I'm working through the details of Deligne and Mumford's 69' paper, "The Irreducibility of the Space of Curves of Given Genus", and I had a few quick questions:


1) On p. 77, they claim that for $x$ a node, $\pi:C'\rightarrow C$ the blowing-up of $x$ and $x\_1,x\_2\in C'$ the two points of $\pi^{-1}(x)$. Then supposedly it's "easy" to show that for an invertible sheaf $\mathcal L$ on $C$, we have $$\rm{Hom}(\mathfrak m\_x,\mathcal L)\cong H^0(C',\pi^\*\mathcal L),\rm{Hom}(\mathfrak m\_x^2,\mathcal L)\cong H^0(C',\pi^\*\mathcal L(x\_1+x\_2)),$$ but I'm having trouble seeing it. I've tried to show it locally by assuming the curve looks like $xy=0$ in $\mathbb A^2$, but then I'm finding that on an affine open $U$ which trivializes $\mathcal L$ and doesn't contain $x$, then the restriction of such a homomorphism is equivalent to giving a section of $\mathcal L$ over $U$, and thus a section of $\pi^\*\mathcal L$ over the isomorphic preimage of this open. But around $x$ I'm finding that such a homomorphism is equivalent to giving two functions, one on each branch, which vanish at $x\_1,x\_2$, which seems to be giving an isomorphism with $H^0(C',\pi^\*\mathcal L(-x\_1-x\_2))$. I find the same thing for the square. Any help? I'm sure this is easy for experts...


On p. 78
2) At the beginning of \textit{Case 3} they claim, and I agree, that the restriction of $\omega\_C^{-n}$ has degree at most -2 on each component $E$ of $C$. Shouldn't this automatically imply it has no sections on the preimage of that component in the blow-up? Why do they go on to speak about when $n=2$, the degree of $\omega\_C$ is 1 on that component, and two poles are allowed? Furthermore, if the whole point is when $n\geq 3$, why is this necessary?


3) Later after the corollary on that page, why can we assume that the projective bundle $\mathbb P(\pi\_\*(\omega^n\_{C/S}))$ is trivializable so we can actually get a morphism from $S$ to the Hilbert Scheme?



Questions on theorem in Deligne-Mumford's '69 Paper: $\omega_C^n$ is very ample $n\geq 3$
Since no one has written an official answer to this question and the bounty is ending in a few hours, I'll provide an answer of my own to the question, though see ulrich's comments for another approach.


1) The crucial observation here is that $\mathfrak m\_x=\pi\_\*(\mathcal O\_{C'}(-x\_1-x\_2))$ where $x\in C$ is a node, $\pi:C'\rightarrow C$ is the partial normalization at $x$ and $x\_1,x\_2$ are the points in $C'$ lying over $x$. This is fairly clear from the description of $\mathcal O\_C(V)$ as the set of $ f\in \mathcal O\_{C'}(\pi^{-1}(V))$ such that $f(x\_1)=f(x\_2)$ for any affine $V$ such that $x$ is its only singular point. Then the claim is seen as follows:


Writing $\mathcal L=(\mathcal L\otimes \omega\_C^{-1})\otimes \omega\_C$, we get that $$\text{Hom}(\mathfrak m\_x,\mathcal L)\cong \text{Hom}(\mathfrak m\_x (\mathcal L^{-1}\otimes \omega\_C),\omega\_C)\cong H^1(C,\mathfrak m\_x(\mathcal L^{-1}\otimes \omega\_C))^{\vee}.$$ Then using the crucial observation above, the projection formula, and the fact that $\pi$ is finite so the higher direct images vanish, we see that 


$H^1(C,\mathfrak m\_x (\mathcal L^{-1}\otimes \omega\_C))^{\vee}\cong H^1(C',\mathcal O\_{C'}(-x\_1-x\_2)\otimes\pi^\*\mathcal L^{-1}\otimes \pi^\*\omega\_C)\cong H^0(C',\pi^\*\mathcal L),$


where I've used the fact that $\pi^\*\omega\_C\cong \omega\_{C'}(x\_1+x\_2)$ and Serre duality on $C'$. The proof of the $\mathfrak m\_x^2$ case is almost identical.


2) They choose $n\geq 2$ since when they used Serre duality an exponent of $1-n$ appeared for $\omega\_C$, and $n$ was originally $n\geq 3$, so they WLOG assume now $-n$ is the exponent of $\omega\_C$ but $n\geq 2$. For the rest, a possibly clearer proof of this result is given as Lemma 10.6.1 in "Geometry of Algebraic Curves, Vol. II" by ACG.


3) The projective bundle is not necessarily trivial, and the Hilbert scheme constructed there parametrizes only those families for which a trivialization exists. However, any family can be pulled back to the principal $PGL$ bundle associated to the projective bundle, and there it does become trivial. Thus there is a map from the associated $PGL$-bundle to the Hilbert scheme.


