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Starting from a curve $C$ (over $\mathbb C$, projective if you like, irreducible, etc), is there a somewhat canonical way of constructing a curve $\tilde C$ and a map $\tilde C\to C$ which only separates the branches of $C$ at singular points but which does not resolve the singularities in the branches themselves?

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1 Answer 1

Suppose $C$ is an integral curve over an algebraically closed field. Let $\tilde{C}\to C$ be the maximal finite birational unramified cover of $C$. I didn't check all the details, but I think it is the curve you are looking for.

First such a maximal cover exists because given two finite birational unramified covers $C_1, C_2$, one can consider $C_1 \times_C C_2$ which is finite and unramified over $C$. An aproriate irreducible component will be a finite birational unramified cover dominating both $C_1$ and $C_2$.

For any $x\in C$ and any $y\in \tilde{C}$ lying over $x$, the complete local ring $\hat{O}_{\tilde{C},y}$ is a quotient of $\hat{O}_{C,x}$ because of the unramified hypothesis and $k(x)=k(y)$. So étale locally we have at best separate the analytic branches without solving the singularities.

So far we don't really use the dimension $1$ hypothesis. It remains to show that $\tilde{C}$ really separates the branches. In other words, suppose that $\tilde{C}=C$, we have to show that the singularities of $C$ are unibranch. Let's work Zariski locally. Let $A$ be the algebra of an affine open subset of $C$. Let $m$ be a maximal ideal of $A$ and suppose there are $r$ maximal ideals $m_1,...,m_r$ in the integral closure $B$ of $A$, lying over $m$ and $r\ge 2$. As $\dim A=1$, it is known that for some $N\ge 1$, $m_1^N...m_r^N\subseteq m$. Let $f\in B$ such that $f\equiv 1 \mod m_1^N$ and $f\equiv 0\mod m_i^N$ for all $i\ge 2$. Then $f^2-f=a\in A$ and $A[f]$ is unramified over $A$ and bigger then $A$. Contradiction.

Finally, the base field can in fact be arbitrary.

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