# Some Notes on Valuations

There are some notes on valuations from the first lecture of Math 223a at Harvard.

## 1. Valuations

Let ${k}$ be a field.

Definition 1

A valuation

$\displaystyle \left\lvert - \right\rvert : k \rightarrow \mathbb R_{\ge 0}$

is a function obeying the axioms

• ${\left\lvert \alpha \right\rvert = 0 \iff \alpha = 0}$.
• ${\left\lvert \alpha\beta \right\rvert = \left\lvert \alpha \right\rvert \left\lvert \beta \right\rvert}$.
• Most importantly: there should exist a real constant ${C}$, such that ${\left\lvert 1+\alpha \right\rvert < C}$ whenever ${\left\lvert \alpha \right\rvert \le 1}$.

The third property is the interesting one. Note in particular it can be rewritten as ${\left\lvert a+b \right\rvert < C\max\{ \left\lvert a \right\rvert, \left\lvert b \right\rvert \}}$.

Note that we can recover ${\left\lvert 1 \right\rvert = \left\lvert 1 \right\rvert \left\lvert 1 \right\rvert \implies \left\lvert 1 \right\rvert = 1}$ immediately.

Example 2 (Examples of Valuations)

If ${k = \mathbb Q}$, we can take the standard absolute value. (Take ${C=2}$.)

Similarly, the usual ${p}$-adic evaluation, ${\nu_p}$, which sends ${p^a t}$ to ${p^{-a}}$. Here ${C = 1}$ is a valid constant.

These are the two examples one should always keep in mind: with number fields, all valuations look like one of these too. In fact, over ${\mathbb Q}$ it turns out that every valuation “is” one of these two valuations (for a suitable definition of equality). To make this precise:

Definition 3

We say ${\left\lvert - \right\rvert_1 \sim \left\lvert - \right\rvert_2}$ (i.e. two valuations on a field ${k}$ are equivalent) if there exists a constant ${k > 0}$ so that ${\left\lvert \alpha \right\rvert_1 = \left\lvert \alpha \right\rvert_2^k}$ for every ${\alpha \in k}$.

In particular, for any valuation we can force ${C = 2}$ to hold by taking an equivalent valuation to a sufficient power.

In that case, we obtain the following:

Lemma 4

In a valuation with ${C = 2}$, the triangle inequality holds.

Proof: First, observe that we can get

$\displaystyle \left\lvert \alpha + \beta \right\rvert \le 2 \max \left\{ \left\lvert \alpha \right\rvert, \left\lvert \beta \right\rvert \right\}.$

Applying this inductively, we obtain

$\displaystyle \left\lvert \sum_{i=1}^{2^r} a_i \right\rvert \le 2^r \max_i \left\lvert a_i \right\rvert$

$\displaystyle \sum_{i=1}^{n} a_i \le 2n\max_i \left\lvert a_i \right\rvert.$

From this, one can obtain

$\displaystyle \left\lvert \alpha+\beta \right\rvert^n \le \left\lvert \sum_{j=0}^n \binom nj \alpha^j \beta^{n-j} \right\rvert \le 2(n+1) \sum_{j=0}^n \left\lvert \binom nj \right\rvert \left\lvert \alpha \right\rvert^j \left\lvert \beta \right\rvert^{n-j} \le 4(n+1)\left( \left\lvert \alpha \right\rvert+\left\lvert \beta \right\rvert \right)^n.$

Letting ${n \rightarrow \infty}$ completes the proof. $\Box$

Next, we prove that

Lemma 5

If ${\omega^n=1}$ for some ${n}$, then${\left\lvert \omega \right\rvert = 1}$. In particular, on any finite field the only valuation is the trivial one which sends ${0}$ to ${0}$ and all elements to ${1}$.

Proof: Immediate, since ${\left\lvert \omega \right\rvert^n = 1}$. $\Box$

## 2. Topological field induced by valuations

Let ${k}$ be a field. Given a valuation on it, we can define a basis of open sets

$\displaystyle \left\{ \alpha \mid \left\lvert \alpha - a \right\rvert < d \right\}$

across all ${a \in K}$, ${d \in \mathbb R_{> 0}}$. One can check that the same valuation gives rise to the same topological spaces, so it is fine to assume ${C = 2}$ as discussed earlier; thus, in fact we can make ${k}$ into a metric space, with the valuation as the metric.

In what follows, we’ll always assume our valuation satisfies the triangle inequality. Then:

Lemma 6

Let ${k}$ be a field with a valuation. Viewing ${k}$ as a metric space, it is in fact a topological field, meaning addition and multiplication are continuous.

Proof: Trivial; let’s just check that multiplication is continuous. Observe that

\displaystyle \begin{aligned} \left\lvert (a+\varepsilon_1)(b+\varepsilon_2) - ab \right\rvert & \le \left\lvert \varepsilon_1\varepsilon_2 \right\rvert + \left\lvert a\varepsilon_2 \right\rvert + \left\lvert b\varepsilon_1 \right\rvert \\ &\rightarrow 0. \end{aligned}

$\Box$

Now, earlier we saw that two valuations which are equivalent induce the same topology. We now prove the following converse:

Proposition 7

If two valuations ${\left\lvert - \right\rvert_1}$ and ${\left\lvert - \right\rvert_2}$ give the same topology, then they are in fact equivalent.

Proof: Again, we may safely assume that both satisfy the triangle inequality. Next, observe that ${\left\lvert a \right\rvert < 1 \iff a^n \rightarrow 0}$ (according to the metric) and by taking reciprocals, ${\left\lvert a \right\rvert > 1 \iff a^{-n} \rightarrow 0}$.

Thus, given any ${\beta}$, ${\gamma}$ and integers ${m}$, ${n}$ we derive that

$\displaystyle \left\lvert \beta^n\gamma^m \right\rvert_1 < 1 \iff \left\lvert \beta^n\gamma^m \right\rvert < 1$

with similar statements holding with “${<}$” replaced by “${=}$”, “${>}$”. Taking logs, we derive that

$\displaystyle n \log\left\lvert \beta \right\rvert_1 + m \log \left\lvert \gamma \right\rvert_1 < 0 \iff n \log\left\lvert \beta \right\rvert_2 + m \log \left\lvert \gamma \right\rvert_1 < 0$

and the analogous statements for “${=}$”, “${>}$”. Now just choose an appropriate sequence of ${m}$, ${n}$ and we can deduce that

$\displaystyle \frac{\log \left\lvert \beta_1 \right\rvert}{\log \left\lvert \beta_2 \right\rvert} = \frac{\log \left\lvert \gamma_1 \right\rvert}{\log \left\lvert \gamma_2 \right\rvert}$

so it equals a fixed constant ${c}$ as desired. $\Box$

## 3. Discrete Valuations

Definition 8

We say a valuation ${\left\lvert - \right\rvert}$ is discrete if its image around ${1}$ is discrete, meaning that if ${\left\lvert a \right\rvert \in [1-\delta,1+\delta] \implies \left\lvert a \right\rvert = 1}$ for some real ${\delta}$. This is equivalent to requiring that ${\{\log\left\lvert a \right\rvert\}}$ is a discrete subgroup of the real numbers.

Thus, the real valuation (absolute value) isn’t discrete, while the ${p}$-adic one is.

## 4. Non-Archimedian Valuations

Most importantly:

Definition 9

A valuation ${\left\lvert - \right\rvert}$ is non-Archimedian if we can take ${C = 1}$ in our requirement that ${\left\lvert a \right\rvert \le 1 \implies \left\lvert 1+a \right\rvert \le C}$. Otherwise we say the valuation is Archimedian.

Thus the real valuation is Archimedian while the ${p}$-adic valuation is non-Archimedian.

Lemma 10

Given a non-Archimedian valuation ${\left\lvert - \right\rvert}$, we have ${\left\lvert b \right\rvert < \left\lvert a \right\rvert \implies \left\lvert a+b \right\rvert = \left\lvert a \right\rvert}$.

Proof: We have that

$\displaystyle \left\lvert a \right\rvert = \left\lvert (a+b)-b \right\rvert \le \max\left\{ \left\lvert a+b \right\rvert, \left\lvert b \right\rvert \right\}.$

On the other hand, ${\left\lvert a+b \right\rvert \le \max \{ \left\lvert a \right\rvert, \left\lvert b \right\rvert\}}$. $\Box$

Given a field ${k}$ and a non-Archimedian valuation on it, we can now consider the set

$\displaystyle \mathcal O = \left\{ a \in k \mid \left\lvert a \right\rvert \le 1 \right\}$

and by the previous lemma, this turns out to be a ring. (This is the point we use the fact that the valuation is non-Archimedian; without that ${\mathcal O}$ need not be closed under addition). Next, we define

$\displaystyle \mathcal P = \left\{ a \in k \mid \left\lvert a \right\rvert < 1 \right\} \subset \mathcal O$

which is an ideal. In fact it is maximal, because ${\mathcal O/\mathcal P}$ is the set of units in ${\mathcal O}$, and is thus necessarily a field.

Lemma 11

Two valuations are equal if they give the same ring ${\mathcal O}$ (as sets, not just up to isomorphism).

Proof: If the valuations are equivalent it’s trivial.

For the interesting converse direction (they have the same ring), the datum of the ring ${\mathcal O}$ lets us detect whether ${\left\lvert a \right\rvert < \left\lvert b \right\rvert}$ by simply checking whether ${\left\lvert ab^{-1} \right\rvert < 1}$. Hence same topology, hence same valuation. $\Box$

We will really only work with valuations which are obviously discrete. On the other hand, to detect non-Archimedian valuations, we have

Lemma 12

${\left\lvert - \right\rvert}$ is Archimedian if ${\left\lvert n \right\rvert \le 1}$ for every ${n = 1 + \dots + 1 \in k}$.

Proof: Clearly Archimedian ${\implies}$ ${\left\lvert n \right\rvert \le 1}$. The converse direction is more interesting; the proof is similar to the analytic trick we used earlier. Given ${\left\lvert a \right\rvert \le 1}$, we wish to prove ${\left\lvert 1+a \right\rvert \le 1}$. To do this, first assume the triangle inequality as usual, then

$\displaystyle \left\lvert 1+a \right\rvert^n < \sum_j \left\lvert \binom nj \right\rvert\left\lvert a \right\rvert^j \le \sum_{j=0}^n \left\lvert a \right\rvert^j \le \sum_{j=0}^n 1 = n+1.$

Finally, let ${n \rightarrow \infty}$ again. $\Box$
In particular, any field of finite characteristic in fact has ${\left\lvert n \right\rvert = 1}$ and thus all valuations are non-Archimedian.

## 5. Completions

We say that a field ${k}$ is complete with respect to a valuation ${\left\lvert - \right\rvert}$ if it is complete in the topological sense.

Theorem 13

Every field ${k}$ is with a valuation ${\left\lvert - \right\rvert}$ can be embedded into a complete field ${\overline{k}}$ in a way which respects the valuation.

For example, the completion of ${\mathbb Q}$ with the Euclidean valuation is ${\mathbb R}$. Proof: Define ${\overline{k}}$ to be the topological completion of ${k}$; then extend by continuity; $\Box$
Given ${k}$ and its completion ${\overline{k}}$ we use the same notation for the valuations of both.

Proposition 14

A valuation ${\left\lvert - \right\rvert}$ on ${\overline{k}}$ is non-Archimedian if and only if the valuation is non-Archimedian on ${k}$.

Proof: We saw non-Archimedian ${\iff}$ ${\left\lvert n \right\rvert \le 1}$ for every ${n = 1 + \dots + 1}$. $\Box$

Proposition 15

Assume ${\left\lvert - \right\rvert}$ is non-Archimedian on ${k}$ and hence ${\overline{k}}$. Then the set of values achieved by ${\left\lvert - \right\rvert}$ coincides for ${k}$ and ${\overline{k}}$, i.e. ${\{ \left\lvert k \right\rvert \} = \{ \left\lvert \overline{k} \right\rvert \}}$.

Not true for Archimedian valuations; consider ${\left\lvert \sqrt2 \right\rvert = \sqrt2 \notin \mathbb Q}$. Proof: Assume ${0 \neq b \in \overline{k}}$; then there is an ${a \in k}$ such that ${\left\lvert b-a \right\rvert < \left\lvert b \right\rvert}$ since ${k}$ is dense in ${\overline{k}}$. Then, ${\left\lvert b \right\rvert \le \max \{ \left\lvert b-a \right\rvert, \left\lvert a \right\rvert \}}$ which implies ${\left\lvert b \right\rvert = \left\lvert a \right\rvert}$. $\Box$

## 6. Weak Approximation Theorem

Proposition 16 (Weak Approximation Theorem)

Let ${\left\lvert-\right\rvert_i}$ be distinct nontrivial valuations of ${k}$ for ${i=1,\dots,n}$. Let ${k_i}$ denote the completion of ${k}$ with respect to ${\left\lvert-\right\rvert_i}$. Then the image

$\displaystyle k \hookrightarrow \prod_{i=1}^n k_i$

is dense.

This means that distinct valuations are as different as possible; for example, if ${\left\lvert-\right\rvert _1 = \left\lvert-\right\rvert _2}$ then we might get, say, a diagonal in ${\mathbb R \times \mathbb R}$ which is as far from dense as one can imagine. Another way to think of this is that this is an analogue of the Chinese Remainder Theorem.

Proof: We claim it suffices to exhibit ${\theta_i \in k}$ such that

$\displaystyle \left\lvert \theta_i \right\rvert_j \begin{cases} > 1 & i = j \\ < 1 & \text{otherwise}. \end{cases}$

Then

$\displaystyle \frac{\theta_i^r}{1+\theta_i^r} \rightarrow \begin{cases} 1 & \text{ in } \left\lvert-\right\rvert_i \\ 0 & \text{ otherwise}. \end{cases}$

Hence for any point ${(a_1, \dots, a_n)}$ we can take the image of ${\sum \frac{\theta_i^r}{1+\theta_i^r} a_i \in k}$. So it would follow that the image is dense.

Now, to construct the ${\theta_i}$ we proceed inductively. We first prove the result for ${n=2}$. Since the topologies are different, we exhibit ${\alpha}$, ${\beta}$ such that ${\left\lvert \alpha_1 \right\rvert < \left\lvert \alpha_2 \right\rvert}$ and ${\left\lvert \beta_1 \right\rvert > \left\lvert \beta_2 \right\rvert}$, and pick ${\theta=\alpha\beta^{-1}}$.

Now assume ${n \ge 3}$; it suffices to construct ${\theta_1}$. By induction, there is a ${\gamma}$ such that

$\displaystyle \left\lvert \gamma \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \gamma \right\rvert_i < 1 \text{ for } i = 2, \dots, n-1.$

Also, there is a ${\psi}$ such that

$\displaystyle \left\lvert \delta \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \delta \right\rvert_n < 1.$

Now we can pick

$\displaystyle \theta_1 = \begin{cases} \gamma & \left\lvert \gamma \right\rvert_n < 1 \\ \phi^r\gamma & \left\lvert \gamma \right\rvert_n = 1 \\ \frac{\gamma^r}{1+\gamma^r} & \left\lvert \gamma \right\rvert_n > 1 \\ \end{cases}$

for sufficiently large ${r}$. $\Box$