Finitely generated Dedekind domains Let $R$ be a Dedekind domain. Suppose that $R$ is finitely generated as a ring, say $R = \mathbb{Z}[x_1,\dots,x_n]/I$ for some ideal $I \subseteq \mathbb{Z}[x_1,\dots,x_n]$.
Is $R$ necessarily the ring of integers of a global field?
Is the ring of fractions of $R$ necessarily a global field? What would be a good way to prove this?
 A: Here is a proof, with the help of the comments above. I tried to be as detailed as possible.
Typically, the definition of Dedekind domains includes fields. The fields that are finitely generated as rings are precisely the finite fields $\mathbb{F}_{p^n}$, and these are not global fields.
If additionally, $\mathrm{dim}(R)=1$, then there are two cases:

*

*$\mathrm{char}(R)=p$. Then $R$ is a finitely generated $\mathbb{F}_p$-algebra, so by the Noether Normalization Lemma there is an injective ring morphism $\mathbb{F}_p[t] \to R$ that turns $R$ into a finitely generated $\mathbb{F}_p[t]$-module. So $R$ is a finite $\mathbb{F}_p[t]$-extension, in particular it is integral. This implies that the fraction field $K$ of $R$ is generated by finitely many algebraic elements, as a field extension of $\mathbb{F}_p(t)$. So it is a finite extension of $\mathbb{F}_p(t)$, see here. So it is a global field.

*$\mathrm{char}(R)=0$. Because $\mathbb{Z}$ is not a field, the standard Noether Normalization Lemma does not apply, but there is a more general version of Noether normalization in the notes by Hochster here. It implies that $R[\tfrac{1}{n}]$ is finitely generated as $\mathbb{Z}[\tfrac{1}{n}]$-module, for some natural number $n$. Now the same proof applies as in characteristic $p$: since $R$ is integral and finitely generated, its fraction field will be generated as a field by finitely many algebraic elements. As a result, it is a finite extension of $\mathbb{Q}$. So it is a global field as well.

