## DVRs II August 30, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, commutative algebra, number theory.
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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 1 If the domain ${R}$ is Noetherian, integrally closed, and has a unique nonzero prime ideal ${\mathfrak{m}}$, then ${R}$ 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 ${\mathfrak{m}}$ is principal, by showing it is invertible (as will be seen below). We divide the proof into steps:

Step One

For an ideal ${I \subset R}$, let ${I^{-1} := \{ x \in F: xI \subset R \}}$, where ${F}$ is the quotient field of ${R}$. Then clearly ${I^{-1} \supset R}$ and ${I^{-1}}$ is an ${R}$-module, but it isn’t clear that ${I^{-1} \neq R}$.

Nevertheless, I claim that ${\mathfrak{m}^{-1} \neq R}$.

The proof runs across a familiar line—show that any maximal element in the set of ${I}$ with ${I^{-1} \neq R}$ is prime. Then, since there is a maximal element (by Noetherianness) containing any ${(a)}$ for ${a \in \mathfrak{m}}$ (in which case ${(a)^{-1} = Ra^{-1} \neq R}$), that maximal element must be ${\mathfrak{m}}$, which proves our claim.

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

Lemma 2 If ${S}$ is a Noetherian domain, any maximal element in the set of ideals ${I \subset S}$ with ${I^{-1} \neq S}$ is prime.

Let ${J}$ be a maximal element, and suppose ${ab \in J}$, with ${a,b \notin J}$. I claim that if ${z \in J^{-1} - S}$, then ${za, zb \in J^{-1} - S}$. The ${J^{-1}}$ part follows since ${J^{-1}}$ is a ${S}$-module.

By symmetry it’s enough to prove the other half for ${a}$; but then if ${za \in S}$, we’d have ${z( (a) + J ) \subset S}$, which implies ${ ( (a) + J)^{-1} \neq S}$, contradiction.

Then it follows that ${z(ab) = (za) b \notin S}$, by applying the claim I just made twice. But ${ab \in J}$, contradiction.

Step Two

I now claim ${\mathfrak{m}\mathfrak{m}^{-1} = R}$. Well, we know that ${\mathfrak{m}\mathfrak{m}^{-1} \subset R}$ by definition of inverses. and ${\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1}}$ too. So ${\mathfrak{m}\mathfrak{m}^{-1}}$ is an ideal sandwiched between ${\mathfrak{m}}$ and ${R}$. Thus we only need to prove that ${\mathfrak{m} \mathfrak{m}^{-1} = \mathfrak{m}}$ is impossible. To do this, choose some ${a \in \mathfrak{m}^{-1} - R}$ which must satisfy ${a \mathfrak{m} \subset \mathfrak{m}}$. Now by the next lemma, ${a}$ is integral over ${R}$

Lemma 3 If ${N \subset F}$ is a finitely generated ${R}$-module satisfying ${b N \subset N}$ for some ${b \in F}$, then ${b}$ is integral over ${R}$

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

So returning back above, ${a \notin R}$ is integral over ${R}$, a contradiction by integrally closedness.

Step Three

Now I claim ${\mathfrak{m}}$ is principal, which by Proposition 6 two days ago proves the claim. Indeed, since ${\mathfrak{m} \mathfrak{m}^{-1} = R}$, write

$\displaystyle 1 = \sum m_i n_i, \quad m_i \in \mathfrak{m}, \ n_i \in \mathfrak{m}^{-1}.$

At least one ${m_j n_j}$ is invertible, since ${R}$ is local. Then I claim ${\mathfrak{m} = (m_j)}$. Indeed, if ${x \in \mathfrak{m}}$, then

$\displaystyle x = m_j (xn_j) (m_jn_j)^{-1} \in R m_j.$

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.

1. Dedekind domains « Delta Epsilons - August 31, 2009

[…] 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 […]

2. Anonymous - September 1, 2009

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

Akhil Mathew - September 1, 2009

We have the mini-lemma: if $z \in J^{-1} - S$ and $a \in S - J$, then $za \in J^{-1} - S$ because otherwise $z \in ( J + (a) )^{-1} - S$.

This means $za \in J^{-1} - S$. Replacing $z$ by $za$ and $a$ by $b$, it follows that $zab \in J^{-1} - S$, contradiction.