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

or zero-padding,

\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}


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}


\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

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