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

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

Applying this inductively, we obtain $$\left\lvert \sum_{i=1}^{2^r} a_i \right\rvert \le 2^r \max_i \left\lvert a_i \right\rvert$$ or zero-padding, $$\sum_{i=1}^{n} a_i \le 2n\max_i \left\lvert a_i \right\rvert.$$ From this, one can obtain

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

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

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

$\Box$

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 $$\left\lvert \beta^n\gamma^m \right\rvert_1 < 1 \iff \left\lvert \beta^n\gamma^m \right\rvert < 1$$ with similar statements holding with “$<$” replaced by “$=$”, “$>$”. Taking logs, we derive that

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

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

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

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-Archimedean Valuations

Most importantly:

Definition 9. A valuation $\left\lvert - \right\rvert$ is non-Archimedean 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 Archimedean.

Thus the real valuation is Archimedean while the $p$-adic valuation is non-Archimedean.

Lemma 10. Given a non-Archimedean 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

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

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-Archimedean valuation on it, we can now consider the set $$\mathcal O = \left\{ a \in k \mid \left\lvert a \right\rvert \le 1 \right\}$$ 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-Archimedean; without that $\mathcal O$ need not be closed under addition). Next, we define $$\mathcal P = \left\{ a \in k \mid \left\lvert a \right\rvert < 1 \right\} \subset \mathcal O$$ 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-Archimedean valuations, we have

Lemma 12. $\left\lvert - \right\rvert$ is Archimedean if $\left\lvert n \right\rvert \le 1$ for every $n = 1 + \dots + 1 \in k$.

Proof: Clearly Archimedean $\implies$ $\left\lvert n \right\rvert \le 1$. 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

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

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-Archimedean.

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-Archimedean if and only if the valuation is non-Archimedean on $k$.

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

Proposition 15. Assume $\left\lvert - \right\rvert$ is non-Archimedean 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 Archimedean 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 $$k \hookrightarrow \prod_{i=1}^n k_i$$ 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 $$\left\lvert \theta_i \right\rvert_j \begin{cases} > 1 & i = j \\ < 1 & \text{otherwise}. \end{cases}$$ Then

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

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

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

Also, there is a $\psi$ such that $$\left\lvert \delta \right\rvert_1 > 1 \quad\text{and}\quad \left\lvert \delta \right\rvert_n < 1.$$ Now we can pick

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

for sufficiently large $r$. $\Box$