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Let $A$ be a ring (commutative with unity), and let $B$ be a regular finite-type $A$-algebra of relative dimension 1 over $A$. (ie, Spec $B$ is a regular curve over $A$).

Let $\mathfrak{p}$ be a prime of $B$ finite over $A$ (ie, it corresponds to a horizontal divisor of Spec $B$ over Spec $A$). In particular, $\mathfrak{p}$ is height 1, since $B$ is a curve over $A$.

Is the completion of $B$ at $\mathfrak{p}$ isomorphic to $A[[X]]$?

(we can assume $A,B$ integral domains if that makes things simpler).

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This is easy if B is an algebraically closed field, isn't it? –  Piotr Pstrągowski Apr 2 '13 at 23:00
    
@PiotrPstragowski: you meant $A$ instead of $B$ I think. –  user18119 Apr 3 '13 at 7:35
    
Please be more precise on your definition of $\mathfrak p$. If you mean $V(\mathfrak p)$ is finite surjective over $\mathrm{Spec}(A)$, then the statement is not true if $A$ is not a field. –  user18119 Apr 3 '13 at 7:40
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