Let $(A,\mathfrak{m})$ be a local domain with integral closure $B$ which is a regular local domain with maximal ideal $\mathfrak{M}$. Assume in addition that $A/\mathfrak{m} = B/\mathfrak{M}$ and $B\mathfrak{m} = \mathfrak{M}$.
Does it follow that $A$ is regular and therefore $A = B$? If not in general, what if $\dim(A) = 1$? It is enough to proof $\mathfrak{M}^2 \cap A = \mathfrak{m}^2$, since then it is $\mathfrak{m}/\mathfrak{m}^2$ a vector subspace of $\mathfrak{M}/\mathfrak{M}^2$ but I wasn't able to show this.
The background is whether one can see a singularity in the case of just one analytic branch in the extended Ideal.
many thanks in advance!