Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know some references where I can find this, but they seem tedious(Both Hartshorne and Ueno cover this).

I am wondering if there is an elegant way to describe these. If this task is too difficult in general, how about just $\mathbb{P}^n$?

Thanks!

share|improve this question

1 Answer 1

up vote 6 down vote accepted

Quasi-coherent sheaves on affine schemes (say $Spec(A)$) are obtained by taking an $A$-module $M$ and the associated sheaf (by localizing $M$). This gives an equivalence of categories between $A$-modules and q-c sheaves on $Spec(A)$.

Let $R$ be a graded ring, $R = R_0 + R_1 + \dots$ (direct sum). Then we can, given a graded $R$-module $M$, consider its associated sheaf $\tilde{M}$. The stalk of this at a homogeneous prime ideal $P$ is defined to be the localization $M_{(P)}$, which is defined as generated by quotients $m/s$ for $s$ homogeneous of the same degree as $m$ and not in $P$.

In short, we get sheaves of modules on the affine scheme just as we get the normal sheaves of rings. We get sheaves of modules on the projective scheme in the same homogeneous localization way as we get the sheaf of rings.

However, it's no longer an equivalence of categories. Why? Say you had a graded module $M= M_0 + M_1 + \dots$ (in general, we allow negative gradings as well). Then it is easy to check that the sheaves associated to $M$ and $M' = M_1 + M_2 + \dots$ are exactly the same. Nevertheless, it is possible to get every sheaf on $Proj(R)$ for $R$ a graded ring in this way. See Proposition II.5.15 in Hartshorne.

share|improve this answer
    
Sorry I didn't respond sooner, nice answer so far, but one more request. Can you tie the last fact(about losing eq of cat to the fact that M and M' have the same sheaves) back to the geometry of Proj. I am thinking it comes from the fact that Proj is isomorphic to shifted proj(I just mean overthe same ring with the grading scaled by some factor.) thanks! –  BBischof Jul 25 '10 at 5:27
    
Yes, $Proj R$ is isomorphic to $Proj R^{(d)}$ where $R^{(d)} = \bigoplus_{n} R_{nd}$; this is a version of the $d$-uple embedding of projective space. Now, $M', M$ agree in sufficiently large dimensions. This means that if you scale the grading by some large $d$, we have $M'^{(d)}, M^{(d)}$ isomorphic as $R^{(d)}$-modules. In particular, their associated sheaves on $Proj(R^{(d)}$ are the same. Pulling back to $Proj(R)$ gives isomorphic sheaves. –  Akhil Mathew Jul 25 '10 at 5:39
    
[Of course, this relies on the fact that pulling back the sheaf associated to $M^{(d)}$ from the $d$-uple embedding gives you the sheaf associated to $M$, which I'm not sure really explains things well.] Maybe one way to think of it is on projective space over a field. Say you have some section of the sheaf associated to $M'$. Then on the $i$th hyperplane multiply by $x_i$ to a high power to get something of $M$. Then divide by $x_i$ to the same power, since it is nonzero on this hyperplane. –  Akhil Mathew Jul 25 '10 at 5:43
    
Alright, I am glad that my intuition was correct there. So, we are safe in Proj to think about R-mod instead of QCsheaves, we should just keep in mind that some of them will yield the same sheaves. And sorry for taking so long to respond, these two comments didn't show up, so I didn't notice you responded. :/ –  BBischof Jul 25 '10 at 19:06
    
Also, I have accepted this answer. I think that seeing them simply as R-mods is simple enough. :D –  BBischof Jul 25 '10 at 19:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.