I have a question concerning normalizations of plane curves, which I know little about.
Consider the simple node $V(f = y^2 - x^3 - x^2)$. Then $t=y/x$ is integral over $k[x,y]$ so that $(k[x,y]/f) \to k[x,y,t]/\langle y-tx, t^2 - (x+1) \rangle$ is the normalization, where the right ring is the coordinate ring from the blow up after setting $s=1$. Consequently it is $S^{-1}( k[x,y,t]/\langle y-tx, t^2 - (x+1) \rangle)$ the integral closure of $([,]/)_{\langle x,y \rangle}$, where $S = A \backslash \langle x,y \rangle$.
My question is whether I can generalize this in the following way: Let $A$ be the coordinate ring of an affine irreducible plane curve $c$, with maximal ideal $\mathfrak{m}$ corresponding to the origin. Blow up $c$ at the origin and we get a strict transform $s$ with coordinate ring $B$ and a ring homomorphism $A \to B$.
Assume now $s$ is nonsingular at all points lying over the origin, i.e. $B_{\nu_i}$ is regular for all maximal ideals $\nu_1, \ldots, \nu_k$ lying over $\mathfrak{m} B$. Now let $S = A \backslash \mathfrak{m}$. Then $S^{-1}B$ is integrally closed, since it is integrally closed at the localization of all maximal ideals.
Is it true that the integral closure $\overline{A_{\mathfrak{m}}}$ is isomorphic to $S^{-1}B$? If not how are those rings related? My problem is with the cases where I can't see from the equation of $A$ that $t = y/x$ is integral over $k[x,y]$. Can I maybe make a coordinate transformation such that this is the case.
Thank you very much.