve the equation x2 + 1 = 2yn over Z, where n ≥ 3 is an odd integer.

Solution

Suppose (x, y) is a solution. By factorising in Z[i], we get xi + 1 = ε(1 + i)(a + bi)n for some
a, b ∈ Z and a unit ε ∈ Z[i]. Note that y = a2 + b2 is odd, since x2 + 1 is never divisible by 4, so
one of a, b is even and the other is odd. Since the units of Z[i] have the form ik for some k by
Exercise 2.2.3∗, they are all nth powers since n is odd, so we can assume ε = 1.

By assumption,

2 = (1 + ix) + (1 − ix)
= (1 + i)(a + bi)n + (1 − i)(a − bi)n
= i(1 − i)(a + bi)n + (1 − i)(a − bi)n
= (1 − i) ((±b ∓ ia)n + (a − bi)n)

where the ±1 sign depends on n modulo 4. Since n is odd, this last expression is divisible by

±b ∓ ia + a − bi = (a − b)(1 ∓ i).

Thus, (1 − i)(a − b)(1 ∓ i) divides 2. This is equivalent to a − b | 1, so a − b = ±1. Without loss
of generality, suppose that a and b are non-zero since {|a|, |b|} = {0, 1} yields y = 1 and thus
x = ±1. Now we distinguish a few cases, depending on which one of a and b is even and whether
a − b is 1 or −1.

1. b is even and a − b = 1. In that case, our equation is

f (n) := (1 + i)(1 + b(1 + i))n + (1 − i)(1 + b(1 − i))n − 2 = 0.

Unlike Exercise 8.7.20†, this is already a dyadic analytic function since |(1 + i)|2 = 2−1/2

which means that |b(1 + i)|2 ≤ 2−3/2 < 2− 1 (we are working with the dyadic i ∈
2−1

Q2). This is a unity root filter, so we are just focusing on the "real dyadic" part of

(1 + i)(1 + b(1 + i))n − 1. When we expand this modulo b3, we get

(1 + i)(1 + b(1 + i)n + b2(1 + i)2n(n − 1)/2) − 1 = i + 2bin + (1 + i)2b2in(n − 1)/2

8.7. EXERCISES 365

since (1 + i)2 = 2i, which has real dyadic part −b2n(n − 1). Clearly, |b2|2 > |b3|2 since

b = 0 so the Strassmann bound is 2. The previous computation in fact shows that the first
∞
two coefficients of f are zero (when written as a Mahler series k=0 ak x ), which means
k

that n = 0 and n = 1 are solutions. In other words, these are the only solutions, which are

ruled out by the statement.

2. b is even and a − b = −1. Since n is odd, we have (−1 + b(1 ± i))n = −(1 − b(1 ± i))n so
our equation is

f (n) := (1 + i)(1 − b(1 + i))n + (1 − i)(1 − b(1 − i))n + 2 = 0.

The same computation as before shows that the coefficient of n is −2bi + 2bi = 0. Thus,
1
modulo 2b2, we have

f (n) ≡ 4 + 0n

so the Strassmann bound is 0 since |2b2|2 < |4|2. There are no solutions in this case.

3. a is even. Then, a + bi = ±i + a(1 + i) so the equation is

(1 + i)(±i + a(1 + i))n + (1 − i)(±i + a(1 − i))n − 2 = 0.

Since (±i + α)n = ±i(1 + ±iα)n for any α ∈ Q2(i), where the ± are independent and
depend on whether n ≡ 1 (mod 4) or n ≡ −1 (mod 4), our equation is

f (n) = (1 + i)(1 ± ia(1 + i))n + (1 − i)(1 ± ia(1 − i))n ± 2i = 0.

where the first two ± signs are the same and the last one is independent. We will prove
that the Strassmann bound is always 0. Modulo 2a (this is a unity root filter so the "real
dyadic" part gets doubled), we have

f (n) ≡ 2(1 + ±i).

Since |2a|2 < |2(1 ± i)|2, we are done.

To conclude there are no solutions to the equations x2 + 1 = 2yn when y is not equal to 1 and
n ≥ 3, i.e. the only solutions to our equation are (±1, 1).

Linear Recurrences

Exercise 8.7.22†. Let (un)n≥0 be a linear recurrence of rational integers given by i fi(n)αin such
that αi/αj is not a root of unity for i = j. If un is not of the form aαn for some a, α ∈ Z, prove that
there are infinitely many prime numbers p such that p | un for some integer n ≥ 0.

Solution

Without loss of generality, suppose that un is not identically zero. By Corollary 8.5.2 and
Corollary 8.5.1, the condition on the αi tells us that |un| → ∞. The idea is that we will bound
the p-adic valuation of un over a subsequence (uan+b)n≥0 to get a contradiction if (un)n≥0 has
finitely many prime divisors (since (uan+b)n≥0 would then be bouded).

We shall analyze the local behaviour of (un)n≥0 for a fixed prime p. Write un = i fi(n)αin.
We wish to factorise αi by a suitable power of p so that maxi |αi|p = 1. Indeed, since |p1/n|pn =
|p|p = 1/p, the absolute values of powers of p take any value which can be taken by | · |p. Thus,
suppose that maxi |αi|p = 1 and consider the sequence vn = i∈I fi(n)αi(n) where I denotes
the set of i such that |αi|p = 1.mb Let Kp be the field generated by the αi. We claim that
that the integers OKp := {|x|p ≤ 1 | x ∈ Kp} of Kp are finite modulo pk, for any fixed k. This
implies (by the pigenhole principle) that (vn)n≥0 is periodic modulo pk for any k. To prove our

366 CHAPTER 8. P -ADIC ANALYSIS

claim, suppose for the sake of a contradiction that there were infinitely many elements of OKp
non-congruent modulo pk, say f is a set of such elements. Pick a primitive element β of Kp/Qp

with conjugates β1, . . . , βd, and consider an element x = d−1 bi βi ∈ OKp . By definition of the
i=0

p-adic absolute value, we also have |xi| ≤ 1, where xi is the image of x under the embedding

β → βi. To conclude, Cramer’s rule (Exercise C.5.7) or the adjugate (Proposition C.3.7) let us

express the bi as linear combinations of the βji and the xi. Then, using the triangle inequality,

we conclude that |b0|p, . . . , |bd−1|p are bounded. As a consequence, the set

d−1

{(b0, . . . , bd−1) | biβi ≤ 1} ⊆ Qpn

i=0

is compact. In particular, OKp is as well, and thus S too. This implies that there are s, r ∈ S
such that |s − r|p is arbitrarily small, but then they will be equal modulo pk since

u≡v (mod pk) ⇐⇒ u−v ⇐⇒ |u − v|p ≤ |pk| = p−k.
pk ∈ OKp

To conclude, note that (vn)n≥0 is non-zero for large n by the Skolem-Mahler-Lech theorem.
Pick any N so that vN is non-zero, and let Tp be the period of (vn)n≥0 modulo p vp(vn) +1.
Then, |vN+kTp |p is greater than some constant c > 0 for any k, and thus |un+kTp |p as well for
sufficiently large n + kTp, since un − vn → 0. If we finally return to the global behaviour and

let p vary among our finitely many prime divisors of (un)n≥0, we get that, for any sufficiently

large N , vp(uN+k p Tp ) is bounded for any p and for any sufficiently large k. This contradicts
the assumption that |un| → ∞.

Remark 8.7.2

Note that, to prove that αn is periodic modulo p for α ∈ OKp , we cannot simply "convert" (with
the fundamental theorem of symmetric polynomials, after having introduced its conjugates) α to
an element of Fp and use the Frobenius morphism. Why? Because the minimal polynomial of
α does not necessarily have coefficients in Zp. Indeed, we only consider the constant coefficient
of the minimal polynmomial of α to compute its p-adic absolute value, and disregard all other
coefficients. For instance, the roots of X2 − X/2 + 1 over Q2 are in the unit ball.

As another remark, it has in fact been proven, using a generalisation of (a p-adic extension of)
the Thue-Siegel-Roth theorem (see Remark 7.4.3) that un either has the form cαn, or its greater
prime factor tends to infinity. See [28].

Exercise 8.7.23†. Does there exists an unbounded linear recurrence (un)n≥0 such that un is prime
for all n?

Solution

Suppose for the sake of a contradiction that (un)n≥0 is such a sequence. Without loss of generality,
suppose that |un| → ∞ by replacing (un)n≥0 by (uNn+m)n≥0 for some suitable N, m, as indicated
after Corollary 8.5.2. Now, let m be sufficiently large so that um = p is a prime which doesn’t
divide the denominator of the norm of any algebraic number appearing in the formula of um
(so that they still make sense modulo p). Finite fields theory (e.g. Theorem 4.2.1) tells us that
there exists a k unpk ≡ un (mod p) for any n. Indeed, if un ≡ i fi(n)αin, with fi ∈ Fp[X] and
αi ∈ Fp, it suffices to pick k so that fi ∈ Fpk [X] and αi ∈ Fpk , by the Frobenius morphism.

In particular, ump k ≡ 0 (mod p) for any . By assumption, this means that ump = p, contra-
dicting the fact that un → ∞.

8.7. EXERCISES 367

Miscellaneous

Exercise 8.7.24†. Which roots of unity are in Qp?

Solution

Let α = (a1, a2, . . .) is a root of unity of order n in Qp. Suppose initially that p is odd. We first
focus on the case where p n. We have akn ≡ 1 (mod pk). However, the group of units modulo pk
is isomorphic to pk−1(p − 1) by Exercise 3.5.18† (in more elementary terms: there is a primitive

root) so we also have

akgcd(p−1,n) ≡ akgcd(pk−1(p−1),n) ≡ 1.

Hence, α has order dividing gcd(p − 1, n), which implies that n | p − 1 since n is the order of α.

Now suppose that p | n. We wish to reach a contradiction, so suppose without loss of generality
that α has order exactly p, by replacing it by αn/p. Then, akp ≡ 1 (mod pk) so ak ≡ 1 (mod p)
which implies that

vp(akp − 1) = 1 + vp(ak − 1)

by LTE. For large k, vp(ak − 1) stabilises since α = 1, which means that this cannot be at least
k.

It remains to provide a construction for (p − 1)th roots of unity. One can do this using the
structure of (Z/pkZ)×, or by means of Hensel’s lemma: the derivative of Xp − X is 1 which
is never zero so we can lift all roots of Xp − X modulo p to roots in Qp. This is called the
Teichmüller character ω which sends x ∈ (Z/pZ)× to the unique root of Xp−1 − 1 congruent to

x modulo p.

It remains to treat the case where p = 2. When n is odd, the same argument as before works:
this time we even have agkcd(2k−2(2−1),n) ≡ 1 (mod 2k) for k ≥ 2. However, unlike the previous
case, there is now a root of unity of order 2: −1. Since the only root of unity of odd order is 1,
the order of any root of unity must be a power of 2, since α2v2(n) is a root of unity of odd order.
Hence, we shall prove that there is no root of unity of order 4. This is easy: we use the LTE for
p = 2 (which simply amounts to the fact that a square is always 1 modulo 4) to get

v2(a4k − 1) = 1 + v2(a2k − 1)

and this stabilises since α2 = 1 by assumption. This time, the Teichmüller character is defined
as ω : (Z/4Z)× : Q2 sending 1 to 1 and −1 to −1.

To conclude, the roots of unity of Qp are all (p − 1)th roots of unity, as well as a root of order 2
when p = 2.

Exercise 8.7.27† (China TST 2010). Let k ≥ 1 be a rational integer. Prove that, for sufficiently

large n, n has at least k distinct prime factors.
k

Solution

The key lemma is that, for any prime p and any positive integer n, pvp((nk)) ≤ n. Suppose that

we have proven this. Then, if n has at most k − 1 prime factors, say p1, . . . , pm, we have
k

n = m pvi pi ((nk)) ≤ nm ≤ nk−1
k
i=1

which is impossible for large n since n is a polynomial of degree k in n.
k

368 CHAPTER 8. P -ADIC ANALYSIS

It remains to prove this key claim. We use Legendre’s formula and the fact that n = n!
k k!(n−k)!

to write

n logp (n) n n−k k

vp k = i=1 pi − pi − pi .

The wanted result now follows from the trivial inequality x + y ≤ x + y + 1: each of the

terms n − n−k − k is at most 1, so the whole sum is less than or equal to logp(n) .
pi pi pi

This gives us vp n ≤ logp(n), i.e. pvp((nk)) ≤ n as claimed.
k

Exercise 8.7.28†. Find all additive functions f : ZN → Z, where addition is defined componentwise.
(To those who have read Section C.2, the fact that there are a nice characterisation of those functions
should come off as a surprise.)

Solution

We claim that the Z-linear functions from ZN → Z are given by linear combinations of the
coordinates, which is surprising since the vectors ei with 1 in the ith coordinate and 0 everywhere
else do not form a basis of ZN: any linear combination of them has finitely many non-zero
coordinates (so (1, 1, . . .) isn’t one for instance)! This problem thus has two parts: proving that
any such function is 0 on all but finitely many ei, and proving that an additive function which is
0 on the ei is identically 0. We will do the second part first.

Suppose that f : ZN → Z is additive and f (ei) = 0 for all i, i.e. f is 0 on the space of vectors
with finitely many non-zero coordinates. The special property of Z is that we can use the theory
of divisibility. More precisely, if the coordinates of x ∈ ZN eventually get all divisible by m, then
m | f (x). Indeed, if x = (x0, x1, . . .) is such that m | xn for any n ≥ N , we have

f (x) = f (0, . . . , 0, xN , xN+1, . . .) = mf (0, . . . , 0, xN /m, xN+1/m, . . .).

Thus, if the xi get eventually all divisible by increasingly large integers, f (x) must be zero! For
instance, f (a0, a1p, a2p2, . . .) is divisible by pn for any n so must be zero. You should now be able
to see the p-adic flavor of this problem (even if we won’t really use any of the theory developped
in this chapter)! In particular, x is congruent modulo p to

f (x0, x1(p + 1), x2(p + 1)2, . . .) = 0.

Since this is true for abritrary p, f (x) must be 0 too. Alternatively, using Bézout’s lemma, there
are 2nyn and 3nzn such that 2nyn + 3nzn = xn. Thus,

f (x) = f (y0, 2y1, 4y2, . . .) + f (z0, 3z1, 9z2, . . .) = 0 + 0 = 0.

Now we prove that any additive function f : ZN → Z is 0 on all but finitely many ei, say i ∈ I.
This implies that the function x → f − i∈I f (ei)xi, where xi denotes the ith coordinate of x, is
zero on every ei so must be identically zero by the previous step. This shows that any additive
function is a linear combinations of the coordinate.

The idea will again be p-adic. We will produce a sequence x = (x0, x1, . . .) such that v2(xn) is
increasing and grows so fast that f (ei) must be 0 for large i, since we have the congruence

n−1

f (x) ≡ xif (ei) (mod 2v2(xn)).

i=0

We can rephrase this as saying that the series ∞ xif (ei) converges dyadically to the rational
i=0

integer f (x). The point is that we have too many degrees of freedom for this to be always a

8.7. EXERCISES 369

rational integer, unless f (ei) = 0 for sufficiently large i. This follows from the fact that, if we write
∞
the dyadic expansion of a dyadic integer as i=0 ai2i with ai ∈ {0, 1}, then the dyadic integers

with a finite dyadic expansion are exactly the rational integers. Indeed, this decomposition is
∞ ∞
unique for the same reason that the base 2 decomposition is: if i=0 ai2i = i=0 bi2i , pick

the smallest n such that an = bn to get an2n ≡ bn2n (mod 2n+1), i.e. an = bn, which is a

contradiction. Thus, the dyadic expansion of a rational integer must be its base 2 expansion,

which is indeed finite.

Hence, we pick xi = 2ni with (ni)i≥0 an increasing sequence which grows sufficiently fast. More
mi
precisely, if we write 2ni f (ei) in base 2 as k=ni ak 2k , we want ni+1 to be larger than mi.
That way, the base 2 expansion of 2ni+1 f (ei+1 new terms to the dyadic expansion of
∞ ) only adds
i=0
f (x) = 2ni f (ei), unless f (ei+1) = 0. Since the dyadic expansion of f (x) ∈ Z is finite, for

sufficiently large i, 2ni f (ei) cannot add new terms to it, which means f (ei) = 0 as wanted. This

concludes the solution.

Exercise 8.7.29†. Prove that the Skolem-Mahler-Lech theorem holds over any field of characteristic
zero.

Solution

The idea is to reduce again the problem to sequences of algebraic numbers. More precisely, let K

be the field generated by the numbers involved in the formula for un and pick a transcendance
basis α1, . . . , αk (see Exercise B.4.9†), i.e. a maximal subset of elements which are algebraically
independent over Q. The primitive element theorem then yields K = Q(α1, . . . , αk)(α) for some
α algebraic over Q(α1, . . . , αk) with minimal polynomial π(α1, . . . , αk) ∈ Z(α1, . . . , αk)[X]. The
key point is that we can "replace" α1, . . . , αn by any rational integers a1, . . . , ak and α by any
root a of π(a1, . . . , ak), since an inequality in Q(α1, . . . , αk) reduces to an equality of algebraic
functions in Q(X1, . . . , Xk) modulo π(X1, . . . , Xk).

Given an A = (a1, . . . , ak) and a root a of π(a1, . . . , ak), we shall denote the image of un under

the substitution αi → ai, α → a by un[A,a] (note that this only makes sense if the denominator of

un[ A, a] is non-zero). Our main claim is the following: there is a T such that, for any uN = 0, there

is an A for which uN[A,a] stays non-zero and such that the common difference of the arithmetic
progressions of zeros of (un[A,a])n∈Z divides T . It is straightforward to see that this implies the

wanted theorem: if (unT +m)n∈Z is not identically zero, then pick an N ∈ T Z+m such that uN = 0
and an A as before to get that (un[AT,a+]m)n∈Z has finitely many zeros and thus (unT +m)n∈Z as
well.

It remains to prove this claim. Write un = i fi(n)rin, where ri = Ri(α1, . . . , αk, α) and let
Li(α1, . . . , αk) and Ci(α1, . . . , αk) be the leading and constant coefficients of the minimal poly-
nomial of ri, as seen as a polynomial Z[α1, . . . , αk][X]. Pick B = (b1, . . . , bk) ∈ Zk such that

Li(B) and Ci(B) are non-zero: this implies that the ri are too (when evaluated at b and any

root of π(B)) since their norm is. Then pick a prime p which divides none of them. Finally,

we choose (a1, . . . , ak) such that ai ≡ bi (mod p) (in particular these coefficients are still non-

zero) and u[NA,a] is non-zero, where A = (a1, . . . , ak, a) and a is always an arbitrary root of

π(α1, ..., αk ). This is possible since ibfyuEN[Ax,ae]rcwiseereAa.1lw.7a∗yswzheircoh, then the norm of uN is zero on
(b1 + pZ) ×... × (bk + pZ) so is zero implies that uN = 0.

Now, we consider the norm (vn)n∈Z of un[A,a] to get a sequence of rational numbers. We will

consider (vn)n∈Z as a union of p-adic analytic functions to deduce information about its zero,
as usual. Hence, we shall abusively consider the ri[A,a] as elements of a finite extension Kp
of Qp. Note that they have zero p-adic valuation, i.e. are units in Kp. Indeed, note that
r = Li(A)ri[A,a] is a root of a monic polynomial with constant coefficient c = Ci(A)Li(A)m−1: if

370 CHAPTER 8. P -ADIC ANALYSIS

Ri(A) = Li(A)Xm + . . . + Ci(A), then

R = Li(A)m−1Ri(A)(X/Li(A)) = Xm + . . . + Ci(A)Li(A)m−1.

Since c is not divisible by p, the norm of r cannot be smaller than 1 by the strong triangle
inequality, otherwise the norm of R(r) would be |c|p = 1. Similarly, if it has norm greater than
1, the norm of this polynomial evaluated at r would be |rm|p. Since p Li(A), we conclude that
|ri[A,a]|p = 1 as wanted.

Finally, we wish to transform (vn)n∈Z into analytic function, i.e. have |rit[A,a] − 1|p ≤ 1/p ≤
1

1/p p−1 . This t will be our bound for the common period of the arithmetic progressions of zeros.
For this, consider the ri[A,a] as algebraic numbers again and then as elements of Fp. Since their
degree is bounded (by [K : Q]), their order in Fp is bounded too, say divides T . Then, we have
riT [A,a] = 1 in Fp, so if we return to our p-adic ri[A,a] ∈ Qp, the fundamental theorem of symmetric
polynomials shows that we also have riT [A,a] ≡ 1 (mod p) there. We conclude that (vnT +m)n∈Z
is analytic over OK for any m ∈ [T ], which finishes the proof of our claim. Indeed, note that T
only depends on [K : Q] and p, which was fixed at the beginning of the proof, so does not depend
on N (our chosen index such that uN = 0).

Appendix A

Polynomials

A.1 Fields and Polynomials

Exercise A.1.1∗. Let K be a field. Prove that 0K a = 0K for any a ∈ K.

Solution
0K a = (0K + 0K )a = 0K a + 0K a so 0K a = 0K .

Exercise A.1.2∗. Let † be a binary (taking two arguments) associative operation on a set M . Suppose
that M has an identity. Prove that it is unique. Similarly, prove that, if an element g ∈ M has an
inverse, then it is unique.1

Solution

If e and e are identities, then e = ee = e so e = e . Similarly, if b and b are two inverses of a,
then

b = (b a)b = b (ab) = b
by associativity.

Exercise A.1.3∗. Prove that multiplication of polynomials is associative and commutative.

Solution
Let f = i aiX , g = j bjXj and h = k ckXk be three polynomials. We have

fg = aibj X = bjaiX = gf

i+j= i+j=

since multiplication is commutative in a field and

(f g)h = (aibj)ckX = ai(bjck)X = f (gh)

i+j+k= i+j+k=

since multiplication is associative in a field. (This also works for formal power series.)

1Such a structure is called a monoid.

371

372 APPENDIX A. POLYNOMIALS
Exercise A.1.4∗. Prove that the gcd of 0 and 0 is 0.

Solution

Any polynomial divides 0 and 0 if and only if it divides 0.

Exercise A.1.5∗. Prove that the Euclidean algorithm produces the gcd. Deduce that the gcd of
two polynomials in K[X] is also in K[X]. (As a consequence, the fundamental theorem of algebra
Theorem A.1.1 implies that two polynomials with rational coefficients are coprime in Q[X] if and only
if they have a common complex root.)

Solution

We need to prove that gcd(f, g) = gcd(f − gq, g) for any f, g. This implies that the steps in the
Euclidean algorithm preserve the gcd. Since deg f + deg g decreases at each step, we eventually
reach a situation where f = 0, and the gcd of 0 and g is clearly g. It is however very trivial that
the gcd is conserved since

h | f, g =⇒ h | f − gq, g

and
h | f − gq, g =⇒ h | g, (f − gq) + gq = f.

Exercise A.1.6∗ (Bézout’s Lemma). Consider two polynomials f, g ∈ K[X]. Prove that there exist
polynomials u, v ∈ K[X] such that uf + vg = gcd(f, g).

Solution

Without loss of generality, suppose that deg g ≤ deg f . We proceed by induction on deg g. When
this is 0, i.e. g is constant, we have 0 + ·f + 1/g · g = 1 as wanted. For the induction step, perform
the Euclidean division of f by g: f = gq + r. Since deg r < deg g, by the induction hypothesis,
there are u and v such that uf + vr = 1. Then,

1 = uf + vr
= uf + v(f − gq)
(u + v)f − (qv)g

as wanted.

Remark A.1.1
One might, at first sight, think that this proof also works for non-coprime f, g (which is impossible
for obvious reasons). However, we used the assumption that they were coprime when we said
the base case was deg g = 1: this is only true because the gcd is 1 so the Euclidean algorithm
eventually yields a pair {1, f } with f = 0, right before the pair {1, 0}. Otherwise, we would have
to do the base case when deg g = −∞ which is clearly impossible.

Exercise A.1.7∗. Let f ∈ K[X1, . . . , Xn] be a polynomial in n variables and suppose S1, . . . , Sn ⊆ K
are subsets of K such that |Si| > degXi f . If f vanishes on S1 × . . . × Sn, prove that f = 0. (This is
the generalisation of Corollary A.1.1 to multivariate polynomials.)

A.1. FIELDS AND POLYNOMIALS 373

Solution

We proceed by induction on n, the base case being the previous proposition. Fix xn ∈ Sn. Then,
the polynomial

g(xn) = f (X1, . . . , Xn−1, xn) ∈ K[X1, . . . , Xn−1]

vanishes on S1 × . . . × Sn−1 and has degree less than |Si| in Xi. Hence, g(xn) = 0. Finally, g is a
polynomial in Xn (with coefficients in the ring K[X1, . . . , Xn−1]) of degree less than |Sn| vanishing
on Sn, which implies that it’s 0 by Corollary A.1.1. (Technically, to use Corollary A.1.1 we would
need to work over a field, while we are only working over a ring: K[X1, . . . , Xn]. However, this
is trivial to fix: this is an integral domain so we can embed it its field of fractions, i.e. work over
the field K(X1, . . . , Xn).)

Exercise A.1.8∗. Prove that (f g) = f g + gf and (f + g) = f + g for any f, g ∈ K[X]. Show
also that (f n) = nf f n−1 for any positive integer n, where f k denotes the kth power and not the kth

iterate. More generally, show that

n n

fi = fi fj .

i=1 i=1 j=i

Solution

Write f = i aiXi and g = j bjXj. We have

(f + g) = k(ak + bk)Xk−1 = iaiXi−1 + jbj Xj−1 = f + g

k ij

which shows additivity. For the multiplication, we have



(f g) =  aibj Xk = kaibj Xk−1

i+j=k i+j=k

and

f g+gf = iaibj Xk−1 + jaibj )Xk−1 = (iaibj + jaibj )Xk−1 = kaibj Xk−1

i+j=k i+j=k i+j=k i+j=k

as wanted. Finally, the last point follows from the (f g) = f g + g f by induction:

n n−1 n−1 n

fi = fn fi + fn fi fj = fi fj .

i=1 i=1 i=1 n=j=i i=1 j=i

The previous point follows is the case f1 = . . . = fn = f .

Exercise A.1.9∗. Prove that every function f : Fp → Fp is polynomial.

Solution
This follows from Lagrange’s interpolation theorem since Fp is finite.

374 APPENDIX A. POLYNOMIALS

Exercise A.1.10∗. Prove that the derivative of a rational function does not depend on its form: i.e.
(f /g) = ((hf )/(hg)) for any f, g, h ∈ K[X] with g, h = 0.

Solution f g−gf
We have (f /g) = g(X)2

and (hf ) (hg) − (hg) hf
g2
(hf /hg) = (hg(X ))2

A.2 Algebraic Structures and Morphisms

Exercise A.2.1∗. Prove that 1R and 0R are unique, and that any element has a unique additive
inverse and a unique multiplicative inverse if it is non-zero.

Solution
This follows from Exercise A.2.9∗.
Exercise A.2.2∗. Let R be a ring. Prove that 0Ra = a0R = 0R for any a ∈ R.

Solution
The proof is the same as for Exercise A.1.1∗.

Exercise A.2.3∗. Prove that char R is the smallest m ≥ 0 such that R contains a copy of Z/mZ

Solution

If R contains a copy of Z/mZ with m ≥ 1 then R has characteristic dividing m which shows the
result when m = 1. If m = 0, then R has characteristic zero since n = 0 for all n ∈ Z. The
converse is clear: the copy Z/mZ is a (mod m) → 1 + . . . + 1 for a ∈ N. (This is well-defined

a times

because the characteristics are the same.)

Exercise A.2.4∗. Prove that the characteristic of a field is either 0 or a prime number p.

Solution

Let c denote the characteristic of a given field K. If c = 0, then c ≥ 2 since the trivial ring is not
a field. Suppose that c = ab. Then, in K, we have ab = 0 which means a = 0 or b = 0 since it’s
an integral domain. By minimality of the characteristic, this means that c = a or c = b.

Exercise A.2.5. Let R be a finite integral domain (i.e. with finitely cardinality). Prove that it is a
field.

A.2. ALGEBRAIC STRUCTURES AND MORPHISMS 375

Solution

Let a ∈ R be non-zero. Consider the powers of a: a, a2, . . .. Since R is finite, there exist i < j
such that ai = aj, i.e. ai(aj−i − 1) = 0. Since a = 0 and R is an integral domain, we get
aj−i − 1 = 0, so that aj−i−1 is the inverse of a.

Exercise A.2.6∗. Prove that a subring of a field is an integral domain.

Solution
If ab = 0 and a = 0 then b = a−1ab = 0.

Exercise A.2.7. What goes wrong if you try to construct the field of fractions of a commutative ring
which isn’t a domain?

Solution

Clearly, if uv = 0, there is something wrong with 1/u. Indeed, we would have 1/u = v/(uv) = v/0
which doesn’t make sense (even formally: 1 · 0 is not equal toThe problem is that a/b = c/d if
ad = bc is not an equivalence relation anymore: we can have a/b = c/d and c/d = x/y but
a/b = x/y. Indeed, this is how the usual proof of transitivity goes: we have ad = bc and cy = dx
so

ady = bcy = bdx
which doesn’t necessarily means ay = bx since d might not be invertible. Here is a concrete
counterexample, if dd = 0, then 1/d = d /0 and d /0 = 1/0 but 1/d = 1/0.

Exercise A.2.8∗. Let R be an integral domain. Prove that R[X] is also one.

Solution

Suppose that f and g are non-zero elements of R[X] with respective leading coefficients a and b.
Then, the leading coefficient of f g is ab since ab = 0 as R is an integral domain, which implies
in particular that f g is non-zero.

Exercise A.2.9∗. Prove that the identity e of a group G is unique, and that any a ∈ G has a unique
inverse. Moreover, prove that (xy)−1 = y−1x−1.

Solution
If e and e are two identities then e = ee = e . The inverse of xy is y−1x−1 since (xy)(y−1x−1) =
xx−1 = e.

Exercise A.2.10∗. Check that (Sn, ◦) is a group.

376 APPENDIX A. POLYNOMIALS

Solution
Since permutations are bijective, they are invertible. Moreover, the identity permutation is the
identity of the group. Finally, it is clear that the operation is associative since composition is.

Exercise A.2.11∗. Prove that a morphism of groups from (G, †) to (H, ) maps the identity of G to
the identity of H.

Solution
Let ϕ be such a morphism and eG, eH be the identities of G and H respectively. We have

ϕ(eG) = ϕ(eG † eG) = ϕ(eG) ϕ(eG)
so ϕ(eG) = eH as wanted (by starring both sides by its inverse).)

Exercise A.2.12∗. Prove that the kernel of a morphism (of rings or groups) is closed under addition.

Solution
If ϕ(a) = 0 and ϕ(b) = 0 then ϕ(a + b) = ϕ(a) + ϕ(b) = 0.

Exercise A.2.13∗. Prove that a morphism of groups is injective iff its kernel is trivial, i.e. consists
of only the identity.

Solution
If it is injective, then the kernel is trivial. Otherwise, suppose that ϕ(a) = ϕ(b) and a = b. Then
ϕ(ab−1) = e so the kernel is non-trivial.

A.3 Exercises

Derivatives

Exercise A.3.1†. Let f, g ∈ K[X] be two polynomials. Prove that the derivative of f ◦ g is g · f ◦ g.

Solution
Write f = i aiXi. Then, (f ◦ g) = i ai(gi) = i iaig gi−1 = g f ◦ g.

Exercise A.3.2†. Let f ∈ K[X] be a non-constant polynomial. Prove that there are a finite number

of g, h ∈ K[X] such that g ◦ h = f , up to affine translation, meaning (g, h) ≡ g(aX + b), h−b .
a

A.3. EXERCISES 377

Solution

By composing with an affine transformation, we may assume that h(0) = 0 and that h is monic.
If we differentiate the equation g ◦ h = f , we get h | f . There is a finite number of such h
since we fixed its leading coefficient and f is non-constant. Since h(0) = 0, there is also a finite
number of h. Since g is uniquely determined from h, we are done.

Exercise A.3.4† (USA TST 2017). Let f, g ∈ R[X] be non-constant coprime polynomials. Prove
that there are at most three real numbers λ such that f + λg is the square of a polynomial.

Solution

The key point is that, if f + λg is a square h2, then h divides f + λg as well as f + λ g = 2hh
so must divide

g (f + λg) − g(f + λg ) = f g − g f

which is independent of f . (Note that this is the determinant of f g which was also used
f g

in Exercise A.3.22†. This explains why it doesn’t depend on λ.)

Also, if f + λg = r2 and f + µg = s2 for µ = λ, r and s are coprime since they two linearly

independent linear conmbinations of f and g, and we know f and g are coprime. Thus, if
f + λig = hi2 for i = 1, . . . , n, we get h1 · . . . · hn | f g − g f as they all divide it and are coprime.
However, when n is large (i.e. greater than 3), the degree of the LHS will be too big so this will
be impossible. Indeed, from f + λig = hi2, we deduce that deg hi is max(deg f, deg g)/2, except
for possibly one value of λ and deg f = deg g. In the first case we are done since

4 max(deg f, deg g)/2 > deg f + deg g − 1,

so we must have f g = g f which is impossible as this would mean f | f since f and g are coprime.
For the second case, if deg(f + λg) is small, note that we can replace f by f + λg (and replace the
λi by other real numbers µi) and this case is now impossible since deg(f + λg) < deg g = deg f .
Note that this doesn’t change the value of f g − g f because we constructed it to be

g (f + λg) − g(f + λg ) = f g − g f.

Exercise A.3.6† (Discrete Derivative). Let f ∈ K[X] be a polynomial of degree n and leading
coefficient a. Define its discrete derivative as ∆f := f (X + 1) − f (X). Prove that, for any g ∈ K[X]
∆f = ∆g if and only if f − g is constant, and that ∆f is a polynomial of degree n − 1 with leading
coefficient an where a is the leading coefficient of f . Deduce the minimal degree of a monic polynomial
f ∈ Z[X] identically zero modulo m, for a given integer m ≥ 1.

Solution

The discrete derivative operator is a morphism (from the space of polynomials of degree at most

n to the space of polynomials of degree at most n − 1), thus it suffices to show that its kernel

consists only of constants. This follows from the second part, that ∆f is a polynomial of degree

n − 1. For this, simply write f = n aiX i. Then,
i=0

n n i−1 i Xj
j
∆f = ai((X + 1)i − Xi) = ai

i=0 i=0 j=0

378 APPENDIX A. POLYNOMIALS

and the term in X n−1 is reached only once for i = n, j = n − 1, with coefficient an n = an.
n−1
Finally, if a polynomial is identically zero modulo m and monic of degree n, ∆nf = n! since the

degree decreases by one every time we apply ∆, while the leading coefficient gets multiplied by

the degree. Thus, m | n!. Conversely, if n is the minimal integer such that m | n!, the polynomial

X = X(X − 1) · . . . · (X − (n − 1))
f = n!
n

works.

Exercise A.3.7†. Let f : R → R be a function. Define its discrete derivative ∆f as x → f (x + 1) −
f (x). Prove that, for any integer n ≥ 0,

∆nf (x) = n n f (x + k).

(−1)n−k k

k=0

Solution

We proceed by induction on n. For n = 0 it is of course trivial. If it’s true for n, then

∆n+1f (x) = ∆(∆nf )(x)

= n n (f (x + k + 1) − f (x + k))

(−1)n−k k

k=0

n+1 n −n f (x + k)
k+1 k
= ((−1)n+1−k

k=0

n+1 nn
+ f (x + k)
= (−1)n+1−k k+1 k

k=0

n+1 n+1 f (x + k).

= (−1)n+1−k k

k=0

Exercise A.3.8†. Let m ≥ 0 be an integer. Prove that there is a polynomial fm ∈ Q[X] of degree
m + 1 such that

n

km = fm(n)

k=0

for any n ∈ N.

A.3. EXERCISES 379

Solution

We proceed by induction on m by noting that n k0 = n + 1 := f0(n) and that
k=0

n

(n + 1)m+1 = (k + 1)m+1 − km+1

k=0

m m+1 n
i
= ki

i=0 k=0

n m−1 m+1
i fi(n)
= (m + 1) km +

k=0 i=0

so that n m−1

fm(n) = km = (n + 1)m+1 − m+1 fi(n)
i m+1
k=0 m+1 i=0

is a polynomial as well. Note also that its leading coefficient is f rac1m + 1.

Roots of Unity

Exercise A.3.9† (Root of Unity Filter). Let f = i aiXi ∈ K[X] be a polynomial, and suppose that
ω1, . . . , ωn ∈ K are distinct nth roots of unity. Prove that

f (ω1) + . . . + f (ωn) = ak .
n
n|k

Deduce that, if K = C,

max |f (z)| ≥ |f (0)|.

|z|=1

(You may assume the existence of a primitive nth root of unity ω, meaning that ωk = 1 for all k < n,

or, equivalently, every nth root of unity are powers of ω. This will be proven in Chapter 3.)

Solution

Let ω be a primitive nth root of unity. Note that, if n m,

n n−1 ωmn − 1
ωm − 1
ωkm = ωkm =

k=1 k=0

since the numerator is zero and the denominator isn’t. When n | m, the sum is simply n 1 =
k=1

n. Thus, we have proven the result for monomials, and the general case follows by taking linear

combinations (if it’s true for f and g it’s also true for af and f + g).

For n > deg f we have f (ω1)+...+f (ωn) = f (0) so
n

max |f (ωk)| ≥ f (ω1) + . . . + f (ωn) = |f (0)|
n
k

by the triangular inequality.

Exercise A.3.10†. Let f = i aiXi ∈ R[X] be a polynomial and ω1, . . . , ωn ∈ C be distinct nth
roots of unity with n > deg f . Prove that

|f (ω1)|2 + . . . + |f (ωn)|2 = ai2.
n
i

380 APPENDIX A. POLYNOMIALS

Denote by S(f ) the sum of the squares of the coefficients of f . Deduce that S(f g) = S(f Xdeg gg(1/X))
for all f, g ∈ R[X]. (Xdeg gg(1/X) is the polynomial obtained by reversing the coefficients of g.)

Solution

Note that

|f (ω)|2 = f (ω)f (ω) = f (ω)f (ω) = f (ω)f (ω−1)

for any ω on the unit circle, since ωω = |ω|2 = 1 for these ω. Thus,

1 n 1 n
n n
|f (ωk)|2 = f (ωk)f (ωk−1)

k=1 k=1

1n aiωki aj ωkj
=
j
n
k=1 i

1n aiaj ωki−j
=

n
k=1 i,j

= a2i

i

by Exercise A.3.9† since n | i − j iff i = j for i, j ∈ [[0, deg f ]], as n > deg f . For the second part,
note that |f (ω)g(ω)| = f (ω)g(1/ω)| for any ω on the unit circle.

Exercise A.3.11†. Let k be an integer. Prove that a∈Fp ak is 0 if p − 1 k and −1 otherwise.
Deduce that any non-constant polynomial f ∈ Fp[X] satisfying f (a) ∈ {0, 1} for all a ∈ Fp must have

degree at least p − 1.

Solution

The first part is Exercise A.3.9† for K = Fp, since non-zero elements of Fp are (p − 1)th roots of
unity by Fermat’s little theorem. For the second, let m be the number of times f (a) = 1. Then,

a∈Fp f (a) ≡ m (mod p). If deg f < p − 1, this sum is zero modulo p by the first part which is
impossible since m ∈ [1, p − 1] (if f is constant over Fp and has degree less than p, f − f (0) has
more roots than its degree so is zero).

Exercise A.3.12†. Let p = 3 be a prime number. Suppose that a and b are integers such that
p | a2 + ab + b2. Prove that (a + b)p ≡ ap + bp (mod p3).

Solution

Note that we can suppose that a, b ≡ 0 (mod p) and reduce the problem to the case where b = 1
by considering x ≡ ab−1 (mod p) so that x2 + x + 1 ≡ 0. In particular, x has order 3 modulo p
since x3 − 1 ≡ (x − 1)(x2 + x + 1) but x ≡ 1 since p = 3. This implies that p ≡ 1 (mod 3) by
Exercise 3.3.4∗. (This is a special case of Theorem 3.3.1.)

The key point is that, since p ≡ 1 (mod 3), we have (X2 + X + 1)2 | (X + 1)p − Xp − 1 := f .

Since (X + 1)p − Xp − 1 ≡ 0 (mod p) by the binomial expansion (see ?? for more details), this

means that (X2 + X + 1)2 divides the polynomial f in Q[X ], and hence in Z[X ] too since it is
p

monic. We conclude that p(X2 + X + 1)2 | f in Z[X] so that

vp(f (x)) ≥ vp(p(x2 + x + 1)) ≥ 3

A.3. EXERCISES 381

as wanted. First, note that X2 + X + 1 is irreducible over Q[X] and that its roots are primitive
third root of unity ω, since X3 −1 = (X2 +X +1)(X −1). Hence, we wish to show that f (ω) = 0,
f (ω) = 0 and f (ω) = 0. We have

f (ω) = (ω + 1)p − ωp − 1 = (−ω2)p − ωp − 1 = 0

since ωp is also a primitive third root of unity. Similarly,

f (ω) = p(ω + 1)p−1 − pωp−1 = pω2(p−1) − pωp−1 = p − p = 0

since 3 | p − 1 and p − 1 is even so we are done.

Remark A.3.1

It has been conjectured that the polynomials (X +1)n −X n −1 where ε = vX2+X+1((X +1)n −Xn −1)
(X 2 +X +1)ε

is 2 if n ≡ 1 (mod 3), 1 if n ≡ −1 (mod 3) and 0 if n ≡ 0 (mod 3) are irreducible. These are

called the Cauchy-Mirimanoff polynomials.

Group Theory

Exercise A.3.14†. Given a group G and a normal subgroup H ⊆ G, i.e. a subgroup such that

x+H −x = H

for any x ∈ G,2 we define the quotient G/H of G by H as G modulo H3, i.e. we say x ≡ y (mod H)
if x − y ∈ H.4 Prove that this indeed a group, and that |G/H| = |G|/|H| for any such G, H.

Solution

G/H is clearly closed under the operation of G and has inverses and an identity. We need however
to check that the operation is well defined: x ≡ y (mod H) and z ∈ G, x + z ≡ y + z (mod H)
and z + x ≡ z + y (mod H). For the former, note that (x + z) − (y + z) = x − y ∈ H since the
inverse of y + z is −z − x, and for the latter note that (z + x) − (z + y) = z + (x − y) − z is in
H because H is normal in G. The second part is obvious: any x ∈ G is equal to exactly |H|
elements modulo H: x + y for y ∈ H.

Exercise A.3.15† (Isomorphism Theorems). Prove the following first, second, and third isomorphism
theorems.

1. Let ϕ : A → B be a morphism of groups. Then, A/ ker ϕ im ϕ. (In particular, ker ϕ is normal
in A and | im ϕ| · | ker ϕ| = |A|.)

2. Let H be a subgroup of a group G, and N a normal subgroup of G. Then, H/H ∩ N HN/N .
(In particular, you need to show that this makes sense: HN is a group and H ∩ N is normal in
H .)

3. Let N ⊆ H be normal subgroups of a group G. Then, (G/N )/(H/N ) G/H.

2In particular, when G is abelian, any subgroup is normal.
3This is where the notation Z/nZ comes from! In fact this shows that, in reality, we should say "modulo nZ" instead
of "modulo n".
4A better formalism is to say that G/H is the set of cosets g + H for g ∈ G. In fact, we will almost always use this
definition in the solutions of exercises (since this is the only place where this will appear), but we introduced it that way
to make the analogy with Z/nZ clearer.

382 APPENDIX A. POLYNOMIALS

Solution

1. Note that ker ϕ is normal in A. Indeed, if ϕ(x) = 1, then ϕ(yxy−1) = ϕ(y)ϕ(x)ϕ(y)−1 = 1
too. Second, note that every element in the image of ϕ has exactly one one preimage in
A/ ker ϕ: indeed, if ϕ(x) = ϕ(y), then xy−1 ∈ ker ϕ so they are equal modulo ker ϕ. This
shows that it is an isomorphism (it is clearly surjective, and we have shown it was injective
too).

2. Note that H ∩ N is normal in H since N is so hH ∩ N h−1 ⊆ N but this is also in H
when h ∈ H so must be equal to H ∩ N . Note also that HN is indeed a group since, if
gm, hn ∈ HN , then mh = h for some ∈ N as N is normal, so

gmhn = gh n ∈ HN.

Similarly, gm = kg for some k ∈ G so (gm)−1 = g−1k−1 ∈ HN . Now, consider the natural
map from H to HN/N , sending h to hN . Its kernel consists of the h such that hN = N , i.e.
h ∈ N . Hence, its kernel is H ∩ N so we get H/H ∩ N HN/N by the first isomorphism
theorem.

3. Consider the surjective map G/N → G/H which sends gN to gH. It is well defined
since N ⊆ H. gN is in the kernel if gH = H, i.e. g ∈ H. Hence, the kernel consists
of hN for h ∈ H, i.e. H/N . We conclude from the first isomorphism theorem that
G/H (G/N )/(H/N ) as wanted.

Exercise A.3.16†. Let G be a finite group, ϕ : G → C× be a non-trivial group morphism (i.e. not
the constant function 1), where (C×, ·) is the group of non-zero complex numbers under multiplication.

Prove that g∈G ϕ(g) = 0.

Solution
Note that, for any h ∈ G, g → hg is a bijection so

ϕ(g) = ϕ(hg) = ϕ(h) ϕ(g)

g∈G g∈G g∈G

which means that g∈G ϕ(g) = 0 by picking an h such that ϕ(h) = 1.

Remark A.3.2

Alternatively, this can be done as follows: the image of ϕ is a subgroup of the group of |G|th roots
of unity by Lagrange, so must be the group of nth roots for some n, greater than 1 by assumption
(this is just the fact that subgroups of cyclic groups are also cyclic). Let ω = exp(2iπ/n) be a
primitive nth root of unity. Hence, we have

x = n−1 = ωn − 1 = 0

ωk ω−1

x∈im ϕ k=0

since the numerator is zero while the denominator isn’t, as n > 1. To conclude, by the first

A.3. EXERCISES 383

isomorphism theorem from Exercise A.3.15†, we have

ϕ(g) = |G| x = 0.

g∈G ker ϕ x∈im ϕ

Exercise A.3.17† (Lagrange’s Theorem). Let G be a group of cardinality n (also called the order of
G). Prove that gn = e for all g ∈ G. In other words, the order of an element divides the order of the

group. More generally, prove that the order of a subgroup divides the order of the group.

Solution
See Theorem 2.5.1 and Exercise 6.3.15∗.

Exercise A.3.18† (5/8 Theorem). Let G be a non-commutative finite group. Prove that the proba-
bility

p(G) = |{(x, y) ∈ G2 | xy = yx}|

|G|2

that two elements commute is at most 5/8.

Solution

Denote by Z the center of the group, i.e. the set of elements which commute with every other

one. For a given x ∈ G, denote also by C(x) the centraliser of x, i.e. the set of y such that x and

y commute. The wanted probability is x∈G |C (x)| . Note that C(x) are subgroups of G (and
|G|2

hence Z is too): if xy = yx and xz = zx then

xyz = yxz = yzx.

First, let’s see how big the center can be. It’s a subgroup of G, so its cardinality divides |G| by

Lagrange’s theorem Exercise A.3.17†. It can’t be |G| since G is non-abelian, it can’t be |G| since
2

G/Z is then isomorphic to Z/2Z so is generated by one element and hence G is generated by Z

and one additional element which means that it’s commutative:

amxany = am+nxy = anyamx

for x, y ∈ Z. For the same reason, it can’t be |G| since G/Z still has prime order so must be
3
|Z |
generated by one element by Lagrange’s theorem. Thus, |G| ≤ 1 .
4

Now, if x ∈ Z, C (x) is a subgroup of G distinct from it so has cardinality at most |G| . To
2

384 APPENDIX A. POLYNOMIALS
conclude,

|{(x, y) ∈ G2 | xy = yx}| |C (x)|
|G|2 = |G|2
x∈G

|G| |C(x)|
= |G|2 +
x∈Z x∈Z |G|2

≤ |Z | + (|G| − |Z |) · |G|/2
|G| |G|2

= |Z | + 1 − |Z |
|G| 2 2|G|

|Z| 1
= 2|G| + 2

≤1+1
82
5

=.
8

Remark A.3.3

One can check that the bound 5/8 is achieved by the quaternion group Q8 consisting of the
elements e, b, b2, b3, a, ab, ab2, ab3 under the presentation a4 = b4 = e, a2 = b2, and ba = ab3.

Exercise A.3.19† (Fundamental Theorem of Finitely Generated Abelian Groups). Let G be an

abelian group which is finitely generated, i.e., if we write its operation as +, there are g1, . . . , gk ∈ G

such that any g ∈ G can be represented as n1g1 + . . . + nkgk for integers ni ∈ Z. Prove that there
is a unique integer n ≥ 0 (called the rank of the group) and a unique sequence of positive integers

d1 | . . . | dm such that

(G, +) (Zn × Z/d1Z × . . . × Z/dkZ, +).

Solution

This problem has two parts: proving that a finite abelian group is isomorphic to a product of
cyclic groups in the wanted way, and proving that the torsion T of a finitely generated abelian
group, i.e. the set of elements with finite order (which is a subgroup here since G is abelian) is
finite and that G Zn × T for some n.

For the first part, pick an element h ∈ G of maximal order m. We claim that the order of any
element g ∈ G divides m. (We know that this must be true by the statement: this m is our dk.
Note however that this is false for non-abelian groups.) Indeed, suppose that x, y ∈ G have order
a, b. We will construct an element of order lcm(a, b). First suppose that a and b are coprime.
Then, a(x + y) = ax has order b since gcd(a, b) = 1, and similarly b(x + y) = by has order a.
Thus, the order of x + y is divisible by a and b, and hence by ab. Conversely, it clearly divides
ab so must be exactly ab.

Now, if a and b are not necessarily coprime, let a = vp(a)≥vp(b) pvp(b) and b = vp(a)<vp(b) pvp(b)
so that a , b are coprime and have product lcm(a, b). The elements (a/a )x and (b/b )y have

respective orders a and b so we are done by the previous step since a and b are coprime.

Now, let H = h be the subgroup generated by g, i.e. {0, g, . . . , (m − 1)g}. This is isomorphic

to Z/mZ. We claim that

G H × G/H.

Continuing in this fashion with G/H (which has a strictly smaller cardinality unless G is already
trivial) yields the wanted decomposition, since we have shown that the di are divisible by the

A.3. EXERCISES 385

previous one (m is divisible by the order of any element). To prve that G H × G/H, we will
find a morphism ϕ from G to H which is the identity H. Indeed, g → (ϕ(g), g (mod H)) will
then be the wanted isomorphism between G and H × G/H: if ϕ(g) = ϕ(g ) and g ≡ g (mod H),
then ϕ(g − g ) = g − g since it is the identity on H so we must have g = g . Thus, our morphism
is injective and hence bijective since |G| = |H| · |G/H|.

We proceed by induction on the minimal number of elements needed to generate G from H.
When H = G it is trivial. Now, suppose ϕ is a morphism from G ⊆ G to G and let g ∈ G \ G .
We will extend ϕ to G , g , the subgroup generated by G and g as desired. Let n be the
order of y in G/G , i.e. the smalles k such that ny ∈ G . Then, ky ∈ G ⇐⇒ n | k. Thus,
ϕ(g + kg) := ϕ(g ) + kϕ(g) is well-defined as long as ϕ(g) is such that

ϕ(ng) = nϕ(g).

Now, note that n divides the order of g which divides m = |H|. Hence, it is always possible to
find such a ϕ(g): if ϕ(ng) = kh, since mg = 0, we have (mk/n)h = 0, i.e. n | k which means that
ϕ(g) = (k/n)h works. Note also that this reasoning shows that the decomposition is unique too.

Now, we prove that torsion-free finitely generated abelian groups are isomorphic to Zn for a
unique n. But first, we show how the problem follows from these two special cases. Note that
G/T is torsion-free: if x (mod T ) has finite order, then nx ∈ T for some n so x has finite order,
i.e. x ∈ T . Pick a basis α1 (mod T ), . . . , αn (mod T ). Now, we claim that

G T × (α1Z + . . . + αnZ) T × Zn

as wanted. This follows from the simple isomorphism (x, y) → x + y. This is surjective by
definition, since α1Z + . . . + αnZ is a system of representatives of G/T . For the injectivity,
note that, if x + y = x + y , then y − y = x − x ∈ T so y = y and thus x = x since
α1Z + . . . + αnZ G/T has trivial torsion. There is one last thing we need to show however:
that T is finite. Pick an isomorphism ϕ : G → T × Zn. Then, the first coordinates of the image
of a generating family of elements of G generate T . Since they all have finite order, they generate
a finite number of elements as wanted.

Hence, we only need to prove that if G has trivial torsion, it is isomorphic to Zn for some n.
Note that this n is unique: if we had an isomorphism from Zm to Zn, we would have one from
(Z/2Z)m → (Z/2Z)n by reducing it modulo 2, and this forces m = n. Pick a generating set of
minimal cardinality α1, . . . , αn. We wish to prove that it is linearly independent. Suppose that
it is not the case, and let N = 0 be the minimum value of the absolute values of the coefficients
of a non-trivial linear combination which is zero. In fact, we shall also pick the generating set
to minimise N . The contradiction will then come from a construction of another generating set
with zero linear combination with smaller coefficients.

Suppose that k1α1 + . . . + knαn = 0 and N = |k1| + . . . + |kn|. Suppose without loss of generality
that 0 < |k1| < |k2|. Say we replace the family α1, . . . , αn by α1 ± α2, α2, . . . , αn. Then,
k1α1 + . . . + knαn = 0 becomes

k1(α1 ± α2) + (k2 ∓ k1)α2 + k3α3 . . . + knαn = 0.

By picking the ±1 sign appropriately, we ensure that |k2 ∓ k1| < |k2| thus leading to a smaller
value of N , which is a contradiction. We are done.

Exercise A.3.20† (Burnside’s Lemma). Let G be a finite group, S a finite set, and · a group action
of G on S, meaning a map · : G × S → S such that e · s = s and (gh) · s = g · (h · s) for any g, h ∈ G
and s ∈ S. Given a g ∈ G, denote by Fix(g) the set of elements of s fixed by g. Prove that

|S/G| = 1 Fix(g),
|G|
g∈G

386 APPENDIX A. POLYNOMIALS

where |S/G| denotes the number of (disjoint) orbits Oi = Gsi. Deduce the number of necklaces that
have p beads which can be of a colours, where p is a prime number and two necklaces are considered
to be the same up to rotation.

Solution

Consider the sum g∈G | Fix(g)|. This is equal to the number of pairs (g, s) such that gs = s.
Hence, this is also equal to s∈S | Stab(s)|, where Stab(s) denotes the elements of G fixing s.
Now consider the orbit Gs of s. We claim that |Gs| = |G/ Stab(s)| = |G|/| Stab(s)|. Indeed,

the map from the left-cosets G/ Stab(s) to Gs sending g Stab(s) to gs is clearly a bijection: if
gs = hs then h−1g ∈ Stab(s) so g Stab(s) = h Stab(s). Hence,

| Fix(g)| = |G| 1
|Gs|
g∈G s∈S

1
= |O|

O∈S/G x∈O

=1

O∈S/G

= |S/G|

as desired.

For the second part, consider the cyclic group Z/pZ acting on the sets of words (necklaces) in
an alphabet (the set of colours) of size a. Why did we choose Z/pZ? Because we consider the
necklaces up to rotation. The action of g ∈ Z/pZ is of course a rotation of g beads, say to
the right. Then, there is one element fixing all words: 0, and all the other ones only fix words
with all letters equal, i.e. monochromatic necklaces. Indeed, if 0 = g ∈ Z/pZ fixes a necklace,
then so does Z/pZ = gZ/pZ which means that the necklace is invariant under all rotations, i.e.
monochromatic. Hence, the number of necklaces is

ap + (p − 1)a
p

by Burnside’s lemma. Notice that this also proves Fermat’s little theorem..

Miscellaneous

Exercise A.3.21† (China TST 2009). Prove that there exists a real number c > 0 such that, for any
prime number p, there are at most cp2/3 positive integers n satisfying n! ≡ −1 (mod p).

Solution

We shall prove that any set S such that a! ≡ b! ≡ 0 (mod p) has cardinality at most 2p2/3.
Consider the polynomial following polynomial

fm = (X + 1) · . . . · (X + m) − 1 ∈ Fp[X].

Since Fp is a field, fm has at most m roots in Fp. Thus, there are at most m integers n such that
n! ≡ (m + n)!, since this is equivalent to fm(n) = 0.
Let k be an integer which we will specify later on. Let N be the set of pairs of elements of S at
a distance less than k, i.e.

N = {{a, b} ⊆ S | a = b, |a − b| < k.}

A.3. EXERCISES 387

By our previous result,

|N | ≤ 1+ 2+ ... + (k − 1) < k2
.
2

Now, let M = {a | ∃b : {a, b} ∈ N }. Consider S \ M . By definition, for any a, b ∈ S \ M , we

have |a − b| ≥ k. Since the elements of S are between 0 and p − 1, by the pigeonhole principle,

we have |S \ M| ≤ p + 1. To conclude,
k

|S| ≤ |S \ M | + |M | ≤ |S \ M | + |N | ≤ p + k2 + 1.
k2

If we now pick k = √ , we get |S| ≤ 2p2/3 as wanted.
3p

Exercise A.3.22† (Mason-Stothers Theorem, ABC conjecture for polynomials). Suppose that A, B, C ∈
C[X] are coprime polynomials such that A + B = C. Prove that

1 + max(deg A, deg B, deg C) ≤ deg(rad ABC)

where rad ABC is the greatest squarefree divisor of ABC (in other words, deg(rad ABC) is the number
of distinct complex roots of ABC). Deduce that the Fermat equation f n + gn = hn for f, g, h ∈ C[X]
does not have non-trivial solutions for n ≥ 2.

Solution

Consider the determinant D = det A B = AB − BA . Note that this is the same up sign
A B

when we replace A and B by two polynomials out of A, B, C: this is because the determinant

is invariant up to sign under column operations (adding certain columns to other columns and

exchanging columns, see Proposition C.3.4). Of course, it can also be proven by computing it

explicitly: (A + B)B − B(A + B) = AB − BA (and the rest follows by symmetry). Thus, r

is a double root of ABC only if it is a root of D: indeed such a root must a double root of one

of A, B, C since they are coprime, say A. It is then a common root of A and A so of D too.

However, a lot more holds. if v is the multiplicity of r in ABC (thus in A in our case), r is a

root of multiplicity v − 1 of D since it’s a root of multiplicity v − 1 of A . Thus,

ABC | rad(ABC)D,

which gives the wanted bound since deg D ≤ deg A + deg B − 1 and the same with B, C and C, A
by symmetry.

Suppose that A = f n, B = gn, C = hn are non-zero and satisfy A + B = C. Then,

1 + n max(deg f, deg g, deg h) = 1 + max(deg A, deg B, deg C)
≤ deg(rad ABC)
= deg(rad f gh) ≤ deg f + deg h + deg h

so n < 3 as wanted.

Exercise A.3.23†. Find all polynomials f ∈ C[X] which send the unit circle to itself.

Solution

As in Exercise A.3.9†, f (z) = f (z−1) for any z on the unit circle. Thus, 1 = |f (z)|2 = f (z)f (z−1).
Hence, f (z)(znf (z−1)) = zn for z on the unit circle, where n = deg f . Note that Xnf (1/X) is

388 APPENDIX A. POLYNOMIALS

indeed a polynomial: if f = n ai X i , then Xnf (1/X) = n an−iX i.
i=0 i=0

Thus, the polynomials f (X)(Xnf (1/X)) and Xn have infinitely many roots in common, which
mean that they are equal. In particular, f | Xn, which implies that f = εXk for some ε and
some k. It is clear that ε must be on the unit circle, and conversely any such ε works (in other

words, the polynomials which send the unit circle to itself contract it and then rotate it).

Exercise A.3.26† (Gauss-Lucas Theorem). Let f ∈ C[X] be a polynomial with roots α1, . . . , αk.
Prove that

f1
=.

f k X − αk

Deduce the Gauss-Lucas theorem: if f ∈ C[X] is non-constant, the roots of f are in the convex hull of
the roots of f , that is, any root β of f is a linear combination i λiαi with i λi = 1 and non-negative
λi ∈ R.

Solution

The identity follows from Exercise A.1.8∗. Let α be a root of f , without loss of generality such

that f (α) = 0. We have

0= n 1 = n α − αk
i=1 α − αk i=1 |α − αk|2

so that n n
i=1
α 1 = |α αk |2 .
|α − αk|2 − αk
i=1

If we now conjugate this equality, we get

α= n αk
i=1 |α−αk|2

n1
i=1 |α−αk|2

which has the desired expression.

Remark A.3.4

You may notice that the first identity is the logarithmic derivative (log f ) . This can be used to

produce an analytic proof of this identity: it holds when X > αk for all k (in particular they are

all real), but is also a polynomial identity in X and the αk, so it must hold polynomially. More

specifically, if we fix the αi ∈ R, it holds for sufficiently large X so it must hold for all X. Thus,
it holds for all αi, X ∈ R which means that it always holds by Exercise A.1.7∗.

Exercise A.3.27† (Sturm’s Theorem). Given a squarefree polynomial f ∈ R[X], define the sequence

f0 = f , f1 = f and fn+2 is minus the remainder of the Euclidean division of fn by fn+1. Define also

V (ξ) as the number of sign changes in the sequence f0(ξ), f1(ξ), . . ., ignoring zeros. Prove that the
number of distinct real roots of f in the interval ]a, b] is V (a) − V (b).5

5If we choose a = −∞, b = +∞, this gives an algorithm to compute the number of real roots of f , by looking at the
signs of the leading coefficients of f0f1, . . ..

A.3. EXERCISES 389

Solution

When x increases from a to b, it may pass through a zero of some fk (otherwise, by the interme-
diate value theorem, V (a) = V (b) and there is clearly no root in the intervall as claimed). We
shall prove that this leaves V (x) invariant if k ≥ 1, and decreases it by 1 precisely when k = 0,
i.e. x is a root of f . Before doing that, note that the important part of the definition of (fn)n≥0
is that fn+1 ≡ −fn−1 (mod f )n for all n. In particular, if fn(x) and fn+1(x) are zero, then so is
fn−1(x), which implies, by induction that x is a root of every fi. This is impossible since f0 = f
and f1 = f have no common root by assumption.

First, suppose that fi(ξ) = 0 for some ξ and i ≥ 1. Then, since fi+1 ≡ −fi−1 (mod fi), fi+1(x)
and fi−1(x) have opposite signs around ξ (and are non-zero by our previous observation). This
means that, before ξ, we had one sign change in (fi−1(x), fi(x), fi+1(x)) since this has the form
(±1, ε, ∓1) for ε ∈ {−1, 1}. After ξ and at ξ, we still have one sign change for the same reason.
Hence, V (x) stays invariant when x passes through a root of some fi with i ≥ 1.

Now, suppose that f (ξ) = 0. Then, around ξ, f (ξ + ε) = εf (ξ) + O(ε2) which means that the
sign of f (x) flips before and after ξ, while the sign of f does not change since ξ is a simple root.
More precisely, before ξ, f (x) and f (x) had opposite sign, while they have the same sign after
ξ. At ξ, we do not count a sign change since f (ξ) = 0 so V (ξ) = V (ξ + ε) for sufficiently small
ε > 0, which finishes the proof.

Exercise A.3.28† (Ehrenfeucht’s Criterion). Let K be a characteristic zero field, let f1, . . . , fk ∈ K[X]
be polynomials and define

f = f1(X1) + . . . + fk(Xk) ∈ K[X1, . . . , Xk].

If k ≥ 3, prove that f is irreducible. In addition, prove that this result still holds if k = 2 and f1 and
f2 have coprime degrees.

Solution

Let us first do the case k = 2. Suppose that f (X) + g(Y ) is reducible, say equal to uv. Let
m = deg f and n = deg g. Consider f (Xn) + g(Y m), which is a polynomial of degree mn in
both X and Y . Let r and s be the homogeneous parts of u(Xn, Y m) and v(Xn, Y m), i.e. the
polynomial formed by the monomials of highest degree of u(Xn, Y m) and v(Xn, Y m). By looking
at the degrees, we must have rs = aXmn + bY mn where a and b are the leading coefficients of u
and v respectively.

Suppose without loss of generality (by symmetry) that r has at least two monomials, i.e. u has
at least two monomials Xi1 Y j1 and Xi2 Y j2 such that

ni1 + mj1 = ni2 + mj2 ⇐⇒ n(i1 − i2) = m(j1 − j2).

Since m and n are coprime, this implies n | j1 − j2 and m | i1 − i2. But then, degX u ≥ m and
degY u ≥ n, which implies that s is constant in both X and Y , i.e. constant, since f (X)+g(Y ) =
uv. This is a contradiction.

Now suppose k ≥ 3 and f = uv. Let ni = deg fi and let ai be the leading coefficient of fi. The
same argument as before shows that

rs = a1X1N + . . . + akXkN ,

where N = n1 · . . . · nk (we replace Xi by XiN/ni and take homogeneous parts). Thus, we have
reduced the problem to the case of monomials. We can however reduce it even further: if we
evaluate this at (X, Y, 1, 0, . . . , 0), we get that aXN + bY N + c is reducible in K[X, Y ] (the

390 APPENDIX A. POLYNOMIALS

factorisation we get is non-trivial since r and s have degree < N so still degree < N when we
evaluate them), say

aXN + bY N + c = (gM XM + . . . + g0)(hN−M XN−M + . . . + h0)

for some polynomials gi, hi in Y of degree < N . Now, substitute y a complex root of bY N + c to
Y . This gives us the polynomial aXN which can only be factored as a product of two monomials,
so

g0(y) = . . . = gM−1(y) = hN−M−1(y) = . . . = h0(y).
But since the roots of bY N +c are distinct (there is no common root with the derivative N bY N−1),
gi and hj for i < M and j < N − M vanish at N distinct points, which is more than their degree.
Thus, they must be zero. This leaves us with the factorisation aXN + bY N + c = gM hN−M XN
which is clearly impossible since XN doesn’t divide the LHS.

Exercise A.3.29† (IMC 2007). Let a1, . . . , an be integers. Suppose f : Z → Z is a function such that

n

f (kai + ) = 0

i=1

for any k, ∈ Z. Prove that f is identically zero.

Solution

Consider the set I of polynomials f = m bi X i ∈ Q[X ] such that
i=0

m

bif (i + x) = 0

i=1

for any x ∈ Z. We claim that this set is an ideal of Q[X], meaning that it’s closed under
addition, and closed under multiplication by any polynomial in Q[X]. The first fact is clear. For
the second, note that multiplication by Xi corresponds to a translation and that multiplication

by a constant is trivial, so we can deduce it from the first fact. Thus, I is closed under gcd: by
Bézout’s lemma, if f, g ∈ I, there are u, v ∈ Q[X] such that

gcd(f, g) = uf + vg ∈ I.

Our goal is to show that I contains the element 1: this gives f (x) = 0 for any x ∈ Z as wanted.

The statement gives us that

n

f = Xkai+ ∈ I

i=1

for any k, such that kai + ≥ 0 for all i. Hence, the problem reduces to proving that these
n
polynomials are coprime, i.e., that for any algebraic number α, i=1 αkai can not always be

zero. This follows from our proof of Theorem C.4.1: this is a linear recurrence, and the only

linear recurrence which is identically zero is the zero recurrence. However, n αkai is clearly
i=1
not the zero recurrence since the coefficient before αkai for every i.

Appendix B

Symmetric Polynomials

B.1 The Fundamental Theorem of Symmetric Polynomials

Exercise B.1.1. Let f ∈ K(X1, . . . , Xn) be a rational function, where K is a field. Suppose f is
symmetric, i.e. invariant under permutations of X1, . . . , Xn. Prove that f = g/h for some symmetric
polynomials g, h ∈ K[X1, . . . , Xn].

Solution
Let r = f /g be a symmetric rational function. We write it as

r = σ∈Sn f (σ(X1, . . . , Xn)) .
g id=σ∈Sn f (σ(X1, . . . , Xn))

The numerator is a symmetric polynomial so the denominator must be too since the quotient
is.

Exercise B.1.2. Prove that the decomposition of a symmetric polynomial f as g(e1, . . . , en) is unique.

Solution
This accounts to proving that f (e1, . . . , en) = 0 if and only if f = 0. This is clear by induction on
n (trivial when n = 1). Let f be such a polynomial and suppose for the sake of a contradiction
that en | f . If we set Xn = 0 we get

f (e1, . . . , en−1, 0) = 0
where the ei are now the elementary symmetric polynomials in X1, . . . , Xn−1. By the induction
hypothesis, this means that f (X1, . . . , Xn−1, 0) = 0, i.e. Xn | f . By symmetry, en = X1 ·. . .·Xn |
f , a contradiction.

B.2 Newton’s Formulas

Exercise B.2.1∗. Prove Corollary B.2.1.

391

392 APPENDIX B. SYMMETRIC POLYNOMIALS

Solution
We have K(p1, . . . , pn) ⊆ K(e1, . . . , en) by the fundamental theorem of symmetric polynomials,
and the reverse inclusion comes from the Newton formulas by induction, as explained before. (We
need the assumption that K is a field because the LHS of the Newton’s formulas has a factor of
k which we need to divide by in the inductive step, and we need K to have characteristic zero
so that k = 0.)

B.3 The Fundamental Theorem of Algebra

Exercise B.3.1∗. Prove Proposition B.3.2.

Solution

By the quadratic formula (or completing the square), solving quadratic equations is equivalent
to finding square roots. Thus, let a + bi ∈ C be a complex number, with a, b ∈ R. We wish to
find a square root x + iy or a + bi, i.e.

x2 − y2 + 2ixy = (x + iy)2 = a + bi.
This means x2 − y2 = a and 2xy = b. This is equivalent to x2 and −y2 being roots of X2 −
aX − b2/4 by Vieta’s formulas. Since the constant coefficient is negative, the roots are real (e.g.
by the intermediate value theorem), and since the product is negative, one is positive and one
negative. Label the positive one as x2 and the negative one as −y2, take the square roots to find
x and y and adjust the sign to have 2xy = b.

B.4 Exercises

Newton’s Formulas

Exercise B.4.2† (Hermite’s Theorem). Prove that a function f : Fp → Fp is a bijection if and only
if a∈Fp f (a)k is 0 for k = 1, . . . , p − 2 and −1 for k = p − 1.

Solution
If f is a bijection, then this is Exercise A.3.11†. Now suppose that this condition holds. Newton’s
formulas (note k = 0 for k < p so Corollary B.2.1 still holds) tell us that there is only one possible
value of ek(f (0), . . . , f (p − 1)) for any fixed k. Hence, we must have

ek(f (0), . . . , f (p − 1)) = ek(0, . . . , p − 1)
since 0, . . . , p − 1 satisfy the condition. This implies that

(X − f (0)) · . . . · (X − f (p − 1)) = (X − 0) · . . . · (X − (p − 1))
so f is a bijection as wanted.

Exercise B.4.3†. Suppose that α1, . . . , αn are such that α1k + . . . + αkn is an algebraic integer for all
n. Prove that α1, . . . , αk are algebraic integers.

B.4. EXERCISES 393

Solution

Newton’s formulas give us k!ei(α1, . . . , αk) ∈ Z for any i. Thus, k!α ∈ Z for any α = αi, by
Exercise 1.5.22†. In particular, since the statement is also true when we replace the αi by αim
for any fixed m, we get k!αm ∈ Z for any m.

Thus, the problem reduces to showing that, if α ∈ Q is algebraic and such that N αn ∈ Z (i.e.
powers of α have bounded denominator) for some non-zero N ∈ Z and any positive integer n,
then α ∈ Z. For large n, the degree of α2n is constant, since the sequence

[Q(α2n ) : Q] = [Q(α2n ) : Q(α2n−1 )][Q(α2n−1 ) : Q]

is a non-increasing sequence of integers. By replacing α by α2m for some large m, we may assume
that this is true for any n ≥ 0. Let β1, . . . , β be the conjugates of α. Consider the minimal
polynomial

f2k = X − βi2k

i=1

of α2k and let Nk = 1/c(f2k )) be the smallest positive integer such that Nkf2k ∈ Z[X]. By
assumption Nk is bounded. However, we have

Nk2f2k+1 (X2) = Nk2 X2 − β 2n+1

i

i=1

= Nk X − βi2n Nk X + βi2n

i=1 i=1

= ±(Nkf2k )(Nkf2k (−X))

which is primitive by Gauss’ lemma 5.1.2. Hence, Nk+1 = Nk2 so N1 must be 1 otherwise
N 2n−1
Nk = → ∞. This means that the minimal polynomial of α has integral coefficients, i.e. α
1

is an algebraic integer.

Remark B.4.1
It is necessary to mention that the key claim admits a very short and intuitive proof if we allows
ourself some ideal theory. The idea is that, if α ∈ Q, we can simply look at the p-adic valuations
to get nvp(α) + vp(N ) ≥ 0 which gives us vp(α) ≥ 0 for large enough n. Hence, α is an integer.
For arbitrary algebraic integers, the same proof works almost verbatim: a number field K is not
always a UFD but always has ideal factorisation. This means that we can this time consider
prime ideals p of K to get nvp(α) + vp(N ) ≥ 0 which implies vp(α) ≥ 0 again. Finally, since this
is true for any prime ideal p, we get α ∈ OK.

Algebraic Geometry

Exercise B.4.4† (Resultant). Let R be a commutative ring, and f, g ∈ R[X] be two polynomials of
respective degrees m and n. For any integer k ≥ 0, denote by Rk[X] the subset of R[X] consisting of
polynomials of degree less than k. The resultant Res(f, g) is defined as the determinant of the linear
map

(u, v) → uf + vg

394 APPENDIX B. SYMMETRIC POLYNOMIALS

from Rm[X] × Rn[X] to Rm+n[X]. Prove that, if f = i aiXi and g = i biXi, we have1

a0 0 · · · 0 b0 0 · · · 0

a1 a0 · · · 0 b1 b0 · · · 0

a2 a1 . . . 0 b2 b1 . . . 0

... ... . . . a0 ... ... . . . b0
Res(f, g) = ... ,
am am−1 · · · ... bn bn−1 · · ·

0 am . . . ... 0 bn . . . ...

... ... . . . am−1 ... ... . . . bn−1

0 0 · · · am 0 0 · · · bn

and, if f = a i X − αi and g = b j X − βj, then2

Res(f, g) = ambn αi − βj.

i,j

In addition, prove that Res(f, g) ∈ (f R[X] + gR[X]).3 Finally, prove that if f, g ∈ Z[X] are monic and
uf +vg = 1 for some u, v ∈ Z[X], Res(f, g) = ±1. (It is not necessarily true that (f R[X]+gR[X])∩R =
Res(f, g)R for specific polynomials f, g, but we always have Res(f, g) ∈ f R[X]+gR[X] by the previous

point.)

Solution

The determinant form of the resultant simply follows from considering the matrix of the linear
function corresponding to the basis 1, X, . . . , Xm+n−1. To prove the explicit formula, consider

the case where A = a, B = b, αi = Ai and βj = Bj are indeterminates. Working over a field
K, the resultant vanishes when Ai = Bj for some i, j since the map is not surjective: it never
reaches 1. Thus, the resultant is divisible by Ai − Bj for all i, j. Looking at the determinant
formula, we see that the degree of Res(f, g) in A1 is n and its leading coefficient is AmBn, which

proves the wanted formula.

For the second part, write the equation uf +vg = r in the monomial basis as RV = (r, 0, . . . , 0) :=

re1, where R is the matrix corresponding to the linear map (u, v) → uf + vf . Hence, we wish to
have rR−1e1 ∈ Rn. ?? tells us that r = det R = Res(f, g) works.

Now let f and g be generic polynomials with integer coefficients of respective degree m and n.

Suppose finally that (f Z[X] + gZ[X]) ∩ Z = Z. Write f and g as m X − αi and n X − βi.
i=1 j=1

We have u(βi)f (βi) = 1 for each i, so

f

(βi) = ± Res(f, g)

i=1

divides 1 as desired.

Exercise B.4.6† (Hilbert’s Nullstellensatz). Let K be an algebraically closed field. Suppose that
f1, . . . , fm ∈ K[X1, . . . , Xn] have no common zeros in K. Prove that there exist polynomials g1, . . . , gm
such that

f1g1 + . . . + fmgm = 1.

1This is an (m + n) × (m + n) matrix, with n times the element a0 and m times the element b0.

n(n−1)

2In particular, the discriminant of f is (−1) a 2 · Res(f, f ).
3In other words, the resultant provides an explicit value of a possible constant in Bézout’s lemma for arbitrary rings
(such as Z).

B.4. EXERCISES 395

Deduce that, more generally, if f is a polynomial which is zero at common roots of polynomials
f1, . . . , fm (we do not assume anymore that they have no common roots), then there is an integer k
and polynomials g1, . . . , gm such that

f k = f1g1 + . . . + fmgm.

Solution

We proceed by induction on the number n of variables. When n = 1 this is just Bézout’s lemma.
Now, if n ≥ 1, we will eliminate one variable with the resultant. Consider the polynomial

g = ResXn (fm, U1f1 + . . . + Um−1fm−1) ∈ K[U1, . . . , Um−1][X1, . . . , Xn−1],

where U1, . . . , Um−1 are formal variables. By Exercise B.4.4†, (x1, . . . , xn−1) is a root of g if and
only if fm and U1f1 + . . . + Um−1fm−1 have a common root xn at (x1, . . . , xn−1), i.e. (x1, . . . , xn)
is a common root of f1, . . . , fm, or if the leading coefficient in Xn of fm and U1f1+. . .+Um−1fm−1
vanish at (x1, . . . , xn−1), i.e. the leading coefficient in Xn of f1, . . . , fm vanish at (x1, . . . , xn−1)
(we say (x1, . . . , xn−1) is a common root at infinity). We wish to rule out the second case. This
is not very hard: perform the change of coordinates Xi → Xi + ciXn for i = 1, . . . , n − 1 and
some ci to get constant leading coefficients in Xn (thus sharing no common root).

Hence, g has no root by assumption since f1, . . . , fm have no common root. However, a root of
g is simply a common root of its coefficients gi when g is seen as a polynomial in U1, . . . , Um−1.
This implies that a linear combination of the gi is 1, by the induction hypothesis. Finally, notice
that

g = ResXm (f, U1f1 + . . . + Um−1fm−1) = uf + v(U1f1 + . . . + Um−1fm−1)
for some u, v ∈ K[X1, . . . , Xn][U1, . . . , Um−1], by Exercise B.4.4†. Hence, the coefficients gi of g
are linear combinations of the fi (with coefficients in K[X1, . . . , Xn]). We conclude that a linear
combination of the fi is 1 as wanted.

For the second part, suppose without loss of generality that f = 0. Use the first part on
f1, . . . , fm, 1−Xn+1f which have no common root by assumption (this is known as Rabinowitsch’s
trick). Thus, there are g1, . . . , gm, g ∈ K[X1, . . . , Xn+1] such that

g1f1 + . . . + gmfm + g(1 − Xn+1f ) = 1.

Now, evaluate this at Xn+1 = 1/f and multiply by a large enough power of f to get the wanted
equality.

Exercise B.4.7† (Weak Bézout’s Theorem). Prove that two coprime polynomials f, g ∈ K[X, Y ] of

respective degrees m and n have at most mn common roots in K. (Bézout’s theorem states that they
have exactly mn common roots counted with multiplicity, possibly at infinity.4)

Solution

We can assume without loss of generality that K has as many elements as we want by iteratively
adding new elements to K using Exercise 4.2.1∗.)

We shall proceed as in Exercise B.4.6†. Consider the resultant h = ResY (f, g). This is a
polynomial of degree at most mn by its matrix expression of Exercise B.4.4†. By the same
exercise, if (x, y) is a common root of f, g, then x is a root of h. Thus, we would be done if
there was at most one possible value of y for each x, since h has degree at most mn and thus

4This requires some care: we need to define the multiplicity of common roots as well as what infinity means. See any
introductory text to algebraic geometry, e.g. Sharevich [shafarevich]. See also the appendix on projective geometry of
Silverman-Tate [26].

396 APPENDIX B. SYMMETRIC POLYNOMIALS

has at most mn roots. Note that we already get that there are finitely many common roots
(although that’s already a consequence of Bézout’s lemma). Here is how we can achieve that: do
a change of coordinates X → X + c Y for some c chosen so that each x appears at most once as
a common root (x, y) of f and g: this is possible because the common roots in this new system
of coordinates are (α + c β, β) and there are finitely many c for which

α+cβ =α +cβ ⇐⇒ c= α−α
β −β.

Exercise B.4.8†. Prove that n + 1 polynomials f1, . . . , fn+1 ∈ K[X1, . . . , Xn] in n variables are
algebraically dependent, meaning that there is some non-zero polynomial f ∈ K[X1, . . . , Xn+1] such
that

f (f1, . . . , fn+1) = 0.

Solution

We present two solutions: one with linear algebra and one with resultants.
For the first solution, consider the linear system of equations in (N + 1)n+1 variables

ai1,...,in+1 f1i1 · . . . · f in+1 = 0. (∗)

n+1

i1 ,...,in+1 ≤N

We wish to find a non-trivial solution to this system. Let us count the number of equations we

have. Set M = maxi(deg fi). Then, the LHS of (∗) is a polynomial of degree (n + 1)M N , when
we consider the ai1,...,in+1 as formal variables. Hence, we have (N + 1)n+1 unknowns and

N ((n + 1)M N )n+1 − 1
(n + 1)M N − 1
((n + 1)M N )k =

k=0

equations, one for each coefficient. For large N , (N + 1)n+1 > ((n+1)M N )n+1 −1 , which means
(n+1)M N −1

that there is a non-trivial solution as wanted (the kernel is non-trivial by e.g. the rank-nullity

theorem ??, or Proposition C.1.2).

To make the idea of the second solution clearer, we treat the case n = 1 first. If f, g ∈ K[X]
are polynomials, the resultant h = ResX (f − S, g − T ) is a non-zero polynomial in S, T with
coefficients in K. Indeed, it is non-zero since when S − f and T − g are coprime it takes a
non-zero value (we can choose T = 0 and S ∈ K to be a large constant for instance). However,
when S = f and T = g, the polynomials f − S and g − T are not coprime anymore so h(f, g) = 0

as wanted.

Now, we construct by backwards induction on k a polynomial with coefficients in K[X1, . . . , Xk]
vanishing at f1, . . . , fn+1. In other words, we eliminate one variable each time. Here is how we
do it: at first, fn,i = fi. Then, we define the polynomials

fk−1,i = ResXk (fk,k+1 − Tk,k+1, fk,i − Tk,i)

for i = 1, . . . , k. At each step we get rid of Xk and introduce k + 1 new variables. Thus,
f0,1 ∈ K[{Ti,j | i ≤ j − 1}. It is clear that it is zero when evaluated at Tn,i = fi for every i and
Tk,i constant for i ≤ k − 1 ≤ n − 2. Indeed, note that Res(A, B)(t) is not in general equal to
Res(A(t), B(t)), since A(t), B(t) do not have the same degree as A, B. If we consider constant
polynomials as polynomials of degree deg(fk,i − Tk,i) > 0, then

ResXk (fk,k+1 − Tk,k+1, fk,i − Tk,i) = 0,

B.4. EXERCISES 397

as can be seen from the matrix expression of Exercise B.4.4†. It remains to prove that there
is some choice of such Tk,i for which f0,1 is not the zero polynomial. This is easy to see: we
can choose Tk,k+1 = 0 for all k and at each step we pick Tk,i so that fk,k+1 and fk,i + Tk,i are
coprime. Indeed, if fk,k+1 has irreducible prime factors, if we pick + 1 values of Tk,i one of
them must work, as otherwise we would have

π | (fk,i + Tk,i) − (fk,i − Tk,i) = Tk,i − Tk,i

for some irreducible π | fk,k+1 and distinct Tk,i, Tk,i ∈ K by the pigeonhole principle. This is
impossible sine it implies Tk,i = Tk,i. There is still one slight technicality: we could have ≥ |K|.
However, we can simply add elements to K to get a sufficiently large K as in Exercise B.4.7†,
and then consider the norm of the polynomial f we obtain (i.e. take the product over each of its
conjugates, exactly like we did in the solution of Exercise 1.5.22†).

Exercise B.4.9† (Transcendence Bases). Let L/K be a field extension. Call a maximal set of K-
algebraically independent elements of L a transcendence basis. Prove that, if L/K has a transcendence
basis of cardinality n, then all transcendence bases have cardinality n. This n is called the transcendence
degree trdegK L. Finally, show that, if L = K(α1, . . . , αn) any maximal algebraically independent
subset of α1, . . . , αn is a transcendence basis. (In particular trdegK L ≤ n.)

Solution

We prove a result analogous to Proposition C.1.2: if α1, . . . , αm ∈ L are K-algebraically inde-
pendent and β1, . . . , βn ∈ L are such that any element of L is algebraic over K(β1, . . . , βn), then
m ≤ n. Since transcendence bases satisfy both conditions, this shows that trdegK L is well-
defined. This almost Exercise B.4.8†: any family of n + 1 elements algebraic over K(β1, . . . , βn)
is algebraically dependent over K. The only difference is that, in our case α1, . . . , αm are not
necessarily in K(β1, . . . , βn). However, the first argument still works perfectly fine, the only
difference is that, if αi has degree di over K(β1, . . . , βn), we get (at most)

m (mM N )n+1 − 1
di mM N − 1

i=1

equations this time, which is still less than (N + 1)m for large N if m > n.

For the second part, note that, by the same argument as Theorem 1.3.2 or by Chapter 6, any
element of K(α1, . . . , αn) is algebraic over K(S), where S ⊆ {α1, . . . , αn} is a maximal subset of
K-algebrically independent element.

Exercise B.4.10†. Let K be an algebraically closed field which is contained in another field L.
Suppose that f1, . . . , fm ∈ K[X1, . . . , Xn] are polynomials with a common root in L. Prove that they
also have a common root in K.

Solution

We present two solutions, one based on Hilbert’s Nullstellensatz from Exercise B.4.6† and one
in characteristic 0 based on transcendence basis from Exercise B.4.9†. For the first sol, note
that f1, . . . , fm have a common root in L if and only if there are no g1, . . . , gm ∈ L[X1, . . . , Xn]
such that f1g1 + . . . + fmgm = 1. In that case, there are no such gi in K[X1, . . . , Xn] either, so
f1, . . . , fm also have a common root in K.

We new present the second solution, which is perhaps more intuitive as it "lifts" (or "reduces"
in our case) the common root over L to a common root over K. Thus, suppose that char K = 0

398 APPENDIX B. SYMMETRIC POLYNOMIALS

and let α1, . . . , αk be a K-transcendence basis for the field generated by K and the common
root. Then, let α be such that this field is equal to K(α1, . . . , αk, α), there exists such an α by
the primitive element theorem 6.2.1. Let

r1(α1, . . . , αk)(α), . . . , rn(α1, . . . , αk)(α))
be the common root, with ri ∈ K(X). The equality

fi(r1(α1, . . . , αk)(α), . . . , rn(α1, . . . , αk)(α)) = 0

is an equality modulo the minimal polynomial π(α1, . . . , αk) of α. Thus, if we replace αi by
ai ∈ K and α by a root a ∈ K of π(a1, . . . , ak), we get a common root in K. We just need to

check that the Ari.(1a.71∗, .: . . , ak)(a) are well-defined, i.e. their denominator is non-zero. This follows
from Exercise the denominator is non-zero so it stays non-zero infinitely many times in

Kn. Note that ri(α) is not necessarily a polynomial, instead it is algebraic over K(α1, . . . , αk),

but by considering its norm (the product with its conjugates over K(α1, . . . , αk)) we can get a
polynomial. Indeed, if the norm of ri(α) is non-zero then so is ri(α). (We also need to be careful
with the leading coefficient of π: if it vanishes α has too few conjugates and things can get weird,

but we can simply pick a1, . . . , ak so that it doesn’t vanish either.)

Miscellaneous

Exercise B.4.11† (ISL 2020 Generalised). Let n ≥ 1 be an integer. Find the maximal N for which
there exists a monomial f of degree N which can not be written as a sum

n

eifi

i=1

with fi ∈ Z[X1, . . . , Xn].

Solution

The answer is N = n(n−1) . First, we prove that X2X32 · . . . · Xnn1 can not be written in the
2
desired form. Suppose for the sake of a contradiction that X2X32 · . . . · Xnn1 = i eifi for some
n(n−1)
polynomials fi, which we suppose without loss of generality to be homogeneous of degree 2 −i

(by ignoring all other monomials). Then, we sum ε(σ)Xσ1(2) · . . . · Xσn(−n1) over all permutations

σ ∈ Sn of [n], where ε denotes the signature (see Definition C.3.2). Since the ei are symmetric,

we have

ε(σ)Xσ1(2) · . . . · Xσn(−n1) = ei ε(σ)fi(Xσ(1), . . . , Xσ(n)).

σ∈Sn i σ∈Sn

Here is the key point: if f has degree less than n(n−1) , σ∈Sn ε(σ)f (Xσ(1), . . . , Xσ(n)) = 0.
2

This is an obvious contradiction as the LHS is a sum of distinct monomials so is non-zero.

To prove this claim, suppose without loss of generality that f is a monomial n Xiai . Since
1), two ai must be equal, i=1 i ↔ j.
n ni=1(i −
i=1 ai < say ai = aj. Denote by τ the transposition

Then, by grouping permutations of [n] by orbits σ, σ ◦ τ , the sum is zero since

f (Xσ(1), . . . , Xσ(n)) = f (Xσ◦τ(1), . . . , Xσ◦τ(n))
but ε(σ ◦ τ ) = −ε(σ) by Exercise C.3.11∗ so the sum over each orbits cancels out.

It remains to prove that X1a1 · . . . · Xnan works when a1 + . . . + an > n(n−1) . When a1, . . . , an ≥ 1
2
it is trivial since the monomial is divisible by e1. We proceed by induction on a12 + . . . + a2n, with

the following base case: a1, . . . , an ≥ 1 the monomial is divisible by e1.

B.4. EXERCISES 399

Now suppose that a1 + . . . + an > n(n−1) and, without loss of generality, 0 = a1 ≤ a2 ≤ ... ≤ an.
2
There must exist some k such that ek+1 ≥ ek +2, since otherwise ek ≤ k −1 for all k contradicting

our initial assumption on the sum. Now consider

X1a1 · . .. · Xnan − X1a1 · ... · X ak−1 Xkak −1 · .. . · Xnan −1 en−k .

k−1

We claim that the sum of the squares of the exponents in any monomial appearing in this

polynomial is less than Xa21ka−k+−11.X. .ka+k−a12n·, thus concluding the inductive step. To see this, express a
monomial of X1a1 ·...· . . . · Xnak en−k as

X1a1 +b1 · ... · X Xak−1+bk−1 ak+bk−1 · ... · Xnan +bn −1
k−1 k

for some bi ∈ {0, 1} with b1 + . . . + bn = n − k. The wanted result then follows from the

convexity of the square function: if bi = 1 for some i < k and bj = 0 for some j ≥ k, then
(ai + 1)2 + (aj − 1)2 < a2i + aj2. Iterating this process to "push" all the ones to the positions
greater than or equal to k, we get

(a1 + b1)2 + . . . + (ak−1 + bk−1)2 + (ak + bk − 1)2 + . . . + (an + bn − 1)2 ≤ a12 + . . . + a2n

with equality if and only if we already had equality in the beginning, i.e. if the monomial is
X1a1 · . . . · Xnan . However, we have ruled that case out by subtracting precisely this monomial, so
we are done.

Exercise B.4.12† (Lagrange). Given a rational function f ∈ K[X1, . . . , Xn], we denote by Gf the
set of permutations σ ∈ Sn such that

f (X1, . . . , Xn) = f (Xσ(1), . . . , Xσ(n)).

Let f, g ∈ K(X1, . . . , Xn) be two rational functions. If Gf ⊆ Gg, prove that there exists a rational
function r ∈ K[e1, . . . , en](X) such that

g = r ◦ f.

Solution

We present the proofs in Prasolov [21]. Partition Gg into disjoint cosets Gf =
h1Gf , h2Gf , . . . , hkGf and write fi = hif and gi = hif for each i, where σf means
f (Xσ(1), . . . , f (Xσ(n)) (we say the group of permutations Sn acts on the field K(X1, . . . , Xn)).
(This is where we use the assumption that Gf ⊆ Gg.)

For the first proof, notice that

k gi
i=1 T − fi

is, by definition, symmetric in X1, . . . , Xn. Since Ω = k T − fi is as well, we get
i=1

k T gi = F (T )
i=1 − fi Ω(T )

for some F ∈ K(e1, . . . , en)[T ] by the fundamental theorem of symmetric polynomials. Notice
3.2.2∗.
that F (f ) = k f − fi is F /(T −f) evaluated at T = f by Exercise Hence, we conclude
i=2

that

F (f ) k gi Ω (f ) = g
=
Ω (f ) i=1 (T − fi)Ω

since Ω vanishes at f = fi.
(T −fi)Ω

400 APPENDIX B. SYMMETRIC POLYNOMIALS

The second proof is perhaps more intuitive. We consider the system of equations

k

fisgi = Ts,

i=1

where the exponent represents powers and not iterates. Cramer’s rule from Exercise C.5.7 and
the Vandermonde determinant from Appendix C and tell us that

D
g=

∆

where 1 ··· 1

f1 · · · fn = fi − fj
∆ = ...
... ... i<j

f1k−1 · · · fkk−1

and T0 1 · · · 1

T1 f2 · · · fk
D = ... ... . . . ... .

Tk−1 f2k−1 · · · fkk−1

Write this as g = D∆ . Notice that ∆2 is symmetric, while D and ∆ both change sign when two
∆2

fi are switched, so D∆ is symmetric in f2, . . . , fk. However, it is easy to see that, for any i,

ei(f2, . . . , fk) can be expressed polynomially in terms of f1 and ej(f1, . . . , fk). Hence, this D∆
∆2

is a rational function in f with symmetric coefficients by the fundamental theorem of symmetric

polynomials.

Exercise B.4.13† (Iran Mathematical Olympiad 2012). Prove that there exists a polynomial f ∈
R[X0, . . . , Xn−1] such that, for all a0, . . . , an−1 ∈ R,

f (a0, . . . , an−1) ≥ 0

is equivalent to the polynomial Xn + an−1Xn−1 + . . . + a0 having only real roots, if and only if
n ∈ {1, 2, 3}.

Solution

If n ≤ 3, the discriminant satisfies the condition. Indeed, the discriminant of f = i X − αi is
the square of i<j αi − αj so is positive if all αi are positive. It remains to prove that, for these
n, i<j αi − αj is real if and only if all αi are (its square is real so it must be real or purely
imaginary). For n = 1, it is trivial since any polynomial of degree 1 with real coefficients splits
in R. For n = 2, if the roots of f are α = α, then α − α is not real since complex conjugation
negates it. For n = 3, if the roots of f are α = α and β ∈ R, then complex conjugation also
negates

(α − α)(β − α)(β − α)

so it isn’t real as desired.

Now, if there exists such a polynomial for n ≥ 4, there exists one for n = 4 by setting g(a, b, c, d) =
f (a, b, c, d, 0, . . . , 0). Thus, it only remains to prove that there doesn’t exist such a polynomial
for n = 4. For this, consider the special polynomial f (0, b, 0, d) since we know precisely when
the roots of X4 + bX2 + d are real. For convenience, we shall in fact consider the polynomial
g(r, s) = f (0, −r − s, 0, rs) which is non-negative iff the roots of

X4 − (r + s)X2 + rs = (X2 − r)(X2 − s)