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Let $C$ be a reduced, reducible curves over an algebraically closed field with at worst nodal singularities. Does there exist an irreducible non-singular curve $\tilde{C}$ and a finite, surjective morphism $\tilde{C} \to C$ which is an isomorphism on the smooth locus of $C$? We know this is true when $C$ is irreducible, namely by taking the normalization.

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It seems like you're not sure exactly what hypotheses you want here.

At present, you ask if you can take a reduced curve $C$ with at worst nodal singularities and find an irreducible nonsingular curve $\widetilde{C}$ and a finite surjective map $f : \widetilde{C} \to C$ which is an isomorphism on the nonsingular locus of $C$.

This can't be done. Suppose $C$ has two irreducible components, $X$ and $Y$. Since $X$ is closed, so is its preimage $f^{-1}(X)$. Since $C$ is an irreducible curve, the only closed subsets of $C$ are individual points and the entire curve. The preimage of $X$ can't be an individual point, since $f$ needs to be an isomorphism onto the smooth locus of $C$. So $f^{-1}(X) = C$. But then no point of $\widetilde{C}$ can map to a point of $Y$ that lies outside of $X$.

One thing you could do would be to take $C$, normalize each component separately, and then take the disjoint union of the components. Of course this gives you a reducible curve, but it's nonsingular and moreover it's the best you're going to do if you insist on $\widetilde{C}$ being nonsingular.

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