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

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?