4
$\begingroup$

I am reading an article in which a proof is based on defining a subsheaf by only giving its global sections.

The exact setting is that, one has a surjective finite morphism $f:Y\to X$ between separated $S$-schemes of finite type, over a Noetherian base $S$. Then, $\mathcal{A}$ is defined to be a subsheaf of the coherent sheaf $f_\ast\mathcal{O}_Y$ of $\mathcal{O}_X$-algebras, such that the ring of its global sections is a given $\Gamma(X,\mathcal{O}_X)$-subalgebra of $\Gamma(X,f_\ast\mathcal{O}_Y)\cong \Gamma(Y,\mathcal{O}_Y)$.

For the proof to work, $\mathcal{A}$ should be the largest coherent subsheaf in $f_\ast\mathcal{O}_Y$ of $\mathcal{O}_X$-algebras with the given global sections.

So, my question:

is the given information sufficient to determine $\mathcal{A}$? I.e. given a morphism $f:Y\to X$, as above, and a $\Gamma(X,\mathcal{O}_X)$-subalgebra $A\subseteq\Gamma(X,f_\ast\mathcal{O}_Y)$, is there a largest coherent subsheaf $\mathcal{A}\subseteq f_\ast\mathcal{O}_Y$ of $\mathcal{O}_X$-algebras, with $\Gamma(X,\mathcal{A})\cong A$? If so, how is it defined?

One has the subsheaf generated by $A$, but that may not be the largest (in terms of inclusions), especially when $f_\ast\mathcal{O}_Y$ is not generated by its global sections. That is, in some cases we need to have $A\cong \Gamma(X,f_\ast\mathcal{O}_Y)$, in which case we want $\mathcal{A}\cong f_\ast\mathcal{O}_Y$.

I need to use the result 'proven' in the article in my research, and I need this line of argument to work, maybe in a more restrictive setting. So, if the answer to the above question is negative, it will be useful to know

what extra 'mild' assumptions $f$ needs to satisfy for the desired subsheaf $\mathcal{A}\subseteq f_\ast\mathcal{O}_Y$ to be determined by its global sections.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ As I said in my MSE comments, most of the time, you have little choice, at least in the standard cases. If $X,Y$ are irreducible varieties over an algebraically closed field $k$, then $\Gamma(X,f_*\mathcal{O}_Y)=k$ and thus the only choice you have for $A$ is $k$. Then $\mathcal{A}$ is forced to be $f_*\mathcal{O}_Y$. $\endgroup$
    – Mohan
    Commented Apr 5, 2017 at 16:54
  • 1
    $\begingroup$ I for got to say, irreducible projective varieties in my comment. $\endgroup$
    – Mohan
    Commented Apr 5, 2017 at 17:38
  • $\begingroup$ Hi. Your schemes are affine OR ``affine over the base''? $\endgroup$
    – Artur Jackson
    Commented Apr 6, 2017 at 0:22
  • $\begingroup$ Hi @ArturJackson, I am working in the category of all schemes that are separated (over the base) and of finite type over the base, as mentioned in the body of the question. If the schemes were to be affine (over $\mathbb{Z}$), there would be no question to ask. In my previous comment, I just wanted to emphasis that I cannot restrict myself to projective varieties. Maybe I did not state my comment clearly, so I will change it in the light of yours. $\endgroup$
    – user24453
    Commented Apr 6, 2017 at 7:18
  • $\begingroup$ @Mohan: thank you for your comment, but I cannot restrict myself to such varieties. That is, I do not have the luxury to assume the base to be a field, and I need to consider all schemes that are separated (over the base) and of finite type over the base. PS Apologies if my previous comment was unnecessarily confusing. $\endgroup$
    – user24453
    Commented Apr 6, 2017 at 7:23

0

Reset to default

You must log in to answer this question.