Quasi-coherent $\mathcal{F} \cong \tilde{M}$ on $\operatorname{Proj}B$ does not determine $M$

Question

I would like to understand in detail the fact that given $\tilde{M}\cong \tilde{N}$ on $\operatorname{Proj}B$, then it is not always true in general that $M \cong N$.

I know this has to do with playing with the grading; so, as I understand, one fixes a $K\in \mathbb{N}$ and considers $N = \sum_{k \geq K} M_k$ and then proves that:

i) $N$ is not isomorphic to $M$, because in degrees $d<K$ one has $N = 0$;

ii) one proves $\tilde{M} \cong \tilde{N}$ because for every homogenous element $f$, one has $$\tilde{M}(D_+(f))=M_{(f)}\cong N_{(f)}=\tilde{N}(D_+(f))$$

Although I can buy it intuitively, I am bit lost on the concrete and specific details of this though.

Can someone explain to me this example?

Ps. This example is taken from Q. Liu's book (remark 1.18, 5.1.4), where he works with two abstract graded modules $M$ and $N$, so the argument should work in this generality, and I would like to understand it this way.

Answer 1

The point is that the projection map $\pi:M\to N$ and inclusion map $i:N\to M$ both become isomorphisms after applying the sheafification functor $\widetilde{-}$. To check this, we can cover $\operatorname{Proj} B$ by open sets of the form $D_+(b)$ for $b$ homogeneous of positive degree and show that the induced maps $\widetilde{\pi}(D_+(b)):\widetilde{M}(D_+(b)) \to \widetilde{N}(D_+(b))$ and $\widetilde{i}(D_+(b)):\widetilde{N}(D_+(b)) \to \widetilde{M}(D_+(b))$ are isomorphisms, as a map of sheaves which restricts to an isomorphism on each element of an open cover is an isomorphism.

To show this, we compute: $\widetilde{M}(D_+(b))\cong M_{(b)}$ and $\widetilde{N}(D_+(b))\cong N_{(b)}$ by definition, where elements of $M_{(b)}$ are of the form $\frac{m}{b^p}$ for $m\in M$ homogeneous with $\deg m = p\deg b$ (similarly for $N_{(b)}$). The map $\widetilde{\pi}(D_+(b))$ is given by $\frac{m}{b^p}\mapsto \frac{\pi(m)}{b^p}$. I claim this is injective and surjective:

• Injective: suppose $\frac{m}{b^p}$ maps to zero. Let $q\in\Bbb Z_{\geq 0}$ be a nonnegative integer so that $\deg b^qm \geq K$. Then $\frac{m}{b^p}=\frac{b^qm}{b^{p+q}}$, and as $\deg b^qm\geq K$, we have that $\pi(b^qm)=b^qm$. So the image of $\frac{m}{b^p}=\frac{b^qm}{b^{p+q}}$ is zero implies that $\frac{m}{b^p}$ is also zero, proving injectivity.
• Surjectivity: suppose $\frac{n}{b^r}$ is an element of $N_{(b)}$. Then $\deg n \geq K$ and $\deg n = r\deg b$. By definition of $N$, $n$ can also be regarded as an element of $M$ of the same degree. So $\frac{n}{b^r}$ is also an element of $M_{(b)}$ and it maps to $\frac{n}{b^r}\in N_{(b)}$, proving surjectivity.

The proof for the map $\widetilde{i}_{(b)}$ is the same, except you use the trick for surjectivity instead of injectivity. Let me know if you need help working that out and I can add to this answer.