I have a few questions regarding the proofs of proposition 7.4 and theorem 7.5 in Hartshorne's Algebraic Geometry 1.7 Intersections in Projective Space.
Here is the beginning of proposition 7.4 and part of the proof
My question is how do you see the isomorphism at the end?
Then the module $N \subseteq M''$ generated by $m$ is isomorphic to $(S/\mathfrak{p})(-l).$
The second is about theorem 7.5. In the proof Hartshorne writes
If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence then $Z(\text{Ann}(M)) = Z(\text{Ann}(M')) \cup Z(\text{Ann}(M''))$
where $M$ is a finitely generated graded $k[x_0, \ldots, x_n]$-module. I thought the equality $$\text{Ann}(M)) = \text{Ann}(M') \cap \text{Ann}(M'')$$
was only true if $M$ was the direct sum of $M'$ and $M''$, is this always the case for the modules in question?
Here is the Theorem and the beginning of its proof
My background in commutative algebra is very introductory, so I would love some help here and or good references. I have looked at Eisenbuds Commutative Algebra and Atiyah-Macdonald, but none of those contain the exact results that Hartshorne presents. For example, Eisenbuds proposition 3.7 proves almost the same thing as Hartshorne proposition 7.4 but doesn't specify the twist. This also sort of addresses my second question, but I don't see how $\mathfrak{p}$ comes to play.
Best Answer
$\mathfrak{p}=I_m$ is the annihilator of the homogenous element $m \in M’’$. Therefore, the ($S$-linear, non-homogenous) map $i: S \rightarrow N$ mapping $a$ to $am$ is onto with kernel $\mathfrak{p}$. But $i$ maps a homogenous element of degree $k$ to an element of degree $k+l$. So for $i: S/\mathfrak{p} \rightarrow N$ to be an isomorphism of graded modules, you need to change the grading on $S/\mathfrak{p}$, so that elements of degree $k$ (for the new grading) correspond to elements of degree $k-l$ for the standard grading. That’s exactly what is achieved by considering $(S/\mathfrak{p})(-l)$ (maybe up to the sign convention?).
if you have a (split or not) exact sequence $0 \rightarrow M’ \rightarrow M \rightarrow M’’ \rightarrow 0$ of finitely generated modules over a Noetherian ring, then you can check that $\mathrm{Ann}(M) \subset \mathrm{Ann}(M’) \cap \mathrm{Ann}(M’’)$. The reverse inclusion is indeed false, but a slightly weaker statement holds: $\mathrm{Ann}(M’) \mathrm{Ann}(M’’) \subset \mathrm{Ann}(M)$. Now all you have to do is prove that for any two ideals $I,J$, $Z(IJ)=Z(I \cap J)$. This is because $IJ \subset I \cap J \subset \sqrt{IJ}$.