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Let $k$ be an algebraically closed field, and let $B$ be a finitely generated $k$ algebra that is also a Domain. Then $B$ is the affine coordinate ring of some affine variety $Y$; this part is straight out of Hartshorne and is not terribly difficult to understand. However, I am having trouble making sense of the following extension of this statement.

If $B$ is Dedekind, then $Y$ is both non-singular and has dimension 1.

This shows up in Hartshorne (page 41, Lemma 6.5, last paragraph).

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up vote 4 down vote accepted

Let $P$ be a non-zero prime ideal of $B$. Since $P$ is a maximal ideal, ht($P$) = 1. Hence dim $B$ = 1.

It remains to prove that $B_P$ is regular. Since $B_P$ is Noetherian, integrally closed and dim $B_P = 1$, $B_P$ is a discrete valuation ring(e.g. Atiyah-MacDonald). Hence $B_P$ is regular.

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Thank you much. That is wonderfully simple. – John Martin Aug 9 '12 at 18:23
Hey John. I'm not trying to be annoying or anything at all, but as i see that you are relatively new here, do you know you can say "thank you" by "accepting" an answer? Just press the checkmark next to your preferred answer.. Gives reputation, says thank you, tells others that they dont need to answer the question anymore... – Joachim Aug 9 '12 at 22:08

These are essentially part of the definition. Namely, one can define a Dedekind domain as an integrally closed, noetherian, dimension one domain. There is an important result that says that since $B$ is normal, the singular points of $Y=\operatorname{Spec}B$ have codimension $\ge 2.$ Since $\dim Y=1,$ the subvariety $\operatorname{Sing}Y=\emptyset$ must be empty.

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