Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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

share|improve this answer
add comment

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, $g(X')=p_a(X)-$ the number of singular points.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.