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