About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain? - MathOverflow

About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain?

2011-10-05T15:06:10Z

Let $K/F$ be a function field with exact field of constants $F$ ($F$ is a finite field of characteristic $p$ prime). A prime in $K$ is a discrete valuation in $K$ containing $F$. It has a unique maximal ideal $P$ which we can refer as our prime of $K$. Now, if I choose any prime $Q$ of $K$, then I construct the ring of $A$ as the ring of all elements of $K$ which are regular at every prime of $K$ different from $Q$. That is a generalisation of polynomial ring in one variable. In all the books I read up to now, they say that "IT is well known that $A$, is a Dedekind domain". I'm trying to find the proof of that but I cannot find it. Can, you tell me please why $A$ is a Dedekind domain?

Answer by Alison Miller for About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain?

2011-10-05T18:00:43Z

There are two standard approaches to this, which I'll sketch. I don't know your background, so I don't know which one will be easier for you to fill in.

(a) Using the following criterion for Dedekind rings, which is part of theorem 5.1 in May's notes:

An integral domain $R$ is Dedekind iff it is Noetherian and all its localizations at prime ideals are DVRs.

If you know enough algebraic geometry (or the underlying commutative algebra), checking this criterion is automatic. If not, you may have to do some work.

(A long time ago, I wrote an expository paper for a class using this approach for the case of affine nonsingular plane curves. It's available online on William Stein's website, with the caveats that it's the first expository paper I ever wrote, I didn't really know algebraic geometry at the time, and if I were to rewrite it, I would do it completely differently. From my experiences writing it, I know how hard it is to find references for this sort of thing in the literature, although I expect they are easier to find now than they were in 2005.)

(b) Using the following theorem, which I'm citing from the Wikipedia article on Dedekind rings, although it should be easily found in commutative algebra books also.

Theorem: Let $R$ be a Dedekind domain with fraction field $K$. Let $L$ be a finite degree field extension of $K$ and denote by $S$ the integral closure of $R$ in $L$. Then $S$ is itself a Dedekind domain.

Now choose some non-constant element $a$ of $A$, and let $R = F[a]$ be the ring generated by $a$. The ring $R$ is a polynomial ring, so Dedekind. Furthermore, our function field $K$ is a finite extension of $\mathop{\mathrm{Frac}}(R)$, and $A$ is the integral closure of $R$ in $K$, so the theorem applies.

Answer by Rahery Jacques for About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain?

2011-11-14T17:17:45Z

Thank you for the answer. I've got another problem. It seems to be obvious but I can't write the proof. Here is the statement:

If $A$ is a Dedekind domain contained in $k$ field, and $a\in A$, then for a prime ideal $I$ of $A$, if the localization of $A$ at $I$ gives a place(or prime) of $k$ with maximal ideal $P$ , we have $v_P (a) = m$, where $m$ is the power of $I$ in the decomposition of $(a)$ as factor of prime ideals of the Dedekind domain $A$.