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I would appreciate help with the following.

Let $k$ be a field that's algebraically closed and let $f$ be a polynomial in $k[X,Y]$. Prove that $R=k[X,Y]/(f)$ is a Dedekind domain if and only if at one of $f(a,b)$ and the two partials at $(a,b) \in k^{2}$ is not zero, for all $(a,b)$ in $k^{2}$.

Thank you all

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«smooth manifold» is usually reserved to the case where $k$ is $\mathbb R$ or $\mathbb C$ or, more generally, when $k$ is a nce enough topological field. –  Mariano Suárez-Alvarez Dec 23 '12 at 19:26
    
@MarianoSuárez-Alvarez: I think smoothness in general is sometimes defined by the statement in the question. –  tomasz Dec 24 '12 at 11:22
    
But I'd be surprised if anyone used the word manifold except when $k$ is a locally compact normed field! –  Mariano Suárez-Alvarez Dec 24 '12 at 15:10
    
got it, thanks a million. i was away for holidays - sorry about that :) –  Dquik Dec 29 '12 at 14:19

1 Answer 1

up vote 2 down vote accepted

By the Jacobian criterion for regularity $R_m$ is regular for all maximal ideals $m\subset R$ if and only if at least one out the two partial derivatives of $f$ at $(a,b) \in k^{2}$ is not zero, for all $(a,b)$ in $k^{2}$. But in this case all maximal ideals have height one, so $R_m$ are DVRs, equivalently $R$ is Dedekind.

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