There are some notes on valuations from the first lecture of Math 223a at Harvard.
Let be a field.
The third property is the interesting one. Note in particular it can be rewritten as .
Note that we can recover immediately.
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:
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:
Proof: First, observe that we can get
Applying this inductively, we obtain
From this, one can obtain
Letting completes the proof.
Next, we prove that
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:
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:
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
Thus, the real valuation (absolute value) isn’t discrete, while the -adic one is.
4. Non-Archimedian Valuations
Thus the real valuation is Archimedian while the -adic valuation is non-Archimedian.
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.
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
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.
We say that a field is complete with respect to a valuation if it is complete in the topological sense.
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.
Proof: We saw non-Archimedian for every .
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
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
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 .