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DVRs II *August 30, 2009*

*Posted by Akhil Mathew in algebra, algebraic number theory, commutative algebra, number theory.*

Tags: discrete valuation rings, Noetherian rings, PIDs, prime ideals, UFDs

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Tags: discrete valuation rings, Noetherian rings, PIDs, prime ideals, UFDs

trackback

Earlier I went over the definition and first properties of a discrete valuation ring. Today, it’s time to say how we can tell a ring is a DVR–it turns out to be not too bad, which is nice because the properties we need in this criterion are often easier to work with than the existence of some discrete valuation.

Today’s result is:

Theorem 1If the domain is Noetherian, integrally closed, and has a unique nonzero prime ideal , then is a DVR. Conversely, any DVR has those properties.

This is equivalent to the fact that Dedekind domains have unique factorization, as we’ll see eventually.

First, let’s do the last part. We already know the part about DVRs being PIDs, which are Noetherian; integrally closedness is true for any UFD (and thus for any PID). The converse is much harder.

We will show that is principal, by showing it is *invertible* (as will be seen below). We divide the proof into steps:

**Step One **

For an ideal , let , where is the quotient field of . Then clearly and is an -module, but it isn’t clear that .

Nevertheless, I claim that .

The proof runs across a familiar line—show that any maximal element in the set of with is prime. Then, since there is a maximal element (by Noetherianness) containing any for (in which case ), that maximal element must be , which proves our claim.

So to fill in the missing link, we must prove:

Lemma 2If is a Noetherian domain, any maximal element in the set of ideals with is prime.

Let be a maximal element, and suppose , with . I claim that if , then . The part follows since is a -module.

By symmetry it’s enough to prove the other half for ; but then if , we’d have , which implies , contradiction.

Then it follows that , by applying the claim I just made twice. But , contradiction.

**Step Two **

I now claim . Well, we know that by definition of inverses. and too. So is an ideal sandwiched between and . Thus we only need to prove that is impossible. To do this, choose some which must satisfy . Now by the next lemma, is integral over .

Lemma 3If is a finitely generated -module satisfying for some , then is integral over .

This follows by the Cayley-Hamilton theorem, and is a standard criterion for integrality. (It can even be generalized to faithful finitely generated -modules.)

So returning back above, is integral over , a contradiction by integrally closedness.

**Step Three **

Now I claim is principal, which by Proposition 6 two days ago proves the claim. Indeed, since , write

At least one is invertible, since is local. Then I claim . Indeed, if , then

So we’re done. Taking stock, we have an effective way to say whether a ring is a DVR. Next up, we’ll discuss the application to Dedekind domains and factorization of ideals. Then, completions for fields, which will be a reprise of what we’ve already said for rings so will go quickly and extensions of valuations. Then, ramification. But I should not try to plan too far ahead, since I usually end up changing my mind anyway.

Oh, and I should do a bit of citation- I’m following Serre’s *Local Fields* and Cassels-Frohlich’s *Algebraic Number Theory* here as primary sources, but I’ll probably use Lang’s *Algebraic Number Theory* as well as some commutative algebra books as I continue into more heavily number-theoretic material.

[…] DVR is a Dedekind domain, and the localization of a Dedekind domain at a nonzero prime is a DVR by this. Another example (Serre) is to take a nonsingular affine variety of dimension 1 and consider the […]

Can you explain exactly how “two applications” of the claim about the non-prime ideal J leads to z(ab)=(za)b not being in S? I see that za is in J^-1 and not in S, and b is not in J… but we don’t yet know (J^{-1})^{-1}=J, so I don’t see how to conclude that the product zab is not in S.

Thanks

We have the mini-lemma: if and , then because otherwise .

This means . Replacing by and by , it follows that , contradiction.