About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:43:18Z http://mathoverflow.net/feeds/question/77241 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77241/about-the-ring-used-in-the-definition-of-drinfeld-modules-why-is-that-ring-a-ded About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain? Rahery Jacques 2011-10-05T15:06:10Z 2011-11-14T17:17:45Z <p>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?</p> http://mathoverflow.net/questions/77241/about-the-ring-used-in-the-definition-of-drinfeld-modules-why-is-that-ring-a-ded/77268#77268 Answer by Alison Miller for About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain? Alison Miller 2011-10-05T18:00:43Z 2011-10-05T18:00:43Z <p>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.</p> <p>(a) Using the following criterion for Dedekind rings, which is part of theorem 5.1 in <a href="http://www.math.uchicago.edu/~may/MISC/Dedekind.pdf" rel="nofollow">May's notes</a>: </p> <blockquote> <p>An integral domain $R$ is Dedekind iff it is Noetherian and all its localizations at prime ideals are DVRs.</p> </blockquote> <p>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.</p> <p>(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 <a href="http://wstein.org/129-05/final_papers/Alison_Miller.pdf" rel="nofollow">on William Stein's website</a>, 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.) </p> <p>(b) Using the following theorem, which I'm citing from the <a href="http://en.wikipedia.org/wiki/Dedekind_ring" rel="nofollow">Wikipedia article on Dedekind rings</a>, although it should be easily found in commutative algebra books also.</p> <blockquote> <p>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.</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/77241/about-the-ring-used-in-the-definition-of-drinfeld-modules-why-is-that-ring-a-ded/80913#80913 Answer by Rahery Jacques for About the ring used in the definition of drinfeld modules. Why is that ring a dedekind domain? Rahery Jacques 2011-11-14T17:17:45Z 2011-11-14T17:17:45Z <p>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:</p> <p>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 $A$.</p>