Premise and main idea: I'm not an expert in the field of commutative algebra and when I encounter problems regarding local rings I try to solve them by following a sort of geometric intuition.
It was not easy to write this question in a succinct way, but basically I'd like to solve the following problem: suppose that you have a singular point $x$ on a curve $X$, for example a simple node, then you desingularize it and obtain two new points $x_1,x_2$ ''above'' $x$. I want to recover the completions of the local rings at $x_1$ and $x_2$ by starting from the complete local ring at $x$.
Notation: Let $X$ be an integral, Noetherian scheme of dimension $1$. Suppose that $x\in X$ is a singular point and consider the normalization $\nu:X'\to X$ such that $\nu^{-1}(x)=\{x_1,\ldots,x_d\}$.
For any integral domain $B$ we denote with $B'$ its normalization in the fraction field.
If $B$ is a local ring, $B^\hat{}$ is its completion with respect to the maximal ideal.
Question: Put by simplicity $A=\mathcal O_{X,x}$. I want to recover the complete rings $\left(\mathcal O_{X',x_i}\right)^\hat{}$ (for every $i$) by starting from $A^\hat{}$. By looking at many books it seems that the points $\{x_1,\ldots,x_d\}$ correspond to the minimal prime ideals of $A^\hat{}$ in the following way:
Suppose that $\mathfrak q$ ranges among all the finitely many prime ideals of $A^{\hat{}}$, then:
$$\bigoplus_\mathfrak q \left(A^{\hat{}}/\mathfrak q\right)^\prime\cong \bigoplus_{i=1}^d \left(\mathcal O_{X',x_i}\right)^\hat{}$$
In other word we should have an isomorphism $\left(\mathcal O_{X',x_i}\right)^\hat{}\cong \left(A^{\hat{}}/\mathfrak q\right)^\prime$ for a certain minimal prime ideal $\mathfrak q$.
Is it true? I'm trying to guess "the relationship" but I have no idea on how to formalize/prove it... please help me.
Here is a road map as you said. Let $B$ denote the integral closure of $A$. In the geometric case, (requires more than just Noetherian integral scheme), one has $B$ a finite type $A$-module and thus one has $\widehat{A}\subset\widehat{B}$. But, by Chinese remainder theorem, $\widehat{B}=\prod \widehat{B_i}$, where the $B_i$s are the finitely many localizations of $B$ at the maximal ideals. All the $\widehat{B_i}$ are integrally closed by Zariski's main theorem and all are finite type modules over $\widehat{A}$, so integral over it. They have the same fraction field also follows, so they are the integral closures in the total quotient ring of $\widehat{A}$.