Algebraic Number Theory — Chapter 5 Valuations and completions

Chapter 5
Valuations and completions

5.1. Global fields

Definition 5.1.1. §

A global function field, or a function field in one variable over a finite field, is a finite extension of $𝔽_{p} (t)$ for some prime $p$ .

Remark 5.1.2. §

A function field $K$ in one variable over the finite field $𝔽_{p}$ is actually isomorphic to a finite separable extension of $𝔽_{p} (u)$ for some $u \in K$ , so we always suppose that such a field is a separable extension of $𝔽_{p} (t)$ in what follows.

Remark 5.1.3. §

In these notes, we will often refer to a function field in one variable over a finite field more simply as a function field.

Definition 5.1.4. §

A field $K$ is said to be a global field if it is either a number field or a function field in one variable over a finite field.

Definition 5.1.5. §

The ring of integers $𝒪_{K}$ of a finite extension $K$ of $𝔽_{p} (t)$ for a prime $p$ is the integral closure of $𝔽_{p} [t]$ in $K$ .

Definition 5.1.6. §

Let $𝔭$ be a nonzero prime ideal in the ring of integers of a global field $K$ . Its ramification index (resp., residue degree) $e_{𝔭}$ (resp., $f_{𝔭}$ ) is its ramification index (resp., residue degree) over $𝔭 \cap ℤ$ if $K$ is a number field and over $𝔭 \cap 𝔽_{p} [t]$ if $K$ is a function field of characteristic $p$ .

In the case of function fields, the choice of ring of integers is not really canonical. For instance, under the field isomorphism $σ : 𝔽_{p} (t) \to 𝔽_{p} (t)$ taking $t$ to $t^{- 1}$ , the ring of integers $𝔽_{p} [t]$ is carried to $𝔽_{p} [t^{- 1}]$ . In particular, if we write $u = t^{- 1}$ , then $σ$ restricts to an isomorphism

σ : 𝔽_{p} [t, t^{- 1}] \to_{}^{\sim} 𝔽_{p} [u, u^{- 1}] .

This gives a one-to-one correspondence between the nonzero prime ideals of $𝔽_{p} [t]$ aside from $(t)$ and the nonzero prime ideals of $𝔽_{p} [t^{- 1}]$ aside from $(t^{- 1})$ .

To phrase this in terms of algebraic geometry, one should view $Spec 𝔽_{p} [t]$ and $Spec 𝔽_{p} [t^{- 1}]$ as two affine open neighborhoods of and covering the projective line $ℙ_{𝔽_{p}}^{1}$ over $𝔽_{p}$ that have intersection $Spec 𝔽_{p} [t, t^{- 1}]$ . The transition, or gluing map, between the two affine spaces along the intersection is that induced by $σ$ . The point of $ℙ_{𝔽_{p}}^{1}$ that is not contained in $Spec 𝔽_{p} [t]$ is the prime ideal $(t^{- 1})$ in $Spec 𝔽_{p} [t^{- 1}]$ , and this is known as the point $\infty$ at infinity.

Definition 5.1.7. §

Let $K$ be a global field of characteristic $p$ . A prime of $K$ over $\infty$ is a prime ideal of the integral closure of $𝔽_{p} [t^{- 1}]$ that lies over $(t^{- 1})$ .

Definition 5.1.8. §

Let $K$ be a global function field. The primes of $K$ are the nonzero prime ideals of $𝒪_{K}$ , which are called finite primes, and the primes above $\infty$ in $K$ , which are also known as infinite primes.

We compare this with the case of number fields.

Definition 5.1.9. §

Let $K$ be a number field. The finite primes of $K$ are the nonzero prime ideals in $𝒪_{K}$ . The infinite primes of $K$ are the archimedean primes of $K$ . The primes of $K$ are the finite and infinite primes of $K$ .

The finite primes of a global field are exactly the nonzero prime ideals of its ring of integers. However, the infinite primes of number fields and function fields are quite different. In the next section, we shall see another classification of the primes that reflects this. For now, note the following.

Every finite prime in a global field $K$ of characteristic $p$ gives rise to a discrete valuation on $K$ , as does every infinite prime, being a prime ideal in a Dedekind ring that is the integral closure of $𝔽_{p} [t^{- 1}]$ in $K$ .

Example 5.1.10. §

The valuation attached to the prime $\infty$ of $𝔽_{p} (t)$ for a prime $p$ takes a quotient $\frac{f}{g}$ of nonzero polynomials in $𝔽_{p} [t]$ to $\deg g - \deg f$ by definition, so it equals $v_{\infty}$ .

Remark 5.1.11. §

As with nonzero prime ideals in $𝒪_{K}$ , we can speak of the ramification index $e_{𝔭}$ and residue degree $f_{𝔭}$ of a prime $𝔭$ of $K$ over $\infty$ . In fact, if $R$ is the integral closure of $𝔽_{p} [t^{- 1}]$ in $K$ , then $𝔭$ is a prime ideal of $R$ lying over some prime ideal $(f) = 𝔭 \cap 𝔽_{p} [t^{- 1}]$ of $𝔽_{p} [t^{- 1}]$ . The residue field of $𝔭$ is $R ∕ 𝔭$ , and so we may speak of its residue degree over $(f)$ . Similarly, the ramification index is the highest power of $𝔭$ dividing $𝑓𝑅$ .

5.2. Valuations

Definition 5.2.1. §

A (multiplicative) valuation (or absolute value) on a field $K$ is a function $| | : K \to ℝ_{\geq 0}$ such that

i.: $| a | = 0$ if and only if $a = 0$ ,
ii.: $| 𝑎𝑏 | = | a | | b |$ , and
iii.: $| a + b | \leq | a | + | b |$

for all $a, b \in K$ .

Remark 5.2.2. §

Every valuation $| |$ on a field $K$ satisfies $| ζ | = 1$ for all roots of unity $ζ \in K^{\times}$ .

Definition 5.2.3. §

We say that a valuation $| |$ on a field $K$ is trivial if $| a | = 1$ for all $a \in K^{\times}$ , and nontrivial otherwise.

Valuations on fields give rise to metrics. That is, given a valuation $| |$ on a field $K$ , it defines a distance function $d$ on $K$ by

d (a, b) = | a - b |

for $a, b \in K$ . We can then give $K$ the topology of the resulting metric space. For instance, the trivial valuation on a field gives rise to the discrete topology on $K$ .

Definition 5.2.4. §

We say that two multiplicative valuations on a field $K$ are equivalent if they define the same topology on $K$ .

Proposition 5.2.5. §

Let $| |_{1}$ and $| |_{2}$ be valuations on a field $K$ . The following are equivalent:

i.: the valuations $| |_{1}$ and $| |_{2}$ are equivalent,
ii.: any $a \in K$ satisfies $| a |_{1} < 1$ if and only if $| a |_{2} < 1$ as well,
iii.: there exists $s > 0$ such that $| a |_{2} = | a |_{1}^{s}$ for all $a \in K$ .

Proof.

If $| |_{1}$ is nontrivial, then there exists $b \in K$ with $| b |_{1} > 1$ , and the sequence $b^{- n}$ converges to $0$ , so the topology $| |_{1}$ induces on $K$ is not discrete. Therefore, the trivial valuation is equivalent only to itself. By a check of conditions (ii) and (iii), we may assume that our two valuations are nontrivial.

If the two valuations are equivalent, then a sequence $a^{n}$ converges to $0$ in the common topology if and only if $| a^{n} |_{i} = | a |_{i}^{n}$ converges to $0$ , which is to say exactly that $| a |_{i} < 1$ . Hence (i) implies (ii).

Suppose that (ii) holds. Let $b \in K$ be such that $| b |_{1} > 1$ , and set

s = \frac{\log | b |_{2}}{\log | b |_{1}} > 0

so that $| b |_{2} = | b |_{1}^{s}$ . Choose $a \in K^{\times}$ , and let

t = \frac{\log | a |_{1}}{\log | b |_{1}} \in ℝ

so that $| a |_{1} = | b |_{1}^{t}$ . Let $m, n \in ℤ$ with $n \neq 0$ be such that $t < q = \frac{m}{n}$ . Then

| a |_{1} = | b |_{1}^{t} < | b |_{1}^{q},

{| \frac{a^{n}}{b^{m}} |}_{1} < 1,

which implies $| a^{n} b^{- m} |_{2} < 1$ and then $| a |_{2} < | b |_{2}^{q}$ as well. Since this holds for all rational $q > t$ , we have $| a |_{2} \leq | b |_{2}^{t}$ . On the other hand, if we assume instead that $q < t$ , then we get $| a |_{1} > | b |_{1}^{q}$ , which implies in turn that $| a^{- n} b^{m} |_{1} < 1$ , that $| a^{- n} b^{m} |_{2} < 1$ , that $| a |_{2} > | b |_{2}^{q}$ , and finally that $| a |_{2} \geq | b |_{2}^{t}$ . We therefore have that

| a |_{2} = | b |_{2}^{t} = | b |_{1}^{𝑠𝑡} = | a |_{1}^{s} .

Hence, (ii) implies (iii).

For $i \in {1, 2}$ , consider the ball

B_{i} (a, 𝜖) = {b \in K ∣ | a - b |_{i} < 𝜖}

of radius $𝜖 > 0$ about $a \in K$ . If $s$ is as in (iii), then $B_{1} (a, 𝜖) = B_{2} (a, 𝜖^{s})$ , so the topologies defined by the two valuations are equivalent. That is, (iii) implies (i). □

Remark 5.2.6. §

If $| |$ is a valuation and $s > 1$ , then $| |^{s}$ need not be a valuation. For instance, let $| |$ be the usual absolute value on $ℝ$ . Then $| |^{2}$ is not a valuation, as $2^{2} > 1^{2} + 1^{2}$ .

Definition 5.2.7. §

A nonarchimedean valuation $| |$ on a field $K$ is a nontrivial multiplicative valuation such that

| a + b | \leq \max (| a |, | b |)

for all $a, b \in K$ .

For our purposes, the following ad-hoc definition of an archimedean valuation will suffice.

Notation 5.2.8. §

An archimedean valuation on $K$ is a nontrivial valuation on $K$ that is not nonarchimedean.

Remark 5.2.9. §

Every valuation in the equivalence class of a nonarchimedean valuation is also nonarchimedean.

The following is a useful equivalent condition for a valuation to be nonarchimedean.

Lemma 5.2.10. §

A nontrivial valuation $| |$ on a field $K$ is nonarchimedean if and only if $| n \cdot 1 | \leq 1$ for all integers $n \geq 2$ .

Proof.

We write $n \cdot 1 \in K$ more simply as $n$ . If $| |$ is nonarchimedean, then

| n | \leq \max (| 1 |, \dots, | 1 |) = 1

for $n \geq 2$ . Conversely, suppose that $| n | \leq 1$ for $n \geq 2$ (and hence for all $n \in ℤ$ ). Let $α, β \in K$ . For an integer $k \geq 1$ , note that

| α + β |^{k} \leq \sum_{i = 0}^{k} | (\binom{k}{i}) α^{i} β^{k - i} | \leq \sum_{i = 0}^{k} | (\binom{k}{i}) | \max {(| α |, | β |)}^{k} \leq (k + 1) \max {(| α |, | β |)}^{k} .

Taking $k$ th roots, we obtain

| α + β | \leq {(1 + k)}^{\frac{1}{k}} \max (| α |, | β |),

and taking the limit as $k$ tends to infinity provides the result. □

As the integer multiples of $1$ in a field of positive characteristic are units and $0$ , Lemma 5.2.10 yields the following.

Corollary 5.2.11. §

Every nontrivial valuation on a field of positive characteristic is nonarchimedean.

We also have the following notion, generalizing that of a discrete valuation.

Definition 5.2.12. §

An additive valuation $v$ on a field $K$ is a function $v : K \to ℝ \cup {\infty}$ satisfying

i.: $v (a) = \infty$ if and only $a = 0$ ,
ii.: $v (𝑎𝑏) = v (a) + v (b)$ , and
iii.: $v (a + b) \geq \min (v (a), v (b))$

for all $a, b \in K$ .

The following gives the comparison between additive valuations and nonarchimedean valuations.

Lemma 5.2.13. §

Let $K$ be a field. Let $v : K \to ℝ \cup {\infty}$ and $| | : K \to ℝ_{\geq 0}$ be such that there exists a real number $c > 1$ such that $| a | = c^{- v (a)}$ for $a \in K$ $($ taking $c^{- \infty} = 0)$ . Then $v$ is an additive valuation on $K$ if and only if $| |$ is a nonarchimedean valuation on $K$ .

Proof.

Note that $f (x) = c^{- x}$ and $g (x) = - \log_{c} (x)$ are inverse functions between $ℝ \cup {\infty}$ and $ℝ_{\geq 0}$ . So, $v (𝑎𝑏) = v (a) + v (b)$ if and only if

c^{- (v (a) + v (b))} = c^{- v (a)} c^{- v (b)},

so if and only if $| 𝑎𝑏 | = | a | | b |$ . Moreover,

v (a + b) \geq \min (v (a), v (b))

if and only if

c^{- v (a + b)} \leq c^{- \min (v (a), v (b))} = \max (c^{- v (a)}, c^{- v (b)}),

so if and only if $| a + b | \leq \max (| a | + | b |)$ . Moreover, $a = \infty$ if and only if $c^{- a} = 0$ , the property that $v (a) = \infty$ if and only if $a = 0$ holds if and only if the property that $| a | = 0$ holds if and only if $a = 0$ . □

Definition 5.2.14. §

The value group $| K^{\times} |$ of a valuation on a field $K$ is the subgroup of $ℝ^{\times}$ consisting of elements $| a |$ for $a \in K^{\times}$ .

Definition 5.2.15. §

A valuation on a field $K$ is discrete if and only if its value group is a discrete subset of $ℝ^{\times}$ (with respect to the subspace topology on $ℝ$ ).

Remark 5.2.16. §

It is not so hard to show that any discrete valuation on a field is nonarchimedean.

Remark 5.2.17. §

A nonarchimedean valuation $| |$ on a field $K$ is discrete if and only if there exists $c \in ℝ_{> 1}$ such that $v : K \to ℝ \cup {\infty}$ defined by $v (a) = \log_{c} (| a |)$ for $a \in K$ is discrete (i.e., has image $ℤ \cup {\infty}$ ). In other words, a discrete (additive) valuation $v$ corresponds to an equivalence class of discrete, nonarchimedean valuations on $K$ .

Remark 5.2.18. §

Since $\log_{c} : ℝ_{> 0} \to ℝ$ is an isomorphism for any $c > 1$ and a subgroup of $ℝ$ is discrete if and only if it is a lattice, hence cyclic, a nonarchimedean valuation is discrete if and only if its value group is cyclic.

As with discrete valuations, nonarchimedean valuations give rise to a number of structures on a field.

Lemma 5.2.19. §

Let $K$ be a field and $| |$ a nonarchimedean valuation on $K$ . The set

𝒪 = {a \in K ∣ | a | \leq 1}

is a subring of $K$ that is local with maximal ideal $𝔪 = {a \in K ∣ | a | < 1}$ .

Proof.

If $a, b \in 𝒪$ , then $| 𝑎𝑏 | = | a | \cdot | b | \leq 1$ and $| a \pm b | \leq \max (| a |, | b |) \leq 1$ , so $𝒪$ is a ring. Similarly, $𝔪$ is an ideal. To see that it is maximal, note that if $a \in 𝒪 - 𝔪$ , then $| a^{- 1} | = | a |^{- 1} = 1$ , so $a^{- 1} \in 𝒪$ . Thus $a$ is a unit, so $𝒪$ is a local ring with maximal ideal $𝔪$ . □

Definition 5.2.20. §

The valuation ring of a nonarchimedean valuation $| |$ on a field $K$ is the subring $𝒪$ of $K$ defined by

𝒪 = {a \in K ∣ | a | \leq 1} .

Remark 5.2.21. §

Similarly, we can speak of the valuation ring of an additive valuation $v$ of $K$ . It is $𝒪 = {a \in K ∣ v (a) \geq 0}$ , and its maximal ideal is $𝔪 = {a \in K ∣ v (a) > 0}$ .

Lemma 5.2.22. §

Let $K$ be a field and $| |$ a nonarchimedean valuation on $K$ . Then the valuation $| |$ is discrete if and only if its valuation ring $𝒪$ is a discrete valuation ring.

Proof.

If $| |$ is discrete, then let $π \in 𝔪$ be an element for which $| π |$ is maximal. Then $| π |$ generates the value group of $| |$ , so any $a \in 𝔪$ may be written as $a = π^{k} u$ for some $k \geq 1$ and $u \in 𝒪^{\times}$ . In particular, $𝔪 = (π)$ , so $𝒪$ is a DVR.

Conversely, if $𝒪$ is a DVR, let $π \in 𝔪$ be a uniformizer. Then any $a \in K^{\times}$ may be written $a = π^{k} u$ for some $k \in ℤ$ and $u \in 𝒪^{\times}$ , and we have $| a | = | π |^{k}$ , so the value group $K^{\times}$ is cyclic, hence discrete. □

Lemma 5.2.23. §

Let $K$ be a field and $| |$ a discrete valuation on $K$ . Let $𝔪$ be the maximal ideal of the valuation ring $𝒪$ of $K$ . Then

𝔪^{n} = {a \in K ∣ | a | \leq r^{n}},

for all $n \geq 1$ , where $r$ is the maximal value of $| |$ with $r < 1$ .

Proof.

That $𝔪$ is as stated follows Lemma 5.2.19 and the fact that $| |$ is discrete. Conversely, let $π \in 𝔪$ be a uniformizer. For any $a \in 𝒪$ , we have $a = π^{k} u$ for some $k \geq 0$ and $u \in 𝒪^{\times}$ , and $| a | = | π |^{k}$ , so $r = | π |$ , and $a \in 𝔪^{n}$ if and only if $| a | \leq r^{n}$ . □

It is useful to pick a canonical, or normalized, multiplicative valuation attached to some of the discrete valuations we have studied. We define the $𝔭$ -adic absolute value for any nonzero finite prime $𝔭$ of a global field.

Definition 5.2.24. §

Let $K$ be a global field, and let $𝔭$ be a finite prime of $K$ , or an infinite prime of $K$ if $K$ is a function field. Let $p$ be the characteristic of the residue field of $𝔭$ , and let $f_{𝔭}$ denote the residue degree of $𝔭$ . The $𝔭$ -adic absolute value on $K$ is the unique multiplicative valuation on $K$ that satisfies

| a |_{𝔭} = p^{- f_{𝔭} v_{𝔭} (a)}

for $a \in K^{\times}$ .

Remark 5.2.25. §

If $𝔭$ is a finite prime of a global field $K$ that is a principal ideal $(π)$ of $𝒪_{K}$ , then we denote $| |_{𝔭}$ by $| |_{π}$ as well.

Example 5.2.26. §

The $p$ -adic absolute value $| |_{p}$ on $ℚ$ is defined by $| 0 |_{p} = 0$ and

| a |_{p} = p^{- v_{p} (a)}

for $a \in ℚ^{\times}$ . Note that it is discrete with valuation ring consisting of reduced fractions with denominator not divisible by $p$ .

Example 5.2.27. §

Let $q$ be a prime power. The absolute value $| |_{\infty}$ at infinity on $𝔽_{q} (t)$ is defined by $| 0 |_{\infty} = 0$ and

{| \frac{f}{g} |}_{\infty} = q^{\deg f - \deg g}

for nonzero $f, g \in 𝔽_{q} [t]$ .

As for archimedean valuations, we have the following.

Definition 5.2.28. §

Let $K$ be a number field, and let $σ : K ↪ ℂ$ be an archimedean embedding of $K$ . Then the absolute value with respect to $σ$ is the multiplicative valuation $| |_{σ} : K \to ℝ_{\geq 0}$ defined by $| a |_{σ} = | σ (a) |$ (the complex absolute value of $σ (a)$ ) for $a \in K$ .

In other words, archimedean primes give rise to archimedean valuations. We now make the following definition.

Definition 5.2.29. §

A place of a global field $K$ is an equivalence class of nontrivial valuations on $K$ .

Definition 5.2.30. §

A finite place (resp., infinite place) of a global field $K$ is the equivalence class of the absolute value attached to a finite (resp., infinite) prime of $K$ .

We will see that every place of a global field is either finite or infinite. At present, let us prove this for $ℚ$ .

Theorem 5.2.31 (Ostrowski). §

The places of $ℚ$ are exactly the equivalence classes of the $p$ -adic absolute values on $ℚ$ for a prime number $p$ and of the usual absolute value on $ℚ$ .

Proof.

Let $| |$ be a nontrivial valuation on $ℚ$ . Let $m, n \geq 2$ be integers, and write $m$ as

m = \sum_{i = 0}^{k} a_{i} n^{i}

for some integers $0 \leq a_{i} < n$ for $0 \leq i \leq k$ and some $k \geq 0$ , with $a_{k} \neq 0$ . Note that $n^{k} \leq m$ , so

k \leq \frac{\log m}{\log n} .

Let $N = \max (1, | n |)$ . We also have $| 1 | = 1$ , so $| a_{i} | < n$ , and we have

| m | < \sum_{i = 0}^{k} n | n |^{i} \leq (1 + k) n N^{k} \leq (1 + \frac{\log m}{\log n}) n N^{\frac{\log m}{\log n}} .

Replacing $m$ by $m^{t}$ for some $t > 0$ and taking $t$ th roots of both sides, we have

| m | < {(1 + t \frac{\log m}{\log n})}^{\frac{1}{t}} n^{\frac{1}{t}} N^{\frac{\log m}{\log n}} .

As we let $t \to \infty$ , we obtain

| m | \leq N^{\frac{\log m}{\log n}} .

(5.2.1)

If $| n | \leq 1$ for some integer $n \geq 2$ , then $N = 1$ , and (5.2.1) with this $n$ implies that $| m | \leq 1$ for all $m \geq 2$ , and hence for all $m \in ℤ$ . Since $| |$ is multiplicative and nontrivial, and $ℤ$ is a unique factorization domain, we must then have $| p | < 1$ for some prime number $p$ . Consider the set

𝔪 = {a \in ℤ ∣ | a | < 1} .

Since $| |$ is nonarchimedean by Lemma 5.2.10, the set $𝔪$ is an ideal of $ℤ$ . Since it contains $p$ and is not $ℤ$ , it must equal $(p)$ . Let $s > 0$ be such that $| p | = p^{- s}$ . Any $q \in ℚ$ can be expressed as $q = p^{v_{p} (q)} \frac{m}{n}$ for some $m, n \in ℤ$ with $p ∤ 𝑚𝑛$ . We then have

| q | = | p |^{v_{p} (q)} = p^{- s v_{p} (q)} = | q |_{p}^{s} .

In other words, $| |$ is equivalent to $| |_{p}$ , and moreover, it is not equivalent to $| |_{ℓ}$ for any prime number $ℓ \neq p$ , since $ℓ \notin 𝔪$ , nor is it equivalent to $| |_{\infty}$ , since $| 2 |_{\infty} = 2 > 1$ .

If $| n | > 1$ for all integers $n \geq 2$ , then (5.2.1) implies that

| m |^{\frac{1}{\log m}} \leq | n |^{\frac{1}{\log n}}

for all $m, n \geq 2$ . Switching their roles of $m$ and $n$ gives the opposite inequality. In other words, $| n |^{\frac{1}{\log n}}$ is constant for $n \geq 2$ , equal to some $s > 1$ . We then have

| n | = s^{\log n} = n^{\log s}

for all $n \geq 2$ , and multiplicativity then forces $| q | = | q |_{\infty}^{\log s}$ for all $q \in ℚ$ , where $| |_{\infty}$ denotes the usual absolute value on $ℚ$ . Proposition 5.2.5 implies that $| |$ is equivalent to $| |_{\infty}$ . □

We also have the following, which we leave as an exercise.

Proposition 5.2.32. §

Let $q$ be a prime power. The places of $𝔽_{q} (t)$ are exactly the equivalence classes of the $f$ -adic absolute values, for $f$ an irreducible polynomial in $𝔽_{q} [t]$ , and the absolute value at $\infty$ .

Remark 5.2.33. §

We often index the set $V_{K}$ of places of a global field $K$ by a subscript, such as $v$ . Each place in $V_{K}$ , as we have seen already for $K = ℚ$ and stated for $K = 𝔽_{q} (t)$ , is represented by the multiplicative valuation attached to a unique prime (as in Definitions 5.2.24 and 5.2.28). We write $| |_{v}$ for for this valuation.

We note the following consequence of Theorem 5.2.31.

Proposition 5.2.34. §

For any $a \in ℚ^{\times}$ , we have $| a |_{p} = 1$ for all but finitely many prime numbers $p$ , and we have

\prod_{v \in V_{ℚ}} | a |_{v} = 1 .

Proof.

Since valuations are multiplicative, it suffices to prove this for $- 1$ and every prime number $p$ . But $| - 1 |_{v} = 1$ for all $v \in V_{ℚ}$ , and $| p |_{ℓ} = 1$ for primes $ℓ \neq p$ , while $| p |_{p} = p^{- 1}$ and $| p |_{\infty} = p$ . □

We have a similar consequence of Proposition 5.2.32.

Proposition 5.2.35. §

Let $q$ be a power of a prime number. Let $h \in 𝔽_{q} {(t)}^{\times}$ . Then $| h |_{f} = 1$ for all but finitely many irreducible polynomials $f \in 𝔽_{q} [t]$ , and we have

\prod_{v \in V_{𝔽_{q} (t)}} | h |_{v} = 1 .

We end the section with the weak approximation theorem, which is an analogue of the Chinese remainder theorem for valuations. This requires a lemma.

Lemma 5.2.36. §

Let $| |_{1}, | |_{2}, \dots, | |_{k}$ be nontrivial, inequivalent valuations on a field $K$ . Then there exists an element $a \in K$ such that $| a |_{1} < 1$ and $| a |_{j} > 1$ for all $j \geq 2$ .

Proof.

In the case $k = 2$ , note that Proposition 5.2.5 provides one with elements $α, β \in K$ with $| α |_{1} < 1$ , $| α |_{2} \geq 1$ , $| β |_{1} \geq 1$ , and $| β |_{2} < 1$ . Then $c = \frac{α}{β}$ satisfies $| c |_{1} < 1$ and $| c |_{2} > 1$ .

For $k > 2$ , suppose by induction that we have found an element $α$ such that $| α |_{1} < 1$ and $| α |_{j} > 1$ for all $2 \leq j \leq k - 1$ and an element $β$ such that $| β |_{1} < 1$ and $| β |_{k} > 1$ . If $| α |_{k} > 1$ , then we simply take $a = α$ . If $| α |_{k} = 1$ , then choose $s > 0$ sufficiently large so that $| α |_{j}^{s} > | β |_{j}^{- 1}$ for all $2 \leq j \leq k - 1$ and $| α |_{1}^{s} < | β |_{1}^{- 1}$ . Then $a = α^{s} β$ works.

Finally, if $| α |_{k} < 1$ , let

c_{m} = β^{- 1} {(1 + α^{m})}^{- 1}

for every integer $m \geq 1$ . The sequence ${(1 + α^{m})}^{- 1}$ has a limit of $1$ under the topology of $| |_{i}$ if $| α |_{i} < 1$ and $0$ if $| α |_{i} > 1$ . So, we have that $| c_{m} |_{1} \to | β |_{1}^{- 1}$ , that $| c_{m} |_{j} \to 0$ for $2 \leq j \leq k - 1$ , and that $| c_{m} |_{k} \to | β |_{k}^{- 1}$ . We then take $a = c_{m}^{- 1}$ for a sufficiently large value of $m$ . □

Theorem 5.2.37 (Weak Approximation). §

Let $| |_{1}, | |_{2}, \dots, | |_{k}$ be nontrivial, inequivalent valuations on a field $K$ , and let $a_{1}, a_{2} \dots, a_{k} \in K$ . For every $𝜖 > 0$ , there exists an element $b \in K$ such that $| a_{i} - b |_{i} < 𝜖$ for all $1 \leq i \leq k$ .

Proof.

It follows from Lemma 5.2.36 that there exists for each $i$ an element $α_{i} \in K$ with $| α_{i} |_{i} < 1$ and $| α_{i} |_{j} > 1$ for all $j \neq i$ . For each $i$ and a chosen $δ > 0$ , let $β_{i} = \frac{1}{1 + α_{i}^{m}}$ for a value of $m$ which is sufficiently large in order that $| β_{i} - 1 |_{i} < δ$ and $| β_{i} |_{j} < δ$ for $j \neq i$ . We then set

b = \sum_{j = 1}^{k} a_{j} β_{j},

which satisfies

| b - a_{i} |_{i} \leq | a_{i} |_{i} | β_{i} - 1 |_{i} + \sum_{\begin{array}{c} j = 1 \\ j \neq i \end{array}}^{k} | a_{j} |_{i} | β_{j} |_{i} < \sum_{j = 1}^{k} | a_{j} |_{i} \cdot δ < 𝜖

for a good choice of $δ$ . □

5.3. Completions

Definition 5.3.1. §

A pair consisting of a field $K$ and a valuation $| |$ on $K$ is called a valued field.

Remark 5.3.2. §

When the valuation is understood, a valued field $(K, | |)$ is often simply denoted $K$ .

Definition 5.3.3. §

A topological field $K$ is a field endowed with a topology with respect to which the binary operations of addition and multiplication are continuous, as are the maps that take an element to its additive inverse and a nonzero element to its multiplicative inverse, the latter with respect to the subspace topology on $K^{\times}$ .

Remark 5.3.4. §

A topological field is in particular a topological group with respect to addition, and its multiplicative group is a topological group with respect to multiplication. Moreover, multiplication is continuous on the entire field.

We leave the proof of the following to the reader.

Proposition 5.3.5. §

A valued field is a topological field with respect to the topology defined by its valuation.

Remark 5.3.6. §

A nonarchimedean valued field $K$ has a valuation ring $𝒪$ . Terminology is often abused between the two. For instance, the unit group of $K$ would usually be taken to mean the unit group of $𝒪$ . Or, if $𝒪$ is discrete, a uniformizer of $K$ would mean a uniformizer of $𝒪$ .

Definition 5.3.7. §

A valued field $K$ is said to be complete if it $K$ is complete with respect to the topology defined by its valuation.

Example 5.3.8. §

The fields $ℝ$ and $ℂ$ are complete with respect to their usual topologies.

Definition 5.3.9. §

A field embedding $ι : K \to K^{'}$ , where $(K, | |)$ and $(K^{'}, | |^{'})$ are valued fields, is an embedding of valued fields if $| ι (α) |^{'} = | α |$ for all $α \in K$ . If $ι : K \to K^{'}$ is an embedding of valued fields, we say that $ι$ preserves the valuation on $K$ .

Definition 5.3.10. §

An isomorphism of valued fields is a field isomorphism that is an embedding of valued fields.

We wish to study completions of a valued field $K$ . The completion is a larger field that essentially consists of the limits of Cauchy sequences in $K$ , with field operations determined by the fact that they should be continuous.

Theorem 5.3.11. §

Let $(K, | |)$ be a valued field. Then there exists a complete valued field $(\hat{K}, | |)$ and a embedding $ι : K \to \hat{K}$ of valued fields such that the image $ι (K)$ is dense in $\hat{K}$ .

Proof.

Let $R$ be the set of Cauchy sequences on $K$ . By definition, if ${(a_{n})}_{n} \in R$ , then for any $𝜖 > 0$ , there exists $N \geq 1$ such that $| a_{n} - a_{m} | < 𝜖$ for all $n, m \geq N$ . Thus, But $| | a_{n} | - | a_{m} | | \leq | a_{n} - a_{m} |$ , so ${(| a_{n} |)}_{n}$ is a Cauchy sequence in $ℝ$ , which therefore converges. In other words, we may define a function $∥ ∥ : R \to ℝ_{\geq 0}$ by

∥ {(a_{n})}_{n} ∥ = \lim_{n \to \infty} | a_{n} | .

Note that, in particular ${(| a_{n} |)}_{n}$ is bounded for any ${(a_{n})}_{n} \in R$ . It is easy to check that $R$ is a ring: in particular, if ${(a_{n})}_{n}$ , ${(b_{n})}_{n} \in R$ , then

| a_{n} b_{n} - a_{m} b_{m} | \leq | a_{n} | \cdot | b_{n} - b_{m} | + | b_{m} | | a_{n} - a_{m} |,

and if $| a_{n} | < M$ and $| b_{n} | < M$ for all $n$ , then given $𝜖 > 0$ we choose $m$ , $n$ sufficiently large so that $| a_{n} - a_{m} |, | b_{n} - b_{m} | < \frac{𝜖}{2 M}$ , and the right-hand side is less than $𝜖$ .

Let $𝔐$ be the set of (Cauchy) sequences on $K$ that converge to $0$ . We check that $𝔐$ is a maximal ideal of $R$ . Clearly, the sum of any two sequences that converges to $0$ does as well. If ${(a_{n})}_{n} \in R$ and ${(b_{n})}_{n} \in 𝔐$ , then for any $𝜖 > 0$ , we have that

| a_{n} b_{n} | = | a_{n} | \cdot | b_{n} | < 𝜖

for $n$ large enough so that $| b_{n} | < \frac{𝜖}{M}$ , where $| a_{n} | < M$ for all $n$ . If ${(a_{n})}_{n} \in R - 𝔐$ , then $a_{n} \neq 0$ for $n$ sufficiently large, so we can add an eventually $0$ sequence (which necessarily lies in $𝔐$ ) to it to make $a_{n} \neq 0$ for all $n$ . The sequence ${(a_{n}^{- 1})}_{n}$ is then defined, and it is Cauchy, as $| a_{n} |$ is bounded below, and

| a_{n}^{- 1} - a_{m}^{- 1} | = \frac{| a_{n} - a_{m} |}{| a_{n} | | a_{m} |} .

In other words, $R ∕ 𝔐$ is a field.

Set $\hat{K} = R ∕ 𝔐$ . We have a natural field embedding $ι : K \to \hat{K}$ that takes $a \in K$ to the coset of the constant sequence ${(a)}_{n}$ .

For ${(a_{n})}_{n}, {(b_{n})}_{n} \in R$ , note that

∥ {(a_{n} \cdot b_{n})}_{n} ∥ = \lim_{n \to \infty} | a_{n} b_{n} | = \lim_{n \to \infty} | a_{n} | \cdot | b_{n} | = ∥ {(a_{n})}_{n} ∥ \cdot ∥ {(b_{n})}_{n} ∥

and

∥ {(a_{n} + b_{n})}_{n} ∥ = \lim_{n \to \infty} | a_{n} + b_{n} | \leq \lim_{n \to \infty} (| a_{n} | + | b_{n} |) = ∥ {(a_{n})}_{n} ∥ + ∥ {(b_{n})}_{n} ∥ .

Moreover, $∥ {(a_{n})}_{n} ∥ = 0$ if and only if ${(a_{n})}_{n} \in 𝔐$ . Thus, $∥ ∥$ induces a valuation $| |$ on $\hat{K}$ , and it clearly preserves the valuation on $K$ .

To see that $\hat{K}$ is complete with respect to $∥ ∥$ , let ${(c_{m})}_{m}$ with $c_{m} = {(c_{m, n})}_{n}$ be a Cauchy sequence in $R$ . If it has a limit in $R$ , its image in $\hat{K}$ has a limit as well. For $𝜖 > 0$ , there exists $N \geq 1$ such that for $m \geq k \geq N$ , we have

∥ c_{m} - c_{k} ∥ = \lim_{n \to \infty} | c_{m, n} - c_{k, n} | < \frac{𝜖}{2} .

There then exists $N_{m} \geq N$ such that for $k \leq m$ , we have $| c_{m, n} - c_{k, n} | < \frac{𝜖}{2}$ for all $n \geq N_{m}$ . On the other hand, since each $c_{m}$ is Cauchy, there exist $l_{m} \geq \max (N_{m}, l_{m - 1})$ (with $l_{1} \geq N_{1}$ ) such that $| c_{m, n} - c_{m, n^{'}} | < \frac{𝜖}{2}$ for $n, n^{'} \geq l_{m}$ . Consider the sequence ${(a_{n})}_{n}$ of elements $a_{n} = c_{n, l_{n}}$ of $K$ . For $m \geq k \geq N$ , we have

| a_{m} - a_{k} | = | c_{m, l_{m}} - c_{k, l_{k}} | \leq | c_{m, l_{m}} - c_{k, l_{m}} | + | c_{k, l_{m}} - c_{k, l_{k}} | < 𝜖

by our condition on $ℓ_{m}$ , so ${(a_{n})}_{n} \in R$ . Moreover,

∥ c_{m} - {(a_{n})}_{n} ∥ = \lim_{n \to \infty} | c_{m, n} - c_{n, l_{n}} |,

and

| c_{m, n} - c_{n, l_{n}} | \leq | c_{m, n} - c_{m, l_{m}} | + | c_{m, l_{m}} - c_{n, l_{n}} | < 𝜖

for $n \geq m \geq N$ and $n \geq l_{m}$ , which means that $∥ c_{m} - {(a_{n})}_{n} ∥ \leq 𝜖$ for $m \geq N^{'}$ . That is, the sequence ${(c_{m})}_{m}$ of sequences converges to ${(a_{n})}_{n}$ in $R$ .

Finally, we show that the image of $ι$ is dense. That is, let ${(a_{n})}_{n} \in R$ . For each $m \geq 1$ , we have

∥ {(a_{m})}_{n} - {(a_{n})}_{n} ∥ = \lim_{n \to \infty} | a_{m} - a_{n} |,

and we saw that $| a_{m} - a_{n} | < 𝜖$ for $m, n \geq N$ . But then

∥ {(a_{m})}_{n} - {(a_{n})}_{n} ∥ \leq 𝜖

so the sequence ${(ι (a_{m}))}_{m}$ in $\hat{K}$ converges to the image of ${(a_{n})}_{n}$ . □

Proposition 5.3.12. §

Let $K$ be a valued field and $\hat{K}$ a complete valued field for which there exists a dense embedding $ι : K \to \hat{K}$ that preserves the valuation on $K$ . If $L$ is a complete valued field and $σ : K \to L$ is an embedding of valued fields, then there is a unique extension of $σ$ to an embedding $\hat{σ} : \hat{K} \to L$ of valued fields.

Proof.

Let $| |$ (resp., $| |^{'})$ denote the valuation on $K$ and $\hat{K}$ (resp., $L$ ). For any Cauchy sequence ${(a_{n})}_{n}$ in $K$ , the sequence $σ {(a_{n})}_{n}$ is Cauchy as $| σ (a_{n}) - σ (a_{m}) |^{'} = | a_{n} - a_{m} |$ , and therefore it is convergent. Define $\hat{σ} : \hat{K} \to L$ by

\hat{σ} ({(a_{n})}_{n}) = \lim_{n \to \infty} σ (a_{n}) .

This is clearly a nonzero ring homomorphism, hence a field embedding, and it extends $σ$ . By definition, it preserves the valuation on $\hat{K}$ . Moreover, if $τ : \hat{K} \to L$ is any field embedding extending $σ$ and preserving the valuation on $\hat{K}$ , then for any $m \geq 1$ , we have

| τ ({(a_{n})}_{n}) - σ (a_{m}) |^{'} = \lim_{n \to \infty} | a_{n} - a_{m} |,

which since ${(a_{n})}_{n}$ is Cauchy implies that the sequence ${(σ (a_{m}))}_{m}$ converges to $τ ({(a_{n})}_{n})$ in $L$ . On the other hand, this limit is by definition $\hat{σ} ({(a_{n})}_{n})$ , so $τ = \hat{σ}$ . □

This allows us to make the following definition.

Definition 5.3.13. §

Let $(K, | |)$ be a valued field. Any complete valued field $(\hat{K}, | |)$ as in Theorem 5.3.11 is called the completion of $K$ .

Remark 5.3.14. §

The completion of a valued field is unique up to unique isomorphism fixing $K$ by Theorem 5.3.11.

Example 5.3.15. §

The completion of $ℚ$ with respect to the usual absolute value is isomorphic to $ℝ$ . To see this, define $\hat{ι} : \hat{ℚ} \to ℝ$ via the map that takes the class of a Cauchy sequence in $ℚ$ to its limit. This is clearly a field embedding preserving the valuation, and it is surjective since, for every real number, there exists a sequence of rationals converging to it.

In fact, any complete archimedean valued field $K$ is topologically isomorphic (i.e., isomorphic via fields via a map which is a homeomorphism) to $ℝ$ or $ℂ$ . In other words, $K$ is isomorphic as a valued field to $ℝ$ or $ℂ$ with valuation given by some power of the usual absolute value.

Theorem 5.3.16 (Ostrowski). §

Let $K$ be a complete valued field with respect to an archimedean valuation. Then $K$ is isomorphic as a valued field to either $(ℝ, | |^{s})$ or $(ℂ, | |^{s})$ for some $s \in (0, 1]$ , where $| |$ denotes the usual absolute value on $ℝ$ or $ℂ$ .

Proof.

By Corollary 5.2.11, the field $K$ must have characteristic zero. Let $| |$ denote the valuation on $K$ , and in the proof let us use $| |_{\infty}$ to denote the absolute value on $ℂ$ . By Ostrowski’s theorem on $ℚ$ , the restriction of $| |$ to $ℚ$ is equivalent to the usual absolute value. Note that $| |_{\infty}^{s}$ satisfies the triangle inequality for a given $s \in ℝ_{> 0}$ if and only $s \leq 1$ . Let $s \in (0, 1]$ be such that $| a | = | a |_{\infty}^{s}$ for all $a \in ℚ$ . As $K$ is complete, it must then contain the completion $(ℝ, | |_{\infty}^{s})$ of $ℚ$ with respect to $| |$ .

If $i = \sqrt{- 1} \in K$ , then for $𝜃 \in ℝ$ , we have

| e^{2 𝜋𝑖𝜃} | \leq | \cos 𝜃 | + | \sin 𝜃 | \leq \sqrt{2},

but since this is true for all $𝜃$ , we get $| e^{2 𝜋𝑖𝜃} |^{n} = | e^{2 𝜋𝑖𝑛𝜃} | \leq \sqrt{2}$ for all $n \in ℤ$ . Thus $| e^{2 𝜋𝑖𝜃} | = 1$ . Since we can write $z \in ℂ$ as $z = r e^{2 𝜋𝑖𝜃}$ with $r \geq 0$ and $𝜃 \in ℝ$ , we have $| z | = | z |_{\infty}^{s}$ for all $z \in ℂ \subseteq K$ .

In general, we can replace $K$ by $K (i)$ and extend $| |$ to $K (i)$ by $| a + 𝑏𝑖 | = \sqrt{| a |^{2} + | b |^{2}}$ , so we may assume that in fact $K$ contains $(ℂ, | |_{\infty}^{s})$ as valued fields. Now let $α \in K$ , and let $w \in ℂ$ be such that $| α - w |$ is minimal: this exists since the infimum $t$ occurs as a limit in the closed ball of radius $| α | + u$ about $0$ in $ℂ$ for any $u > t$ .

Now suppose $α \notin ℂ$ , which is to say that $t > 0$ . Replacing $α$ by $α - w$ , we may as well assume that $w = 0$ . Then $t = | α | \leq | α - z |$ for all $z \in ℂ$ . For $z \in ℂ$ and $n \geq 1$ , note that

| z |^{n} + t^{n} \geq | α^{n} - z^{n} | = \prod_{j = 0}^{n - 1} | α - ζ_{n}^{j} z | \geq t^{n - 1} | α - z |

where $ζ_{n} = e^{2 𝜋𝑖 ∕ n}$ . We then have

| α - z | \leq t (\frac{| z |^{n}}{t^{n}} + 1) .

Thus, taking $z$ such that $| z | < t$ and the limit as $n$ tends to $\infty$ , we obtain that $| α - z | \leq t$ . By minimality of $t$ , this forces $| α - z | = t$ .

By the same argument with $α - z$ replacing $α$ , we see then that $| α - z - w | = t$ for all $w \in ℂ$ with $| w | < t$ . Recursively, we then see in particular that $| α - 𝑚𝑧 | = t$ for all $m \geq 1$ and $z \in ℂ$ with $| z | \leq α$ . The set of all such $𝑚𝑧$ being $ℂ$ , we see that $| α - z | = t$ for all $z \in ℂ$ . But then $| z | \leq | z - α | + | α | = 2 t$ for all $z \in ℂ$ which contradicts $| z | = | z |_{\infty}^{s}$ for any $z \in ℂ$ with sufficiently large $| z |_{\infty}$ . In other words, $α$ does not exist. □

Lemma 5.3.17. §

Let $(K, | |)$ be a nonarchimedean valued field, and let $(\hat{K}, | |)$ be its completion. Then $| |$ is a nonarchimedean valuation on $\hat{K}$ with the same value group as its restriction to $K$ . If $𝒪$ (resp., $\hat{𝒪}$ ) denotes the valuation ring of $K$ (resp., $\hat{K}$ ) and $𝔪$ (resp., $\hat{𝔪}$ ) denotes its the maximal ideal, then the canonical map

\bar{ι} : 𝒪 ∕ 𝔪 \to \hat{𝒪} ∕ \hat{𝔪}

is an isomorphism. Moreover, if $| |$ is discrete on $K$ , then it is on $\hat{K}$ as well, and

{\bar{ι}}_{n} : 𝒪 ∕ 𝔪^{n} \to \hat{𝒪} ∕ {\hat{𝔪}}^{n}

is an isomorphism for every $n \geq 1$ .

Proof.

That $\hat{K}$ is nonarchimedean is an immediate corollary of Lemma 5.2.10 (and can also be seen directly). If $a$ is nonzero, then $| a_{n} - a | < | a |$ for all sufficiently large $n$ , so $| a_{n} | = | (a_{n} - a) + a | = | a |$ , and thus the value groups of $| |$ on $K$ and $\hat{K}$ are equal.

Since the embedding $K \to \hat{K}$ preserves the valuation, we have $𝔪 = 𝒪 \cap \hat{𝔪}$ , so $\bar{ι}$ is injective. If $a \in \hat{𝒪}$ , then since $K$ is dense in $\hat{K}$ , there exists $b \in K$ with $| b - a | < 1$ , so $b - a \in \hat{𝔪}$ , which in particular implies $b \in \hat{𝒪} \cap K = 𝒪$ with $\bar{ι} (b + 𝔪) = a + \hat{𝔪}$ . In other words, $\bar{ι}$ is surjective.

If $| |$ is discrete on $K$ , then any Cauchy sequence in $K$ has valuation that is eventually constant or heads to $0$ . As a uniformizer $π$ of $K$ is also one of $\hat{K}$ , it follows immediately that ${\bar{ι}}_{n}$ is injective. We have

𝔪^{n} = {a \in K | | a | \leq | π |^{n}},

and so if $a \in {\hat{𝔪}}^{n}$ and we choose $b \in K$ with $| b - a | < | π |^{n}$ , then $| b | \leq | π |^{n}$ , so $b \in 𝔪^{n}$ . That is, ${\bar{ι}}_{n}$ is surjective. □

Remark 5.3.18. §

A discrete additive valuation $v$ on a field $K$ extends to a discrete valuation on $\hat{K}$ , usually denoted $v$ as well.

Definition 5.3.19. §

A valued field is said to be discretely valued if its valuation is discrete. A complete discrete valuation field is a complete discretely valued field.

Proposition 5.3.20. §

Let $K$ be a complete discrete valuation field. Let $𝒪$ be its valuation ring, and $𝔪$ the maximal ideal of $𝒪$ . Let $T$ be a set of representatives of $𝒪 ∕ 𝔪$ that includes $0$ , and let $π$ be a uniformizer of $𝒪$ . Every element $a \in K$ is a limit of a unique sequence of partial sums of the form

a_{n} = \sum_{k = m}^{n} c_{k} π^{k}

for $m \in ℤ$ and $c_{k} \in T$ for all $k \geq m$ , with $c_{m} \neq 0$ . Moreover, the additive valuation of such an element $a$ is $m$ .

Proof.

Since each $a_{n}$ must have valuation $m$ and the $a_{n}$ must converge to $a$ , we must have $m = v (a)$ . So, we take $a_{m - 1} = 0$ and, inductively, for any $n \geq m$ , we write $a - a_{n - 1} = b_{n} π^{n}$ for some $b_{n} \in 𝒪$ , and let $c_{n} \in T$ be the unique element such that $c_{n} \equiv b_{n} mod 𝔪$ . (Note that $c_{m} \neq 0$ .) Then

a - a_{n} = b_{n} π^{n} - c_{n} π^{n} \in 𝔪^{n + 1},

and $c_{n} \in T$ is unique such that this holds. By definition, $a$ is then the limit of the $a_{n}$ , and the choice of each $c_{n}$ is the only possibility for which this happens, as if $a ≢ a_{n} mod 𝔪^{n + 1}$ , then since $a ≢ a_{k} mod 𝔪^{n + 1}$ for all $k > n$ , the sequence ${(a_{n})}_{n}$ would not converge to $a$ . □

Notation 5.3.21. §

Let $(K, | |)$ be a complete discrete valuation field. The element

\sum_{k = m}^{\infty} c_{k} π^{k} \in \hat{K}

with $c_{k} \in 𝒪$ is the limit of the corresponding sequence of partial sums.

Example 5.3.22. §

By Proposition 5.3.20, the completion a field $K (t)$ with respect to the $t$ -adic valuation on $K$ is isomorphic to the field $K ((t))$ of Laurent series in $t$ . The valuation ring of $K ((t))$ is the ring $K ⟦ t ⟧$ of power series in $K$ .

Definition 5.3.23. §

The field $ℚ_{p}$ of $p$ -adic numbers is the completion of $ℚ$ with respect to its $p$ -adic valuation. Its valuation ring $ℤ_{p}$ is the ring of $p$ -adic integers.

Remark 5.3.24. §

An arbitrary element of $ℚ_{p}$ has the unique form

\sum_{i = m}^{\infty} c_{i} p^{i},

where $m \in ℤ$ and $0 \leq c_{i} \leq p - 1$ for each $i \geq m$ , with $c_{m} \neq 0$ . It is a $p$ -adic integer (resp., unit) if and only if $m \geq 0$ (resp., $m = 0$ ).

Example 5.3.25. §

The element $α = 1 + p + p^{2} + p^{3} + \dots \in ℤ_{p}$ is $\frac{1}{1 - p}$ . To see this, note that

(1 + p + p^{2} + \dots + p^{n}) (1 - p) = 1 - p^{n + 1},

and the sequence ${(1 - p^{n + 1})}_{n}$ converges to $1$ . In particular,

- 1 = \sum_{n = 0}^{\infty} (p - 1) p^{n} \in ℤ_{p} .

Taking into account Lemma 5.3.17, the following gives an alternate description of the valuation ring of the completion of a discrete valuation field.

Proposition 5.3.26. §

Let $K$ be a complete discrete valuation field, let $𝒪$ be its valuation ring, and let $𝔪$ be the maximal ideal of $𝒪$ . Then the map

ϕ : 𝒪 \to \underset{n}{lim_{\leftarrow}} 𝒪 ∕ 𝔪^{n}

that takes $a \in 𝒪$ to the compatible sequence ${(a + 𝔪^{n})}_{n}$ is an isomorphism of rings.

Proof.

This is actually a corollary of Proposition 5.3.20, in that $a_{n} + 𝔪^{n}$ for $a_{n} \in 𝒪$ has a unique representative of the form

\sum_{k = 0}^{n - 1} c_{k} π^{k}

with $c_{i} \in T$ , where $T$ is a set of representatives of $𝒪 ∕ 𝔪$ and $π$ is a fixed uniformizer of $𝒪$ , and the $c_{i}$ are independent of $n \geq i$ . The element

a = \sum_{k = 0}^{\infty} c_{k} π^{k} \in 𝒪

is the unique element of $𝒪$ mapping to ${(a_{n} + 𝔪^{n})}_{n}$ . □

Definition 5.3.27. §

Let $A$ be a discrete valuation ring, and let $𝔭$ be its maximal ideal. We say that $A$ is complete if the canonical map

A \to \underset{n}{lim_{\leftarrow}} A ∕ 𝔭^{n} .

is an isomorphism.

The reader will verify the following.

Lemma 5.3.28. §

Let $A$ be a DVR, and let $K$ be its quotient field. Then $K$ is complete with respect to the discrete valuation induced by the valuation on $A$ if and only if $A$ is complete.

Definition 5.3.29. §

Let $K$ be a field, and let $f = \sum_{n = 0}^{\infty} a_{n} x^{n} \in K ⟦ x ⟧$ . For $k \geq 1$ , the $k$ th derivative of $f$ is the power series $f^{(k)} \in K ⟦ x ⟧$ defined by

f^{(k)} = \sum_{n = 0}^{\infty} (n + 1) \dots (n + k) a_{n + k} x^{n} .

Theorem 5.3.30 (Hensel’s Lemma). §

Let $K$ be a complete nonarchimedean valuation field with valuation ring $𝒪$ having maximal ideal $𝔪$ . Let $f \in 𝒪 [x]$ , and let $\bar{f} \in 𝒪 ∕ 𝔪 [x]$ be the image of $f$ . Suppose that $\bar{α} \in 𝒪 ∕ 𝔪$ is a simple root of $\bar{f}$ . Then there exists a unique root $α$ of $f$ in $𝒪$ that reduces to $\bar{α}$ modulo $𝔪$ .

Proof.

Let $α_{0} \in 𝒪$ be any lifting of $\bar{α}$ , and let $π = f (α_{0}) \in 𝔪$ . (If $𝒪$ is a DVR, we can instead take $π$ to be a uniformizer.) Suppose by induction that we have found $α_{k} \in 𝒪$ for $0 \leq k \leq n$ such that $α_{n} \equiv α_{k} mod π^{2^{k}}$ for all such $k$ and $f (α_{n}) \equiv 0 mod π^{2^{n}}$ . Writing $f = \sum_{i = 0}^{\deg f} a_{i} x^{i}$ with $a_{i} \in 𝒪$ , we see that

f (α_{n} + x) - (f (α_{n}) + f^{'} (α_{n}) x) = \sum_{i = 0}^{\deg f} a_{i} {(α_{n} + x)}^{i} - \sum_{i = 0}^{\deg f} a_{i} α_{n}^{i} - \sum_{i = 0}^{\deg f} i a_{i} α_{n}^{i - 1} x

is an element of $(x^{2})$ inside $𝒪 [x]$ . We therefore have that

f (α_{n} + β π^{2^{n}}) \equiv f (α_{n}) + f^{'} (α_{n}) \cdot β π^{2^{n}} mod π^{2^{n + 1}}

for any $β \in 𝒪$ . Note that $f^{'} (α_{n}) ≢ 0 mod π$ (and in fact is a unit), since $\bar{α}$ is a simple foot of $f$ in $𝒪 ∕ 𝔪$ . As $f^{'} (α_{n})$ is invertible and $π^{2^{n}}$ divides $f (α_{n})$ , we may choose $β \in 𝒪$ such that $f (α_{n} + β π^{2^{n}}) \equiv 0 mod π^{2^{n + 1}}$ , and this choice is unique modulo $π^{2^{n}}$ . We then set $α_{n + 1} = α_{n} + β π^{2^{n}}$ so that $α_{n + 1} \equiv α_{n} mod π^{2^{n}}$ , and again we have $f (α_{n + 1}) \equiv 0 mod π^{2^{n + 1}}$ . Note that $α_{n + 1}$ is unique modulo $π^{2^{n + 1}}$ with this property: in fact,

α_{n + 1} \equiv α_{n} - \frac{f (α_{n})}{f^{'} (α_{n})} mod π^{2^{n + 1}} .

(5.3.1)

Finally, letting

α = \lim_{n \to \infty} α_{n},

we note that $f$ defines a continuous function on $𝒪$ , so

f (α) = \lim_{n \to \infty} f (α_{n}) = 0,

and $α$ is by construction unique with this property among roots reducing to $\bar{α}$ . □

Example 5.3.31. §

The polynomial $f = x^{2} - 2$ has two simple roots in $𝔽_{7}$ , which are $3$ and $4$ . Hensel’s Lemma tells us that it has two roots in $ℤ_{7}$ as well. We may approximate such a root recursively using (5.3.1) in the proof of said result. For instance,

3 - \frac{3^{2} - 2}{2 \cdot 3} = 3 - \frac{1}{6} 7 \equiv 10 mod 49

is a root of $f$ modulo $49$ , and

10 - \frac{1 0^{2} - 2}{2 \cdot 10} = 10 - \frac{1}{10} 7^{2} \equiv 3 + 7 + 2 \cdot 7^{2} + 6 \cdot 7^{3} \equiv 2166 mod 7^{4}

is a root of $f$ modulo $2401$ .

The following lemma provides another nice application of Hensel’s Lemma.

Lemma 5.3.32. §

The group of roots of unity in $ℚ_{p}$ has order $p - 1$ for an odd prime $p$ and $2$ for $p = 2$ .

Proof.

The polynomial $x^{p - 1} - 1$ splits completely into distinct linear factors $𝔽_{p} [x]$ , since $𝔽_{p}^{\times}$ is cyclic of order $p - 1$ . By Hensel’s Lemma, we see that each root of $x^{p - 1} - 1$ in $𝔽_{p}$ lifts uniquely to a root of $x^{p - 1} - 1$ in $ℤ_{p}$ . That is, $μ_{p - 1} (ℚ_{p})$ contains $p - 1$ elements.

Suppose that $ζ_{n}$ is a primitive $n$ th root of unity $ζ_{n}$ in $ℚ_{p}$ (hence in $ℤ_{p}$ ) for $n \geq 1$ . Then $ζ_{n}$ reduces to a root of unity in $𝔽_{p}$ . If the order $m$ of this root of unity is less than $n$ , then $ζ_{n}^{m}$ is trivial in $𝔽_{p}$ , so $ζ_{n}^{m} - 1 \in p ℤ_{p}$ . In particular, there exists a prime $ℓ$ such that $ℤ [μ_{ℓ}] \subseteq ℤ_{p}$ and $ζ_{ℓ} - 1 \in p ℤ_{p}$ . Since $ζ_{ℓ} - 1$ divides $ℓ$ , this would imply $ℓ \in p ℤ_{p}$ , forcing $ℓ = p$ . On the other hand, if $ζ_{2 p} \in ℤ_{p}$ , then $ℤ_{p}$ contains $ℤ_{p} [μ_{2 p}]$ , and so $p^{- 1} {(ζ_{2 p} - 1)}^{φ (2 p)} \in ℤ_{p}^{\times}$ which contradicts the fact that $p$ is a uniformizer in $ℤ_{p}$ . □

The following is a strong form of Hensel’s lemma (without the uniqueness statement) that is sometimes also referred to as Hensel’s lemma.

Theorem 5.3.33 (Hensel). §

Let $K$ be a complete nonarchimedean valuation field with valuation ring $𝒪$ and maximal ideal $𝔪$ . If $f \in 𝒪 [x]$ is primitive and its image $\bar{f} \in 𝒪 ∕ 𝔪 [x]$ factors as $\bar{f} = \bar{g} \bar{h}$ , where $\bar{g}$ and $\bar{h}$ are relatively prime, then $f$ factors as $f = 𝑔h$ in $𝒪 [x]$ , where $g$ and $h$ reduce to $\bar{g}$ and $\bar{h}$ , and $\deg g = \deg \bar{g}$ .

Moreover, if $g^{'}, h^{'} \in 𝒪 [x]$ with $\deg g^{'} = \deg \bar{g}$ satisfy $f \equiv g^{'} h^{'} mod 𝔟$ for some ideal $𝔟 \subseteq 𝔪$ and reduce to $\bar{g}$ and $\bar{h}$ respectively, then $g$ and $h$ can be chosen so that $g \equiv g^{'} mod 𝔟$ and $h \equiv h^{'} mod 𝔟$ .

Proof.

Note that $f \in 𝒪 [x]$ is primitive if and only if $f ≢ 0 mod 𝔪$ . Let $k$ be the degree of $\deg \bar{g}$ , and let $d$ be the degree of $f$ . Let $g_{0}, h_{0} \in 𝒪 [x]$ be lifts of $\bar{g}$ and $\bar{h}$ , respectively, such that $\deg g_{0} = k$ and $\deg h_{0} \leq d - k$ , so

f \equiv g_{0} h_{0} mod 𝔪 .

Since $\bar{g}$ and $\bar{h}$ are relatively prime, there exist $\bar{a}, \bar{b} \in 𝒪 ∕ 𝔪 [x]$ such that $\bar{a} \bar{g} + \bar{b} \bar{h} = 1$ . Let $a, b \in 𝒪 [x]$ be lifts of $\bar{a}$ and $\bar{b}$ , respectively, so we have

a g_{0} + b h_{0} \equiv 1 mod 𝔪 .

Let $𝔞 \subseteq 𝔪$ be the ideal of $𝒪$ generated by the coefficients of $a g_{0} + b h_{0} - 1$ , which will be generated by an element $π \in 𝔞$ that can be taken as the coefficient of maximal valuation. (We use $𝔞$ in the argument below to deal with the possibility that the valuation on $K$ is not discrete.)

Suppose by induction that for $n \geq 1$ and $m \leq n - 1$ , we have found polynomials $g_{m}$ and $h_{m}$ with $\deg (g_{m} - g_{0}) < k$ and and $\deg h_{m} \leq d - k$ such that

f \equiv g_{m} h_{m} mod 𝔞^{m + 1},

for $m \leq n - 1$ and both

\begin{array}{rcl} g_{m + 1} \equiv g_{m} mod 𝔞^{m + 1} & and & h_{m + 1} \equiv h_{m} mod 𝔞^{m + 1} \end{array}

for $m \leq n - 2$ . Let

f_{n} = π^{- n} (f - g_{n - 1} h_{n - 1}) \in 𝒪 [x] .

Since $g_{0}$ is a lift of $\bar{g}$ with $\deg g_{0} = k$ , its leading coefficient is a unit. Hence, using the division algorithm, we may write

b f_{n} = q_{n} g_{0} + r_{n},

where $q_{n}, r_{n} \in 𝒪 [x]$ and $\deg r_{n} < k$ . Then

(a f_{n} + q_{n} h_{0}) g_{0} + r_{n} h_{0} = a f_{n} g_{0} + b f_{n} h_{0} \equiv f_{n} mod 𝔞 .

(5.3.2)

Let $s_{n} \in 𝒪 [x]$ be the polynomial with coefficients that agree with those coefficients of $a f_{n} + q_{n} h_{0}$ that have nonzero reduction modulo $𝔞$ and which are $0$ otherwise. Then set

\begin{array}{rcl} g_{n} = g_{n - 1} + π^{n} r_{n} & and & h_{n} = h_{n - 1} + π^{n} s_{n} . \end{array}

Note that

\begin{array}{l} g_{n} h_{n} & \equiv g_{n - 1} h_{n - 1} + π^{n} (r_{n} h_{n - 1} + s_{n} g_{n - 1}) \\ \equiv g_{n - 1} h_{n - 1} + π^{n} (r_{n} h_{0} + s_{n} g_{0}) \\ \equiv g_{n - 1} h_{n - 1} + π^{n} f_{n} \\ \equiv f mod 𝔞^{n + 1} . \end{array}

Since $\deg (g_{n - 1} - g_{0}) < k$ and $\deg r_{n} < k$ , we have $\deg (g_{n} - g_{0}) < k$ . Since $\deg (r_{n} h_{0}) < d$ and $\deg f_{n} \leq d$ , we have by (5.3.2) that the reduction of $(a f_{n} + q_{n} h_{0}) g_{0}$ modulo $𝔞$ has degree at most $d$ . Since $\deg g_{0} = k$ , we therefore have that the reduction of $a f_{n} + q_{n} h_{0}$ has degree at most $d - k$ . As the nonzero coefficients of $s_{n}$ , which is congruent to $a_{n} f_{n} + q_{n} h_{0}$ modulo $𝔞$ , are all units, we then have $\deg (s_{n}) \leq d - k$ . Hence, we have completed the induction.

Now, since the degree of $g_{n}$ is $k$ , the degree of $h_{n}$ is bounded by $d - k$ , and $⋂_{n \geq 1} 𝔞^{n} = (0)$ , it makes sense to consider the limits of these sequences of polynomials by taking the limits of their coefficients, with the resulting quantity an actual polynomial. Defining $g$ and $h$ to be the limits of the sequences ${(g_{n})}_{n}$ and ${(h_{n})}_{n}$ respectively, we obtain $f = 𝑔h$ , as desired. The last statement follows easily from the above argument. □

Remark 5.3.34. §

A valuation ring satisfying Hensel’s lemma (without the uniqueness statement) is called a Henselian ring. One may check that any Henselian ring satisfies the strong form of Hensel’s lemma as well.

5.4. Extension of valuations

In this section, we study the extension of a valuation on a (complete) field to a larger field.

Definition 5.4.1. §

Let $K$ be a field, and let $| |_{K}$ be a valuation on $K$ . If $L$ is a field extension of $K$ , then an extension $| |_{L}$ of $| |_{K}$ to $L$ is a valuation on $L$ such that $| α |_{L} = | α |_{K}$ for all $α \in K$ .

In the case of global fields, we note the following.

Remark 5.4.2. §

Let $L ∕ K$ be an extension of global fields, let $𝔭$ be a nonarchimedean prime of $K$ and $𝔓$ a prime lying above it. The normalized $𝔓$ -adic valuation is equivalent, but not always equal to, an extension of the normalized $𝔭$ -adic valuation. That is, for $α \in K$ , we have

| α |_{𝔓} = p^{- f_{𝔓} v_{𝔓} (α)} = p^{- f_{𝔓} e_{𝔓 ∕ 𝔭} v_{𝔭} (α)} = | α |_{𝔭}^{e_{𝔓 ∕ 𝔭} f_{𝔓 ∕ 𝔭}} .

In the case that the extension field is of finite degree, the following proposition restricts the possibilities for an extension of the valuation below to the extension field, up to equivalence.

Proposition 5.4.3. §

Let $(K, | |)$ be a complete valuation field, and let $(V, | |)$ be a finite-dimensional normed vector space over $K$ such that $| 𝛼𝑣 | = | α | | v |$ for all $α \in K$ and $v \in V$ . Then $V$ is complete with respect to $| |$ , and if $v_{1}, \dots, v_{n}$ is an ordered basis of $V$ , then the isomorphism $ϕ : K^{n} \to V$ with

ϕ (a_{1}, \dots, a_{n}) = \sum_{i = 1}^{n} a_{i} v_{i}

is a homemorphism.

Proof.

The topology defined on $K^{n}$ by the maximum norm

∥ (a_{1}, \dots, a_{n}) ∥ = \max (| a_{1} |, \dots, | a_{n} |)

for $a_{1}, \dots, a_{n} \in K$ agrees with the product topology. Via the map $ϕ$ , this induces a norm

∥ a_{1} v_{1} + \dots + a_{n} v_{n} ∥ = \max (| a_{1} |, \dots, | a_{n} |)

on $V$ that we must show agrees with topology defined by the original norm $| |$ on $V$ .

It suffices to show that there exists real numbers $c_{1}, c_{2} > 0$ such that

c_{1} ∥ v ∥ \leq | v | \leq c_{2} ∥ v ∥

for all $v \in V$ . Take $c_{2} = | v_{1} | + \dots + | v_{n} |$ . Then we have

| a_{1} v_{1} + \dots + a_{n} v_{n} | \leq \sum_{i = 1}^{n} | a_{i} | | v_{i} | \leq \max (| a_{1} |, \dots, | a_{n} |) \cdot \sum_{i = 1}^{n} | v_{i} | = c_{2} ∥ a_{1} v_{1} + \dots + a_{n} v_{n} ∥ .

Suppose by the induction that we have the existence of $c_{1}$ for all vector spaces of dimension less than $n$ . The case $n = 1$ is covered by taking $c_{1} = | v_{1} |$ . In general, let $W_{i}$ be the $K$ -span of ${v_{1}, \dots, v_{i - 1}, v_{i + 1}, \dots, v_{n}}$ . Then each $W_{i}$ is complete with respect to $| |$ , hence is a closed subspace of $V$ . Let $B$ be an open ball of radius $𝜖 > 0$ about $0 \in V$ such that $B \cap (v_{i} + W_{i}) = \emptyset$ for all $1 \leq i \leq n$ . Let $v = \sum_{i = 1}^{n} a_{i} v_{i} \in V$ with $v \neq 0$ . For any $1 \leq j \leq n$ with $a_{j} \neq 0$ , we have $a_{j}^{- 1} v \in v_{j} + W_{j}$ , so $| a_{j}^{- 1} v | \geq 𝜖$ . In particular, we have $| v | \geq 𝜖 ∥ (a_{1}, \dots, a_{n}) ∥ = 𝜖 ∥ v ∥$ , so we may take $c_{1} = 𝜖$ . □

We will require the following lemma.

Lemma 5.4.4. §

Let $(K, | |)$ be a complete nonarchimedean valuation field. Let

f = \sum_{i = 0}^{n} a_{i} x^{i} \in K [x]

be irreducible with $a_{n} \neq 0$ . Then either $| a_{0} |$ or $| a_{n} |$ is maximal among the values $| a_{i} |$ with $0 \leq i \leq n$ .

Proof.

By multiplying $f$ by an element of $K$ , we may assume that $f \in 𝒪 [x]$ , where $𝒪$ is the valuation ring of $K$ , and at least one coefficient of $f$ is a unit. Let $j$ be minimal such that $a_{j} \in 𝒪^{\times}$ . If $𝔪$ denotes the maximal ideal of $𝒪$ , then

f \equiv (a_{j} + a_{j + 1} x + \dots + a_{n} x^{n - j}) x^{j} mod 𝔪 .

Unless $j = 0$ or $j = n$ , this contradicts Theorem 5.3.33, since $f$ would be reducible in $𝒪 [x]$ . □

The following corollary of Lemma 5.4.4 is immediate.

Corollary 5.4.5. §

Let $(K, | |)$ be a complete nonarchimedean valuation field. Let $f$ be a monic, irreducible polynomial in $K [x]$ such that $f (0)$ lies in the valuation ring $𝒪$ of $| |$ . Then $f \in 𝒪 [x]$ .

This in turn, has the following corollary.

Corollary 5.4.6. §

Let $(K, | |)$ be a complete nonarchimedean valuation field. Let $L$ be a finite extension of $K$ . Let $𝒪$ be the valuation ring of $K$ . Then the integral closure of $𝒪$ in $L$ is equal to

{β \in L ∣ N_{L ∕ K} (β) \in 𝒪} .

Proof.

Let $n = [L : K]$ . Let $β \in L^{\times}$ , and let $f \in K [x]$ be its minimal polynomial. Lemma 1.3.14 tells us that $N_{L ∕ K} (β) \in 𝒪$ for every integral $β \in L$ . On the other hand, we have

N_{L ∕ K} (β) = {(- 1)}^{n} f {(0)}^{n ∕ d},

where $d = [K (β) : K]$ . So, if $N_{L ∕ K} (β) \in 𝒪$ , then $f (0) \in 𝒪$ , and Corollary 5.4.5 tells us that $f \in 𝒪 [x]$ , which means that $β$ is integral. □

We now prove that an extension of a valuation in an algebraic extension of a complete field exists and is unique.

Theorem 5.4.7. §

Let $(K, | |_{K})$ be a complete valuation field, and let $L$ be an algebraic extension of $K$ . Then there is a unique extension of $| |_{K}$ to a valuation $| |_{L}$ on $L$ . The valuation $| |_{L}$ is nonarchimedean if and only if $| |_{K}$ is. If $L ∕ K$ is finite, then $L$ is complete with respect to $| |_{L}$ , and this extension satisfies

| β |_{L} = | N_{L ∕ K} (β) |_{K}^{1 ∕ [L : K]} .

Proof.

If the valuation on $K$ is archimedean, then by Theorem 5.3.16, we have that $(K, | |)$ is isomorphic to $(ℝ, | |^{s})$ or $(ℂ, | |^{s})$ with $s \in (0, 1]$ , and the only extension of $| |^{s}$ on $ℝ$ to $ℂ$ is $| |^{s}$ .

So, suppose that the valuation on $K$ is nonarchimedean. First, we note that it suffices to assume that the degree of $L ∕ K$ is finite, as any algebraic extension is the union of its finite subextensions. Let $n = [L : K]$ , and for $β \in L$ , define

| β |_{L} = | N_{L ∕ K} (β) |_{K}^{1 ∕ n} .

Clearly, $| β |_{L} = 0$ if and only if $β = 0$ , and $| 𝛼𝛽 |_{L} = | α |_{L} | β |_{L}$ for $α, β \in L$ .

Let $A$ be the valuation ring of $K$ , and let $B$ be the integral closure of $A$ in $L$ . Let $α \in L$ . We obviously have $α \in B$ if and only if $α + 1 \in B$ . By Corollary 5.4.6, this tells us that $N_{L ∕ K} (α) \in A$ if and only if $N_{L ∕ K} (α + 1) \in A$ , which says that by definition that $| α |_{L} \leq 1$ if and only if $| α + 1 |_{L} \leq 1$ . If $β \in L^{\times}$ with (without loss of generality), $| α |_{L} \leq | β |_{L}$ , then $| α β^{- 1} |_{L} \leq 1$ , so

| α + β |_{L} = | β |_{L} | α β^{- 1} + 1 |_{L} \leq | β |_{L} = \max (| α |_{L}, | β |_{L}) .

Hence $| |_{L}$ is a nonarchimedean valuation, and it clearly extends $| |_{K}$ . Moreover, $L$ is complete with respect to this valuation by Proposition 5.4.3.

If $∥ ∥$ is any other valuation on $L$ extending that on $K$ , then let us let $C$ be its valuation ring and $𝔫$ be its maximal ideal. Note that the norm of any element of $C$ lies in $A$ , so $C \subseteq B$ . Suppose that $γ \in B - C$ . Let $f$ be the minimal polynomial of $γ$ over $A$ , so $f \in A [x]$ . Moreover, $γ^{- 1} \in 𝔫$ since $C$ is a valuation ring. But then $- 1 = γ^{- \deg f} f (γ) - 1$ is an $A$ -linear polynomial in $γ^{- 1}$ with no constant coefficient that therefore lies in $𝔫$ , a contradiction. That is, $B = C$ . By Proposition 5.2.5, we have that $∥ ∥$ and $| |_{L}$ are equivalent. □

We have the following immediate corollary of the definition of the extended valuation in Theorem 5.4.7.

Corollary 5.4.8. §

Let $K$ be a complete discrete valuation field, and let $L$ be an finite extension of $K$ . Then the extension to $L$ of the valuation on $K$ is discrete.

Let us consider the specific case of global fields.

Proposition 5.4.9. §

The places of a global field are exactly its finite and infinite places.

Proof.

Let $K$ be a global field. Theorem 5.3.16 tells us that any archimedean prime on $L$ must arise from a real or complex embedding of $L$ , so represents an infinite place. So, suppose $| |$ is a nonarchimedean valuation of $K$ and note that its restriction to $ℚ$ in the case that $K$ has characterstic $0$ or $𝔽_{ℓ} (t)$ in the case that $K$ has characteristic a prime $ℓ$ must be equivalent to $| |_{p}$ for some prime $p$ in the former case and to either $| |_{f}$ for some irreducible $f \in 𝔽_{ℓ} (t)$ or $| |_{\infty}$ in the latter case. So, if the latter restriction yields a finite place, coming from a finite prime $𝔭$ , consider $𝔓 = {x \in 𝒪_{K} ∣ | x | < 1}$ , and otherwise consider $𝔓 = {x \in A ∣ | x | < 1}$ , where $A$ is the integral closure of $𝔽_{ℓ} [t^{- 1}]$ in $K$ . Then $𝔓$ is a prime lying over $𝔭 = (p)$ , $(f)$ , or $(t^{- 1})$ in the respective cases. Note that the valuation $| |_{𝔭}$ extends uniquely to the completion of $ℚ$ or $𝔽_{ℓ} (t)$ at $𝔭$ by continuity and then to a valuation on $K_{𝔓}$ that is equivalent to $| |_{𝔓}$ by Theorem 5.4.7. But then the latter valuation is equivalent to $| |$ by uniqueness of the extension, as desired. □

We mention in passing the useful notion of a Newton polygon, as it relates to Lemma 5.4.4.

Definition 5.4.10. §

Let $K$ be a complete nonarchimedean valuation field with additive valuation $v$ , and let $f = \sum_{i = 0}^{n} a_{i} x^{i} \in K [x]$ with $a_{n} \neq 0$ . The Newton polygon of $f$ is the lower convex hull of the points $(i, v (a_{i}))$ .

We omit the proof of the following.

Proposition 5.4.11. §

Let $K$ be a complete nonarchimedean valuation field with additive valuation $v$ , and let $f = \sum_{i = 0}^{n} a_{i} x^{i} \in K [x]$ with $a_{n} \neq 0$ . Let $- \infty \leq m_{1} < m_{2} < \dots < m_{r}$ be the slopes of the line segments of the Newton polygon of $f$ , and let $t_{1}, \dots, t_{r}$ be their respective horizontal lengths. Then for each $j$ with $1 \leq j \leq r$ , the polynomial $f$ has exactly $t_{j}$ roots in an algebraic closure of $K$ with valuation $- m_{j}$ under the extension of $v$ .

Proof.

Let $μ_{1} < \dots < μ_{s}$ be the valuations of the roots of $f$ , and let $k_{i}$ be the number of roots of valuation $μ_{i}$ for $1 \leq i \leq s$ . For such $i$ , set $ℓ_{i} = \sum_{t = 1}^{i} k_{t}$ , and set $ℓ_{0} = 0$ .

Label the roots of $f$ with multiplicity $α_{1}, \dots, α_{n}$ in order of increasing valuation. Since $a_{n - j}$ for $0 \leq j \leq n$ is, up to sign, the sum of all products of $j$ distinct roots of $f$ , its additive valuation is at least that of $α_{1} \dots α_{j}$ . The latter valuation will be less than all other valuations of products of $j$ distinct roots if and only if $j = ℓ_{i}$ for some $1 \leq i \leq s$ . In other words, if $ℓ_{i - 1} < j \leq ℓ_{i}$ , then

v (a_{n - j}) \geq \sum_{t = 1}^{i - 1} k_{t} μ_{t} + (j - ℓ_{i - 1}) μ_{i},

with equality guaranteed if $j = ℓ_{i}$ . It then follows that the lower convex hull of the Newton polygon consists of the line segments between the points $(n - ℓ_{s - i}, \sum_{t = 1}^{s - i} k_{t} μ_{t})$ for $0 \leq i \leq s$ . It follows then that $r = s$ , and the $i$ th line segment has length

t_{i} = (n - ℓ_{s - i - 1}) - (n - ℓ_{s - i}) = ℓ_{s - i} - ℓ_{s - i - 1} = k_{s - i}

and slope

m_{i} = \frac{\sum_{t = 1}^{s - i - 1} k_{t} μ_{t} - \sum_{t = 1}^{s - i} k_{t} μ_{t}}{t_{i}} = \frac{- k_{s - i} μ_{s - i}}{k_{s - i}} = - μ_{s - i} .

□

Example 5.4.12. §

Consider the polynomial $f = 8 x^{4} + 30 x^{3} - 4 x^{2} + 7 x - 2 \in ℚ_{2} [x]$ . Its Newton polygon is the lower convex hull of the points $(0, 3)$ , $(1, 1)$ , $(2, 2)$ , $(3, 0)$ , and $(4, 1)$ , which means the area above the piecewise linear function on $[0, 4]$ consisting of the three line segments between the points $(0, 3)$ , $(1, 1)$ , $(3, 0)$ , and $(4, 1)$ . The line segments have lengths $1$ , $2$ , and $1$ and slopes $- 2$ , $- \frac{1}{2}$ , and $1$ , respectively, so $f$ has one root of $2$ -adic valuation $2$ , two roots of valuation $\frac{1}{2}$ , and one root of valuation $- 1$ .

Example 5.4.13. §

For $n \geq 1$ and a prime number $p$ , the function $x^{n} - p \in ℚ_{p} [x]$ has a Newton polygon with lower boundary the single line segment from $(0, 1)$ to $(n, 0)$ of length $n$ and slope $- \frac{1}{n}$ . Thus, $x^{n} - p$ has $n$ roots of $p$ -adic valuation $\frac{1}{n}$ in an algebraic closure of $ℚ_{p}$ , and these are of course $ζ_{n}^{i} \sqrt[n]{p}$ for $0 \leq i < n$ , where $ζ_{n}$ is a primitive $n$ th root of unity.

We provide some useful corollaries.

Corollary 5.4.14. §

In the notation of Proposition 5.4.11, the polynomial $f$ factors as $f = f_{1} \dots f_{r}$ , where $f_{i} \in K [x]$ has degree $t_{i}$ and the valuations of its roots are all $- m_{i}$ .

Proof.

By uniqueness of the extension $v$ of the valuation on $K$ to the splitting field field $L$ of $K$ , we have that $v \circ σ = v$ for any $σ \in Gal (L ∕ K)$ . Therefore, any two roots of an irreducible factor of $f$ must have the same valuation, hence the corollary. □

Corollary 5.4.15. §

Suppose that $K$ is a complete discrete valuation field with corresponding discrete additive valuation $v$ . If $f \in K [x]$ is monic of degree $n$ and has a Newton polygon with lower boundary a single line segment of slope $- \frac{c}{n}$ , where $c \geq 1$ is relatively prime to $n$ , then $f$ is irreducible.

Proof.

Let $α$ be a root of $f$ in an algebraic closure of $K$ . Since $n$ is the minimal integer such that $𝑛𝑣 (α) \in ℤ$ , we have that $α^{j} \notin K$ for $1 \leq j < n$ , and therefore $L ∕ K$ has degree $n$ . □

This gives a less-standard proof of the following well-known result.

Corollary 5.4.16. §

Let $K$ be a global field, and let $f \in K [x]$ be an Eisenstein polynomial for a nonarchimedean prime $𝔭$ of $K$ . Then $f$ is irreducible. Moreover, the prime $𝔭$ is totally ramified in the extension of $K$ generated by a root of $f$ .

Proof.

Suppose that $f$ is Eisenstein for $𝔭$ , and consider the completion $K_{𝔭}$ of $K$ at $𝔭$ . Then $f$ is still Eisenstein for the ideal generated by $𝔭$ in the valuation ring of $K_{𝔭}$ , and therefore irreducible by the previous corollary. Since $f$ is irreducible over $K_{𝔭}$ , it is irreducible in $K [x]$ . The last statement follows as every root of $f$ has valuation $\frac{1}{n}$ , as in the proof of Corollary 5.4.15. □

5.5. Local fields

Definition 5.5.1. §

A Hausdorff topological space $X$ is locally compact if for every $x \in X$ , there exists an open neighborhood $U_{x}$ of $x$ such that the closure of $U_{x}$ is compact.

Let us make the following definition.

Definition 5.5.2. §

A local field is a valuation field that is locally compact with respect to the topology defined by the valuation.

Lemma 5.5.3. §

Local fields are complete valuation fields.

Proof.

Let $(K, | |)$ be a local field. Let $𝜖 > 0$ be such that the closed ball of radius $𝜖$ around $0$ is compact, and note that by translation this applies to balls around every point. If ${(a_{n})}_{n}$ is a Cauchy sequence in $K$ , then of course there exists $N > 0$ such that $| a_{n} - a_{N} | < 𝜖$ for all $n \geq N$ . Therefore all $a_{n}$ with $n \geq N$ lie in a compact set, and the Cauchy sequence ${(a_{n})}_{n \geq N}$ has a limit. □

Remark 5.5.4. §

If $K$ is an archimedean local field, then being that it is complete, Theorem 5.3.16 tells us that $K$ is isomorphic to $ℝ$ or $ℂ$ , and the resulting valuation on $ℝ$ or $ℂ$ is equivalent to the standard absolute value.

Remark 5.5.5. §

The term “local field” is often used to refer more specifically only to nonarchimedean local fields.

The definition we have given for a local field may not be that most familiar to algebraic number theorists, so let us work to classify such fields.

Proposition 5.5.6. §

Let $K$ be a complete discrete valuation field with valuation ring $𝒪$ and maximal ideal $𝔪$ . The following are equivalent:

i.: $K$ is a local field,
ii.: $𝒪$ is compact, and
iii.: $𝒪 ∕ 𝔪$ is finite.

Proof.

Let $π$ be a unfiormizer of $𝒪$ . If $K$ is locally compact, then $𝔪^{n}$ , being an open and closed neighborhood of $0$ in $𝒪$ , must be compact for some $n \geq 0$ . On the other hand, the map $𝒪 \to 𝔪^{n}$ given by multiplication by $π^{n}$ is a homeomorphism, since it is continuous with an apparent continuous inverse. So (i) implies (ii). Conversely, (ii) implies (i) since the neighborhood $a + 𝒪$ of any $a \in K$ will be compact if $𝒪$ is.

If $𝒪$ is compact, then since $𝒪$ is the disjoint union of its open subsets $a + 𝔪$ for $a$ in a set of coset representatives of $𝒪 ∕ 𝔪$ , we have that the number of such representatives must be finite, so (ii) implies (iii). Conversely, if $𝒪 ∕ 𝔪$ is finite, then there exists a finite set $T$ of coset representatives of it in $𝒪$ . Suppose we have a sequence ${(α_{n})}_{n}$ in $𝒪$ , which we write for each $n$ as

α_{n} = \sum_{i = 0}^{\infty} a_{n, i} π^{i}

for some $a_{n, i} \in T$ for all $i \geq 0$ . Among the coefficients $a_{n, 0}$ , some element of $T$ must occur infinitely many times, so we may choose a subsequence ${(α_{k_{n, 0}})}_{n}$ of ${(α_{n})}_{n}$ such that the $a_{k_{n, 0}, 0}$ are all constant. We then repeat, choosing a subsequence ${(α_{k_{n, 1}})}_{n}$ of ${(α_{k_{n, 0}})}_{n}$ such that the $a_{k_{n, 1}, 1}$ are all constant, and so forth. Then the subsequence ${(α_{k_{n, n}})}_{n}$ of $α_{n}$ converges to

\sum_{i = 0}^{\infty} a_{k_{i, i}, i} π^{i} .

Therefore, $𝒪$ is a sequentially compact metric space, and so it is compact. □

Proposition 5.5.7 (Krasner’s Lemma). §

Let $K$ be a complete nonarchimedean valuation field. We use $| |$ denote the unique extension of the valuation on $K$ to an algebraic closure $\bar{K}$ of $K$ . Let $α, β \in \bar{K}$ . If $α$ is separable over $K (β)$ and

| β - α | < | σ (α) - α |

for every embedding $σ : K (α) ↪ \bar{K}$ fixing $K$ but not $α$ , then $K (α) \subseteq K (β)$ .

Proof.

We must show that $K (α, β) = K (β)$ . So let $σ : K (α, β) ↪ \bar{K}$ be a field embedding fixing $K (β)$ . We have

| σ (α) - β | = | σ (α) - σ (β) | = | α - β |,

the latter equality by the uniqueness of the extension, so

| σ (α) - α | = | σ (α) - β + β - α | \leq \max (| σ (α) - β |, | β - α |) = | β - α | .

By assumption, this forces $σ$ to fix $K (α)$ , hence the result. □

We can derive the following from Krasner’s lemma.

Proposition 5.5.8. §

Let $K$ be a complete nonarchimedean valuation field with valuation ring $𝒪$ . Let $f \in 𝒪 [x]$ be monic, irreducible, and separable of degree $n \geq 1$ . There exists an ideal $𝔞$ of $𝒪$ such that if $g \in 𝒪 [x]$ is monic of $\deg g = n$ and satisfies $f \equiv g mod 𝔞 𝒪 [x]$ , and if $β$ is a root of $g$ in an algebraic closure $\bar{K}$ of $K$ , then $f$ has a root $α$ in $\bar{K}$ such that $K (α) = K (β)$ . In particular, any such $g$ is irreducible.

Proof.

Write $f = \sum_{i = 0}^{n} a_{i} x^{i}$ and $g = \sum_{i = 0}^{n} b_{i} x^{i}$ . Our assumption is that for some positive $δ < 1$ , we have $| a_{i} - b_{i} | \leq δ$ for all $0 \leq i \leq n$ , where $| |$ is the valuation on $K$ (and its unique extension to $\bar{K}$ ). By choosing $δ$ small enough, we may insure that either $| a_{i} | = | b_{i} |$ or $| b_{i} | \leq δ$ (if $a_{i} = 0$ or $b_{i} = 0$ ) for each $i$ . So, there exists $C > 0$ with $| b_{i} | \leq C$ independent of the choice of $g$ . If $β$ is a root of $g$ , then

| β |^{n} \leq \max {| b_{i} | | β |^{i} ∣ 0 \leq i < n} \leq C \max {1, | β |^{n - 1}},

so $| β |$ is bounded independent of $g$ , say by $D$ , which we take to be $\geq 1$ . We then have

| f (β) | = | f (β) - g (β) | \leq \max {| a_{i} - b_{i} | | β |^{i} ∣ 0 \leq i < n} \leq δ \max {1, | β |}^{n - 1} \leq δ D^{n - 1},

and so by choosing $δ$ sufficiently small, we may make $| f (β) |$ arbitrarily small, independent of $β$ , say less than $𝜖^{n}$ for some $𝜖 > 0$ . Note that

| f (β) | = \prod_{i = 1}^{n} | β - α_{i} | < 𝜖^{n},

where $α_{1}, \dots, α_{n}$ are the roots of $f$ . One must then have $| β - α_{i} | < 𝜖$ for some $i$ . If we take $δ$ , and hence $𝜖$ , small enough so that $𝜖 < | α_{i} - α_{j} |$ for all $j \neq i$ , and Krasner’s lemma tells us that $K (α_{i}) \subseteq K (β)$ , which tells us that $g$ is irreducible and $K (α_{i}) = K (β)$ . □

Theorem 5.5.9. §

The following are equivalent for a nonarchimedean valuation field $K$ :

i.: $K$ is a local field,
ii.: $K$ is complete, the valuation on $K$ is discrete, and the residue field of $K$ is finite,
iii.: $K$ is isomorphic to a completion of a global field, and
iv.: $K$ is isomorphic to a finite extension of $ℚ_{p}$ or $𝔽_{p} ((t))$ for some prime $p$ .

Proof.

That (ii) implies (i) is part of Proposition 5.5.6. That (iii) implies (ii) is a consequence of Proposition 5.3.12 and Lemma 5.3.17.

Suppose that (iv) holds. Suppose that $K$ is a finite extension of $ℚ_{p}$ for some prime $p$ . Then $K = ℚ_{p} (α)$ for some $α \in K$ , and let $f$ be its minimal polynomial. Choose $g \in ℚ [x]$ monic of degree that of $f$ and sufficiently close to $f$ so that we may apply Proposition 5.5.8 to see that $g$ is irreducible over $ℚ_{p}$ , so $ℚ_{p} (β) = ℚ_{p} (α)$ for some root $β$ of $g$ . Since $β$ is algebraic over $ℚ$ , we have that $K = ℚ_{p} (β)$ is the completion of $ℚ (β)$ in $K$ . Similarly, if $K$ is a finite extension of $𝔽_{p} ((t))$ , then it is a finite, separable extension of $𝔽_{p} ((t^{1 ∕ p^{k}}))$ for some $k \geq 1$ , which is itself isomorphic to $𝔽_{p} ((t))$ . Hence, we may assume that $K = 𝔽_{p} ((t)) (α)$ for some $α \in K$ , and the above argument with $𝔽_{p} ((t))$ replacing $ℚ_{p}$ yields (iii).

To see that (i) implies (iv), suppose that $K$ is a local field with residue field of characteristic a prime $p$ . If $K$ has characteristic $0$ , then the restriction of the valuation to $ℚ$ cannot be trivial as it would otherwise extend to the trivial valuation on $K$ by Theorem 6.1.4. It must therefore be a nonarchimedean valuation on $ℚ$ with residue characteristic $p$ , and Theorem 5.2.31 tells us that this valuation is equivalent to the $p$ -adic valuation. But then the completion $ℚ_{p}$ embeds canonically into $K$ , so $K$ is an extension of $ℚ_{p}$ . If $K$ has characteristic $p$ , then it cannot be an algebraic extension of $𝔽_{p}$ since the valuation is nontrivial, so it must contain an element $T$ that is transcendental over $𝔽_{p}$ . We have that $K$ is an extension of $𝔽_{p} (T)$ , and by Proposition 5.2.32, the restriction of the valuation on $K$ to $𝔽_{p} (T)$ is the $f$ -adic valuation for some irreducible $f \in 𝔽_{p} [T]$ or the $\infty$ -adic valuation. The completion of $𝔽_{p} (T)$ with respect to this valuation is isomorphic to $𝔽_{q} ((t))$ for some $q$ and embeds in $K$ , and the valuation on $K$ is the unique extension of this valuation to $K$ .

Next, suppose that $K ∕ F$ with $F = ℚ_{p}$ or $𝔽_{p} ((t))$ were an infinite extension. If $K$ contains a transcendental element $x$ over $F$ , then since the residue field of $K$ is finite, $x$ is still transcendental over the largest extension $E$ of $F$ in $K$ in which the valuation of $F$ is unramified. By Theorem 2.5.11, the field extensions $E (x) ∕ E (x^{n})$ all have ramification index $n$ at the unique prime of the valuation ring of $E (x)$ . Let $h < 1$ be the valuation of a uniformizer of $E (x) \subseteq K$ under the unique extension of the valuation on $F$ to $E (x)$ . Then the valuation of a uniformizer of $E (x^{n})$ is $h^{n}$ . Since the valuation of a uniformizer of $E$ is then less than $h^{n}$ for all $n$ , it must be $0$ , which is impossible (in fact, it is $p^{- 1}$ ).

If $K ∕ F$ is algebraic, we can let ${(K_{n})}_{n}$ be an infinite tower of distinct subfields of $K$ with union equal to $K$ . As $K$ is a local field, its residue field is finite by Proposition 5.5.6. Therefore, the extension of residue fields for $K_{n + 1} ∕ K_{n}$ is trivial for sufficiently large $n$ . Since there is only one nonzero valuation on $K_{n + 1}$ extending that of $K_{n}$ , the degree formula then tells us that the ramification degree of the prime of the valuation ring is $[K_{n + 1} : K_{n}]$ , and in particular nontrivial. Consider any sequence ${(π_{n})}_{n}$ , with $π_{n} \in K_{n}$ a uniformizer for each $n$ . If $| |$ is the valuation on $K$ , then we have that $| π_{n} - π_{m} | = | π_{n} |$ for $n > m$ , with $n$ sufficiently large (independent of the choice of $m$ ). But $| π_{n} |$ has a limit of $1$ as $n$ increases (as follows from Theorem 5.4.7), which means that the sequence ${(π_{n})}_{n}$ has no convergent subsequence. Therefore $K$ is not compact, and therefore the extension had to be finite. □

Definition 5.5.10. §

A $p$ -adic field, or $p$ -adic local field, is a finite extension of $ℚ_{p}$ for some prime $p$ .

Definition 5.5.11. §

A Laurent series field (over a finite field) is a finite extension of $𝔽_{p} ((t))$ .

Remark 5.5.12. §

In fact, every finite extension $K$ of $𝔽_{p} ((t))$ is isomorphic to $𝔽_{q} ((y))$ for some power $q$ of $p$ under a map that takes a uniformizer of $K$ to $y$ .