# The moduli scheme of smooth curves of given genus is irreducible

I've heard that Deligne-Mumford's "the irreducibility..." showed this first. But I think that Mumford's "Geometric invariant theory" has its proof.

The proof is as follows: Let $$H$$ be the scheme that represents the functor of 3-canonical embeded smooth curves (as in proposition 5.1. of GIT). Then the moduli scheme $$M$$ is the geometric quotient $$H/PGL$$ (by proposition 5.4. of GIT). And by chapter 0, section 2 of GIT, to show the irreducibility of $$M$$, it's sufficient to show that of $$H$$. And by 5.3, $$H$$ is irreducible. So the moduli $$M$$ is also irreducible.

Is this wrong? Or, is Deligne-Mumford the first one that shows this only using algebraic tools? (GIT uses transcendental technique.)

Thank you very much!

• I recommend that you actually read the introduction to the article by Deligne and Mumford, which reviews the history. Over $\mathbb{C}$ (and thus over any algebraically closed field of characteristic $0$), there are proof of irreducibility of $M_g$ that go back to the work of Hurwitz. Teichmueller proved more -- the universal cover of the (orbifold) moduli space is a ball in complex space, thus contractible. The proof(s) by Deligne and Mumford extend this to fields of arbitrary characteristic. – Jason Starr Mar 15 at 21:49
• The work of Harris and Mumford leads to a purely algebraic proof of irreducibility in characteristic $0$; see the appendix by Fulton to their paper. – ulrich Mar 16 at 8:19
• Thanks for your comments. I'm completely beginner and going to read the paper, but I understand that the moduli $M$ is a scheme over $\operatorname{Spec}\mathbb{Z}$, and that "the irreducibility of the moduli over a field of characteristic $p$" is "the irreducibility of $M \times_\mathbb{Z} k$". And it seems that GIT shows the irreducibility of $M$ by showing that of $H \times \mathbb{C}$. That is, I think that GIT shows the irreducibility of $M$ over any characteristic. – k.j. Mar 16 at 16:34
• @k.j. "That is, I think that GIT shows the irreducibility of $M$ over any characteristic." That is not correct. There are many irreducible schemes over $\text{Spec} \ \mathbb{Z}$ whose fibers over some finite fields $\mathbb{Z}/p\mathbb{Z}$ are reducible. – Jason Starr Mar 17 at 0:17
• @JasonStarr Thank you very much! It's very fundamental, I didn't notice that... – k.j. Mar 17 at 10:08