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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.

  • you may be aware of this exercise (Ex HH.II.5.9): Two graded $B$ modules $M,N$ define the same $\mathcal{O}{Proj(B)}$-module iff $M{\geq d} \cong N_{\geq d}$. – hm2020 Jan 09 '23 at 11:14
  • Hence if $M:=kx_0,x_1$ and if $N$ is the module with $N_l=0$ for $l \leq d-1, N_l:=kx_0,x_1_l$ for $l \geq d$, it follows $M,N$ define the same sheaf on the projective line. – hm2020 Jan 09 '23 at 11:16
  • You find some calculations here. This exercise is similar. https://math.stackexchange.com/questions/4614209/localization-of-graded-rings-prove-kx-0-x-n-x-i-is-isomorphic-to-k – hm2020 Jan 09 '23 at 12:12
  • @hm2020 Thank you, but I'd prefer to understand the example in the abstract, that is in the general case, to see what's really going on! –  Jan 10 '23 at 02:07
  • sure. As I suggested: You must choose an open afine cover $U_i$ of $X:=Proj(B)$ and prove $\tilde{M}(U_i) \cong \tilde{N}(U_i)$ for all $i$. The above example generalize easily. – hm2020 Jan 10 '23 at 10:30
  • Note: You should never go directly for the general case unless you have understood all details in an explicit example. Hence the best approach is to study the case of invertible sheaves on projective space and then to generalize to an arbitrary projective scheme. – hm2020 Jan 10 '23 at 11:06
  • @hm2020 Thank you for your suggestion, but I disagree. One should study an example that contains everything that is needed for what you want to understand, and nothing more, so as to no be bogged down by irrelevant details. In this case, the general example gives a much more clear view of what is really going on. –  Jan 10 '23 at 11:09
  • it depends on what is your aim - if you want to pass an exam and want to be able to "memorize" a particular proof, this approach is sufficient. If you want to understand the structure and the construction of the $Proj(-)$ you should first understand the special case, then the general case. If you want to do research in math this is the only approach. – hm2020 Jan 10 '23 at 11:13
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    @hm2020 Your example is easier to memorize and is the typical thing that a student would learn just for an exam, but does not really say what is the important point of the phenomenon one wants to understand; on the other hand, the abstract example contains exactly what is needed and gives a more clear picture. So it's exactly the opposite of what you are saying. I do not know why you think you are in the position of lecturing anonymous people what they need to do for this or that... –  Jan 10 '23 at 11:23
  • My supervisor (when I was a PhD student in algebraic geometry) gave me the advice that I should always go from the special to the general, and I believe it is correct. My comment was not only meant for you, it was also meant for other readers of this thread. – hm2020 Jan 10 '23 at 11:34
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    @hm2020 We agree to disagree; I think one should be eclectic and work in the concrete or the abstract according to whatever is more convenient for his specific purpose. In any case, I thank you very much for commenting and answering my question and being of help. ;) –  Jan 10 '23 at 11:41
  • good to hear some of my answers are of help to others. – hm2020 Jan 11 '23 at 16:16

1 Answers1

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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.

KReiser
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