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

## 1. Valuations

Let be a field.

**Definition 1**

A **valuation**

is a function obeying the axioms

- .
- .
- Most importantly: there should exist a real constant , such that whenever .

The third property is the interesting one. Note in particular it can be rewritten as .

Note that we can recover immediately.

**Example 2** **(Examples of Valuations)**

If , we can take the standard absolute value. (Take .)

Similarly, the usual -adic evaluation, , which sends to . Here 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 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 (i.e. two valuations on a field are equivalent) if there exists a constant so that for every .

In particular, for any valuation we can force to hold by taking an equivalent valuation to a sufficient power.

In that case, we obtain the following:

**Lemma 4**

In a valuation with , the triangle inequality holds.

*Proof:* First, observe that we can get

Applying this inductively, we obtain

or zero-padding,

From this, one can obtain

Letting completes the proof.

Next, we prove that

**Lemma 5**

If for some , then. In particular, on any finite field the only valuation is the trivial one which sends to and all elements to .

*Proof:* Immediate, since .

## 2. Topological field induced by valuations

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

across all , . One can check that the same valuation gives rise to the same topological spaces, so it is fine to assume as discussed earlier; thus, in fact we can make 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 be a field with a valuation. Viewing 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

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 and 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 (according to the metric) and by taking reciprocals, .

Thus, given any , and integers , we derive that

with similar statements holding with “” replaced by “”, “”. Taking logs, we derive that

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

so it equals a fixed constant as desired.

## 3. Discrete Valuations

**Definition 8**

We say a valuation is **discrete** if its image around is discrete, meaning that if for some real . This is equivalent to requiring that is a discrete subgroup of the real numbers.

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

## 4. Non-Archimedian Valuations

Most importantly:

**Definition 9**

A valuation is **non-Archimedian** if we can take in our requirement that . Otherwise we say the valuation is **Archimedian**.

Thus the real valuation is Archimedian while the -adic valuation is non-Archimedian.

**Lemma 10**

Given a non-Archimedian valuation , we have .

*Proof:* We have that

On the other hand, .

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

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 need not be closed under addition). Next, we define

which is an ideal. In fact it is maximal, because is the set of units in , and is thus necessarily a field.

**Lemma 11**

Two valuations are equal if they give the same ring (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 lets us detect whether by simply checking whether . Hence same topology, hence same valuation.

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

**Lemma 12**

is Archimedian if for every .

*Proof:* Clearly Archimedian . The converse direction is more interesting; the proof is similar to the analytic trick we used earlier. Given , we wish to prove . To do this, first assume the triangle inequality as usual, then

Finally, let again.

In particular, any field of finite characteristic in fact has and thus all valuations are non-Archimedian.

## 5. Completions

We say that a field is **complete** with respect to a valuation if it is complete in the topological sense.

**Theorem 13**

Every field is with a valuation can be embedded into a complete field in a way which respects the valuation.

For example, the completion of with the Euclidean valuation is . *Proof:* Define to be the topological completion of ; then extend by continuity;

Given and its completion we use the same notation for the valuations of both.

**Proposition 14**

A valuation on is non-Archimedian if and only if the valuation is non-Archimedian on .

*Proof:* We saw non-Archimedian for every .

**Proposition 15**

Assume is non-Archimedian on and hence . Then the set of values achieved by coincides for and , i.e. .

Not true for Archimedian valuations; consider . *Proof:* Assume ; then there is an such that since is dense in . Then, which implies .

## 6. Weak Approximation Theorem

**Proposition 16** **(Weak Approximation Theorem)**

Let be distinct nontrivial valuations of for . Let denote the completion of with respect to . Then the image

is dense.

This means that distinct valuations are as different as possible; for example, if then we might get, say, a diagonal in 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 such that

Then

Hence for any point we can take the image of . So it would follow that the image is dense.

Now, to construct the we proceed inductively. We first prove the result for . Since the topologies are different, we exhibit , such that and , and pick .

Now assume ; it suffices to construct . By induction, there is a such that

Also, there is a such that

Now we can pick

for sufficiently large .

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I could not refrain from commenting. Exceptionally well written!

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