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Background

I have been considering the following question. Let $k$ be an algebraically closed field and consider a curve $C \subset \mathbb{P}^2$. Compute its genus, that is, the genus of its normalization, using data "as local" as possible. I understand that desingularization of plane curves is an old topic, but I would like to follow through the following approach.

My idea was to use that for non-singular curves, $g = dim_k H^1(C, \mathcal{O}_C)$, and that for arbitrary plane curves $dim_k H^1(C, \mathcal{O}_C) = (d-1)(d-2)/2$, where $d$ is the degree. That is, we consider the normalization morphism $\pi: C' \to C$ and try to relate the first cohomology groups. This yields the following formula, for a (singular) plane curve $C$ with normaization $\pi: C' \to C$:

$g(C) = (d-1)(d-2)/2 - \sum_{P \in C} dim_k \frac{\mathcal{O}_{C',P}}{\mathcal{O}_{C,P}}$.

Here $\mathcal{O}_{C',P}$ denotes $(\pi_* \mathcal{O}_{C'})_P$, i.e. a certain semilocal dedekind domain.

Actual Question

Let as above $k$ be algebraically closed and $A$ be the local ring of a plane singularity - that is, $A=(k[x,y]/(F))_{(x,y)}$ for some $F \in k[x, y]$ without constant term. Let $B$ be its normalization, that is its integral closure inside its field of fractions. Denote by $p$ the maximal ideal of $A$. In a number of examples I have worked out, there exists an integer $n$ such that $p^n B \subset A$. Is this always true?

Remarks

This would have the desirable consequence that if we set $M = B/A$, then $\hat{M} = M$, where hat denotes completion. Hence the quantity of interest can be obtained as $dim_k \hat{B}/\hat{A}$, which suggests that it is "very local". Here is an even bolder question: can $\hat{B}$ be obtained from $\hat{A}$? That is, if $A_1$, $A_2$ are two local rings of plane singularities and $\hat{A_2} \approx \hat{A_2}$, do they contribute the same term to the genus? If not, what if we require the isomorphism between $\hat{A_1}$ and $\hat{A_2}$ to come from an automorphism of $k[[x, y]]$?

That's a lot of questions. I'm mostly interested in the first (italic) one, the rest is follow-up ramblings.

Thanks, Tom

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1 Answer 1

up vote 2 down vote accepted

The answer for all questions are yes.

First, let $f\in p$ be a non-zero element. Then $A_f=\mathrm{Frac}(A)$ because its only prime ideal if $0$ ($A$ has two prime ideals, one of them, the maximal one, contains $f$, hence disappears in $A_f)$. So $B_f=A_f$ and for any $b\in B$, $f^nb\in A$ for some power of $f$. As $B$ is finitely generated over $A$ (as module), we can choose the same $n$ for all elements of $B$.

For the second question, note that the normalization of $\hat{A}$ is $\hat{B}$.

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That was quick. Thank you very much. –  Tom Bachmann Mar 19 '12 at 23:11

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