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Let $X$ be a stable curve of genus $g$ over the field $k$, i.e., a $k$-rational point of the Deligne-Mumford stack $M_g$.

What is the genus of the normalization of $X$? Does it depend on the number of singularities of $X$?

Note that the normalization of $X$ is a smooth curve. It is still geometrically connected and it is projective.

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The normalization $X'$ of $X$ needs not be conneced, because a stable curve is not necessarily irreducible. So $X'$ is the disjoint union of the normalizations of the irreducible components of $X$. It genus (as sum of genus of its components or its arithmetic genus) can be computed in terms of the arithmetic genus and the number of singular points of $X$. If this is what you are looking for, I will write the formula. In the irreducible case:

$$g(X')=p_a(X)- \text{the number of singular points.}$$

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The genus is a birational invariant, and a curve is birational to its normalization. For example, see this discussion on MO.

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    $\begingroup$ The geometric genus is indeed a birational invariant. But if one talks about a stable curve of genus $g$, then one normally means the arithmetic genus. Indeed all point on $\overline{M_g}$ all have arithmetic genus $g$, but differing geometric genus. $\endgroup$ – M.D. Jun 27 '17 at 16:27

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