Math 129: Algebraic Number Theory Lecture 17: Topology, Completeness

William Stein

Thursday, April 8, 2004

Before starting Section 4, I discuss a question from last time...

Remark 2.1. This definition differs from the one on page 46 of [Cassels-Frohlich, Ch. 2] in two ways. First, we assume that c > 1 instead of c < 1, since otherwise | · |_{p} does not satisfy Axiom 3 of a valuation. Here’s why: Recall that Axiom 3 for a non-archimedean valuation on K asserts that whenever a ∈ K and |a| ≤ 1, then |a + 1| ≤ 1. Set a = p−1, where p = p(t) ∈ K[t] is an irreducible polynomial. Then |a| = c^{0} = 1, since ord_{p}(p − 1) = 0. However, |a + 1| = |p − 1 + 1| = |p| = c^{1} < 1, since ord_{p}(p) = 1. If we take c > 1 instead of c < 1, as I propose, then |p| = c^{1} > 1, as required.

4 Topology

A valuation | · | on a field K induces a topology in which a basis for the neighborhoods of a are the open balls

B(a, d) = {x ∈ K : |x − a| < d}

for d > 0.

Lemma 4.1. Equivalent valuations induce the same topology.

Proof. If | · |_{1} = | · |_{2}^{r}, then |x − a|_{1} < d if and only if |x − a|_{2}^{r} < d if and only if |x − a|_{2} < d^{1/r} so B_{1}(a, d) = B_{2}(a, d^{1/r}). Thus the basis of open neighborhoods of a for | · |_{1} and | · |_{2} are identical.

A valuation satisfying the triangle inequality gives a metric for the topology on defining the distance from a to b to be |a − b|. Assume for the rest of this section that we only consider valuations that satisfy the triangle inequality.

Lemma 4.2. A field with the topology induced by a valuation is a topological field, i.e., the operations sum, product, and reciprocal are continuous.

Proof. For example (product) the triangle inequality implies that

|(a + ε)(b + δ) − ab| ≤ |ε| |δ| + |a| |δ| + |b| |ε|

is small when |ε| and |δ| are small (for fixed a, b).

Lemma 4.3. Suppose two valuations | · |_{1} and | · |_{2} on the same field K induce the same topology. Then for any sequence {x_{n}} in K we have

|x_{n}|_{1} → 0 ⟺ |x_{n}|_{2} → 0.

Proof. It suffices to prove that if |x_{n}|_{1} → 0 then |x_{n}|_{2} → 0, since the proof of the other implication is the same. Let ε > 0. The topologies induced by the two absolute values are the same, so B_{2}(0, ε) can be covered by open balls B_{1}(a_{i}, r_{i}). One of these open balls B_{1}(a, r) contains 0. There is ε′ > 0 such that

B_{1}(0, ε′) ⊂ B_{1}(a, r) ⊂ B_{2}(0, ε).

Since |x_{n}|_{1} → 0, there exists N such that for n ≥ N we have |x_{n}|_{1} < ε′. For such n, we have x_{n} ∈ B_{1}(0, ε′), so x_{n} ∈ B_{2}(0, ε), so |x_{n}|_{2} < ε. Thus |x_{n}|_{2} → 0.

Proposition 4.4. If two valuations | · |_{1} and | · |_{2} on the same field induce the same topology, then they are equivalent in the sense that there is a positive real α such that | · |_{1} = | · |_{2}^{α}.

Proof. If x ∈ K and i = 1, 2, then |x^{n}|_{i} → 0 if and only if |x|_{i}^{n} → 0, which is the case if and only if |x|_{i} < 1. Thus Lemma 4.3 implies that |x|_{1} < 1 if and only if |x|_{2} < 1. On taking reciprocals we see that |x|_{1} > 1 if and only if |x|_{2} > 1, so finally |x|_{1} = 1 if and only if |x|_{2} = 1.

Let now w, z ∈ K be nonzero elements with |w|_{i} ≠ 1 and |z|_{i} ≠ 1. On applying the foregoing to

x = w^{m}z^{n} (m, n ∈ Z)

we see that

m log |w|_{1} + n log |z|_{1} ≥ 0

if and only if

m log |w|_{2} + n log |z|_{2} ≥ 0.

Dividing through by log |z|_{i}, and rearranging, we see that for every rational number α = −n/m,

\begin{matrix}log & |w| & 1 \\ log & |z| & 1 \end{matrix} ≥ α ⟺ \begin{matrix}log & |w| & 2 \\ log & |z| & 2 \end{matrix} ≥ α.

Thus

\begin{matrix}log & |w| & 1 \\ log & |z| & 1 \end{matrix} = \begin{matrix}log & |w| & 2 \\ log & |z| & 2 \end{matrix} ,

so

\begin{matrix}log & |w| & 1 \\ log & |w| & 2 \end{matrix} = \begin{matrix}log & |z| & 1 \\ log & |z| & 2 \end{matrix} .

Since this equality does not depend on the choice of z, we see that there is a constant c (= log |z|_{1} / log |z|_{2}) such that log |w|_{1} / log |w|_{2} = c for all w. Thus log |w|_{1} = c·log |w|_{2}, so |w|_{1} = |w|_{2}^{c}, which implies that | · |_{1} is equivalent to | · |_{2}.

5 Completeness

We recall the definition of metric on a set X.

Definition 5.1 (Metric). A metric on a set X is a map

d : X × X → R

such that for all x, y, z ∈ X,

1. d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y,

2. d(x, y) = d(y, x), and

3. d(x, z) ≤ d(x, y) + d(y, z).

A Cauchy sequence is a sequence (x_{n}) in X such that for all ε > 0 there exists M such that for all n, m > M we have d(x_{n}, x_{m}) < ε. The completion of X is the set of Cauchy sequences (x_{n}) in X modulo the equivalence relation in which two Cauchy sequences (x_{n}) and (y_{n}) are equivalent if lim_{n→∞} d(x_{n}, y_{n}) = 0. A metric space is complete if every Cauchy sequence converges, and one can show that the completion of X with respect to a metric is complete.

For example, d(x, y) = |x − y| (usual archimedean absolute value) defines a metric on Q. The completion of Q with respect to this metric is the field R of real numbers. More generally, whenever | · | is a valuation on a field K that satisfies the triangle inequality, then d(x, y) = |x − y| defines a metric on K. Consider for the rest of this section only valuations that satisfy the triangle inequality.

Definition 5.2 (Complete). A field K is complete with respect to a valuation | · | if given any Cauchy sequence a_{n}, (n = 1, 2, …), i.e., one for which

|a_{m} − a_{n}| → 0 (m, n → ∞, ∞),

there is an a^{∗} ∈ K such that

a_{n} → a^{∗} w.r.t. | · |

(i.e., |a_{n} − a^{∗}| → 0).

Theorem 5.3. Every field K with valuation v = | · | can be embedded in a complete field K_{v} with a valuation | · | extending the original one in such a way that K_{v} is the closure of K with respect to | · |. Further K_{v} is unique up to a unique isomorphism fixing K.

Proof. Define K_{v} to be the completion of K with respect to the metric defined by | · |. Thus K_{v} is the set of equivalence classes of Cauchy sequences, and there is a natural injective map from K to K_{v} sending an element a ∈ K to the constant Cauchy sequence (a). Because the field operations on K are continuous, they induce welldefined field operations on equivalence classes of Cauchy sequences componentwise. Also, define a valuation on K_{v} by

|(a_{n})_{n=1}^{∞}| = lim_{n→∞} |a_{n}| ,

and note that this is well defined and extends the valuation on K.

To see that K_{v} is unique up to a unique isomorphism fixing K, we observe that there are no nontrivial continuous automorphisms K_{v} → K_{v} that fix K. This is because, by denseness, a continuous automorphism σ : K_{v} → K_{v} is determined by what it does to K, and by assumption σ is the identity map on K. More precisely, suppose a ∈ K_{v} and n is a positive integer. Then by continuity there is δ > 0 (with δ < 1/n) such that if a_{n} ∈ K_{v} and |a − a_{n}| < δ then |σ(a) − σ(a_{n})| < 1/n. Since K is dense in K_{v}, we can choose the a_{n} above to be an element of K. Then by hypothesis σ(a_{n}) = a_{n}, so |σ(a) − a_{n}| < 1/n. Thus σ(a) = lim_{n→∞} a_{n} = a.

Corollary 5.4. The valuation | · | is non-archimedean on K_{v} if and only if it is so on K. If | · | is non-archimedean, then the set of values taken by | · | on K and K_{v} are the same.

Proof. The first part follows from the fact proved earlier that a valuation is nonarchimedean if and only if |n| < 1 for all integers n. Since the valuation on K_{v} extends the valuation on K, and all n are in K, the first statement follows.

For the second, suppose that | · | is non-archimedean (but not necessarily discrete). Suppose b ∈ K_{v} with b ≠ 0. First I claim that there is c ∈ K such that |b − c| < |b|. To see this, let c′ = b − \frac{b}{a}, where a is some element of K_{v} with |a| > 1, note that |b − c′| = |\frac{b}{a} | < |b|, and choose c ∈ K such that |c − c′| < |b − c′|, so

|b − c| = |b − c′ − (c − c′)| ≤ max (|b − c′| , |c − c′|) = |b − c′| < |b| .

Since | · | is non-archimedean, we have

|b| = |(b − c) + c| ≤ max (|b − c| , |c|) = |c| ,

where in the last equality we use that |b − c| < |b|. Also,

|c| = |b + (c − b)| ≤ max (|b| , |c − b|) = |b| ,

so |b| = |c|, which is in the set of values of | · | on K.

5.1 p-adic Numbers

This section is about the p-adic numbers Q_{p}, which are the completion of Q with respect to the p-adic valuation. Alternatively, to give a p-adic integer in Z_{p} is the same as giving for every prime power p^{r} an element a_{r} ∈ Z/p^{r}Z such that if s ≤ r then a_{s} is the reduction of a_{r} modulo p^{s}. The field Q_{p} is then the field of fractions of Z_{p}.

We begin with the definition of the N-adic numbers for any positive integer N. Section 5.1.2 is about the N-adics in the special case N = 10; these are fun because they can be represented as decimal expansions that go off infinitely far to the left. Section 5.3 is about how the topology of Q_{N} is nothing like the topology of R. Finally, in Section 5.4 we state the Hasse-Minkowski theorem, which shows how to use p-adic numbers to decide whether or not a quadratic equation in n variables has a rational zero.

5.1.1 The N-adic Numbers

Lemma 5.5. Let N be a positive integer. Then for any nonzero rational number α there exists a unique e ∈ Z and integers a, b, with b positive, such that α = N^{e} · \frac{a}{b} with N ∤ a, gcd(a, b) = 1, and gcd(N, b) = 1.

Proof. Write α = c/d with c, d ∈ Z and d > 0. First suppose d is exactly divisible by a power of N, so for some r we have N^{r} | d but gcd(N, d/N^{r}) = 1. Then

\frac{c}{d} = N^{−r} \frac{c}{d/N^{r}} .

If N^{s} is the largest power of N that divides c, then e = s − r, a = c/N^{s}, b = d/N^{r} satisfy the conclusion of the lemma.

By unique factorization of integers, there is a smallest multiple f of d such that fd is exactly divisible by N. Now apply the above argument with c and d replaced by cf and df.

Definition 5.6 (N-adic valuation). Let N be a positive integer. For any positive α ∈ Q, the N-adic valuation of α is e, where e is as in Lemma 5.5. The N-adic valuation of 0 is ∞.

We denote the N-adic valuation of α by ord_{N}(α). (Note: Here we are using “valuation” in a different way than in the rest of the text. This valuation is not an absolute value, but the logarithm of one.)

Definition 5.7 (N-adic metric). For x, y ∈ Q the N-adic distance between x and y is

d_{N}(x, y) = N^{− ord_{N} (x−y)}.

We let d_{N}(x, x) = 0, since ord_{N}(x − x) = ord_{N}(0) = ∞.

For example, x, y ∈ Z are close in the N-adic metric if their difference is divisible by a large power of N. E.g., if N = 10 then 93427 and 13427 are close because their difference is 80000, which is divisible by a large power of 10.

Proposition 5.8. The distance d_{N} on Q defined above is a metric. Moreover, for all x, y, z ∈ Q we have

d(x, z) ≤ max(d(x, y), d(y, z)).

(This is the “nonarchimedean” triangle inequality.)

Proof. The first two properties of Definition 5.1 are immediate. For the third, we first prove that if α, β ∈ Q then

ord_{N}(α + β) ≥ min(ord_{N}(α), ord_{N}(β)).

Assume, without loss, that ord_{N}(α) ≤ ord_{N}(β) and that both α and β are nonzero. Using Lemma 5.5 write α = N^{e}(a/b) and β = N^{f}(c/d) with a or c possibly negative. Then

α + β = N^{e} (\frac{a}{b} + N^{f−e} \frac{c}{d}) = N^{e} (\frac{ad + bcN^{f−e}}{bd} ) .

Since gcd(N, bd) = 1 it follows that ord_{N}(α + β) ≥ e. Now suppose x, y, z ∈ Q. Then

x − z = (x − y) + (y − z),

so

ord_{N}(x − z) ≥ min(ord_{N}(x − y), ord_{N}(y − z)),

hence d_{N}(x, z) ≤ max(d_{N}(x, y), d_{N}(y, z)).

We can finally define the N-adic numbers.

Definition 5.9 (The N-adic Numbers). The set of N-adic numbers, denoted Q_{N}, is the completion of Q with respect to the metric d_{N}.

The set Q_{N} is a ring, but it need not be a field as you will show in Exercises 4 and 5. It is a field if and only if N is prime. Also, Q_{N} has a “bizarre” topology, as we will see in Section 5.3.

5.1.2 The 10-adic Numbers

It’s a familiar fact that every real number can be written in the form

d_{n} … d_{1}d_{0}.d_{−1}d_{−2} … = d_{n}10^{n} + ⋯ + d_{1}10 + d_{0} + d_{−1}10^{−1} + d_{−2}10^{−2} + ⋯

where each digit d_{i} is between 0 and 9, and the sequence can continue indefinitely to the right.

The 10-adic numbers also have decimal expansions, but everything is backward! To get a feeling for why this might be the case, we consider Euler’s nonsensical series

∑_{n=1}^{∞}(−1)^{n+1}n! = 1! − 2! + 3! − 4! + 5! − 6! + · · · .

You will prove in Exercise 2 that this series converges in Q_{10} to some element α ∈ Q_{10}.

What is α? How can we write it down? First note that for all M ≥ 5, the terms of the sum are divisible by 10, so the difference between α and 1! − 2! + 3! − 4! is divisible by 10. Thus we can compute α modulo 10 by computing 1! − 2! + 3! − 4! modulo 10. Likewise, we can compute α modulo 100 by compute 1!−2!+⋯+9!−10!, etc. We obtain the following table:

α mod 10^{r}
1 mod 10
81 mod 10^{2}
981 mod 10^{3}
2981 mod 10^{4}
22981 mod 10^{5}
422981 mod 10^{6}

Continuing we see that

1! − 2! + 3! − 4! + ⋯ = … 637838364422981 in Q_{10} !

Here’s another example. Reducing 1/7 modulo larger and larger powers of 10 we see that

\frac{1}{7} = … 857142857143 in Q_{10}.

Here’s another example, but with a decimal point.

\frac{1}{70} = \frac{1}{10} · \frac{1}{7} = … 85714285714.3

We have

\frac{1}{3} + \frac{1}{7} = … 66667 + … 57143 = \frac{10}{21} = … 23810,

which illustrates that addition with carrying works as usual.

5.1.3 Fermat’s Last Theorem in Z_{10}

An amusing observation, which people often argued about on USENET news back in the 1990s, is that Fermat’s last theorem is false in Z_{10}. For example, x^{3} +y^{3} = z^{3} has a nontrivial solution, namely x = 1, y = 2, and z = … 60569. Here z is a cube root of 9 in Z_{10}. Note that it takes some work to prove that there is a cube root of 9 in Z_{10} (see Exercise 3).

5.2 The Field of p-adic Numbers

The ring Q_{10} of 10-adic numbers is isomorphic to Q_{2} × Q_{5} (see Exercise 5), so it is not a field. For example, the element … 8212890625 corresponding to (1, 0) under this isomorphism has no inverse. (To compute n digits of (1, 0) use the Chinese remainder theorem to find a number that is 1 modulo 2^{n} and 0 modulo 5^{n}.)

If p is prime then Q_{p} is a field (see Exercise 4). Since p ≠ 10 it is a little more complicated to write p-adic numbers down. People typically write p-adic numbers in the form

\frac{a_{−d}}{p^{d}} + ⋯ + \frac{a_{−1}}{p} + a_{0} + a_{1}p + a_{2}p^{2} + a_{3}p^{3} + ⋯

where 0 ≤ a_{i} < p for each i.

5.3 The Topology of Q_{N} (is Weird)

Definition 5.10 (Connected). Let X be a topological space. A subset S of X is disconnected if there exist open subsets U_{1}, U_{2} ⊂ X with U_{1} ∩ U_{2} ∩ S = ∅ and S = (S ∩ U_{1}) ∪ (S ∩ U_{2}) with S ∩ U_{1} and S ∩ U_{2} nonempty. If S is not disconnected it is connected.

The topology on Q_{N} is induced by d_{N}, so every open set is a union of open balls

B(x, r) = {y ∈ Q_{N} : d_{N}(x, y) < r}.

Recall Proposition 5.8, which asserts that for all x, y, z,

d(x, z) ≤ max(d(x, y), d(y, z)).

This translates into the following shocking and bizarre lemma:

Lemma 5.11. Suppose x ∈ Q_{N} and r > 0. If y ∈ Q_{N} and d_{N}(x, y) ≥ r, then B(x, r) ∩ B(y, r) = ∅.

Proof. Suppose z ∈ B(x, r) and z ∈ B(y, r). Then

r ≤ d_{N}(x, y) ≤ max(d_{N}(x, z), d_{N}(z, y)) < r,

a contradiction.

You should draw a picture to illustrates Lemma 5.11.

Lemma 5.12. The open ball B(x, r) is also closed.

Proof. Suppose y ∉ B(x, r). Then r ≤ d(x, y) so

B(y, d(x, y)) ∩ B(x, r) ⊂ B(y, d(x, y)) ∩ B(x, d(x, y)) = ∅.

Thus the complement of B(x, r) is a union of open balls.

The lemmas imply that Q_{N} is totally disconnected, in the following sense.

Proposition 5.13. The only connected subsets of Q_{N} are the singleton sets {x} for x ∈ Q_{N} and the empty set.

Proof. Suppose S ⊂ Q_{N} is a nonempty connected set and x, y are distinct elements of S. Let r = d_{N}(x, y) > 0. Let U_{1} = B(x, r) and U_{2} be the complement of U_{1}, which is open by Lemma 5.12. Then U_{1} and U_{2} satisfies the conditions of Definition 5.10, so S is not connected, a contradiction.

5.4 The Local-to-Global Principle of Hasse and Minkowski

Section 5.3 might have convinced you that Q_{N} is a bizarre pathology. In fact, Q_{N} is omnipresent in number theory, as the following two fundamental examples illustrate.

In the statement of the following theorem, a nontrivial solution to a homogeneous polynomial equation is a solution where not all indeterminates are 0.

Theorem 5.14 (Hasse-Minkowski). The quadratic equation

a_{1}x_{1}^{2} + a_{2}x_{2}^{2} + ⋯ + a_{n}x_{n}^{2} = 0,  (5.1)

with a_{i} ∈ Q^{×}, has a nontrivial solution with x_{1}, …, x_{n} in Q if and only if (5.1) has a solution in R and in Q_{p} for all primes p.

This theorem is very useful in practice because the p-adic condition turns out to be easy to check. For more details, including a complete proof, see [Serre, A Course in Arithmetic, IV.3.2]].

The analogue of Theorem 5.14 for cubic equations is false. For example, Selmer proved that the cubic

3x^{3} + 4y^{3} + 5z^{3} = 0

has a solution other than (0, 0, 0) in R and in Q_{p} for all primes p but has no solution other than (0, 0, 0) in Q (for a proof see [Cassels, Lectures on Elliptic Curves, §18]).

Open Problem. Give an algorithm that decides whether or not a cubic

ax^{3} + by^{3} + cz^{3} = 0

has a nontrivial solution in Q.

This open problem is closely related to the Birch and Swinnerton-Dyer Conjecture for elliptic curves. The truth of the conjecture would follow if we knew that “Shafarevich-Tate Groups” of elliptic curves were finite.

5.5 Exercises

The following are optional exercises, which you may want to do if you would like to familiarize yourself further with p-adic numbers. They are not part of the formal homework sets and you do not need to hand them in.

1. Compute the first 5 digits of the 10-adic expansions of the following rational

numbers:

\frac{13}{2}, \frac{1}{389}, \frac{17}{19}, the 4 square roots of 41.

2. Let N > 1 be an integer. Prove that the series

∑_{n=1}^{∞}(−1)^{n+1}n! = 1! − 2! + 3! − 4! + 5! − 6! + · · · .

converges in Q_{N}.

3. Prove that −9 has a cube root in Q_{10} using the following strategy (this is a special case of “Hensel’s Lemma”).

(a) Show that there is α ∈ Z such that α^{3} ≡ 9 (mod 10^{3}).

(b) Suppose n ≥ 3. Use induction to show that if α_{1} ∈ Z and α^{3} ≡ 9 (mod 10^{n}), then there exists α_{2} ∈ Z such that α_{2}^{3} ≡ 9 (mod 10^{n+1}). (Hint: Show that there is an integer b such that (α_{1} + b10^{n})^{3} ≡ 9 (mod 10^{n+1}).)

(c) Conclude that 9 has a cube root in Q_{10}.

4. Let N > 1 be an integer.

(a) Prove that Q_{N} is equipped with a natural ring structure.

(b) If N is prime, prove that Q_{N} is a field.

5. (a) Let p and q be distinct primes. Prove that Q_{pq} ≅ Q_{p} × Q_{q}.

(b) Is Q_{p^{2}} isomorphic to either of Q_{p} × Q_{p} or Q_{p}?