2
$\begingroup$

Let $X$ be a projective variety and $\mathcal{I} \subset \mathcal{O}_X$ is an ideal sheaf on $X$ not equal to $\mathcal{O}_X$. I'm supposed to show $\Gamma(X,\mathcal{I}) = 0$.

My first thought was that $\mathcal{I}(X)$ is an ideal of $\mathcal{O}_X(X)$, and since $X$ is a projective variety, $\mathcal{O}_X(X) = k$, the ground field, and of course the only ideals of a field are $k$ itself and $0$. So $\mathcal{I}(X) \in \{k,0\}$.

Since $\mathcal{I} \subsetneq \mathcal{O}_X$, my first thought was that of course $\mathcal{I}(X) \subsetneq \mathcal{O}_X(X)$, so that $\mathcal{I}(X) = 0$, but I don't think this is true anymore.

Just in terms of sheaves, a proper subsheaf can agree at the level of global sections. For instance, take the sheaf of holomorphic functions on a complex domain. A proper subsheaf would be the sheaf of constant functions, but at the level of global sections they definitely agree. I'm thinking we have to use the fact that $\mathcal{O}_X$ and $\mathcal{I}$ are coherent, but I don't really see how it applies.

This is for homework, so I'd prefer a hint over a full answer.

$\endgroup$
6
  • 2
    $\begingroup$ Hint: Look at 1 of $\mathcal{O}_X(X)$ $\endgroup$
    – Arun Kumar
    Commented Mar 13, 2016 at 5:55
  • $\begingroup$ Consider the map $\mathcal{O}\to \mathcal{I}$ given by the section of $\Gamma(X, \mathcal{I})$. $\endgroup$
    – Chen Jiang
    Commented Mar 13, 2016 at 6:23
  • 3
    $\begingroup$ The key observation should be that the restriction to any open subset or even to any stalk of $1$ is again $1$, since the restriction maps are ring homomorphisms. $\endgroup$
    – MooS
    Commented Mar 13, 2016 at 8:59
  • $\begingroup$ I'm not sure what you're referring to as 1, @ArunKumar and Moos. I'll try looking at restrictions though, thanks. $\endgroup$
    – walkar
    Commented Mar 13, 2016 at 14:36
  • 1
    $\begingroup$ If the global sections are equal to $k$, there is a $1$, isn't there? $\endgroup$
    – MooS
    Commented Mar 13, 2016 at 15:06

1 Answer 1

4
$\begingroup$

Since $\mathcal{I}(X) \subset \mathcal{O}_X(X)$ is an ideal, we have $\mathcal{I}(X) \in \{k,0\}$. Suppose $\mathcal{I}(X) = k$. But then for any open set $U \subset X$, we have $\mathcal{I}(U) \subset \mathcal{O}_X(U)$ is an ideal.

Since $1_k = 1 \in \mathcal{I}(X)$ and $\operatorname{res}^{\mathcal{I}}_{X,U}: \mathcal{I}(X) \rightarrow \mathcal{I}(U)$ is a ring homomorphism, we have $\operatorname{res}^{\mathcal{I}}_{X,U}(1) = 1_{\mathcal{O}_X(U)} \ni \mathcal{I}(U)$. So then $\mathcal{I}(U)$ is an ideal containing a unit, so namely $\mathcal{I}(U) = \mathcal{O}_X(U)$ for any open set $U \subset X$.

But we assumed $\mathcal{I} \subsetneq \mathcal{O}_X$, a contradiction. So $\mathcal{I}(X) = 0$, as required.

$\endgroup$
3
  • 2
    $\begingroup$ Note that the restriction map of ideals are no ring homomorphisms, since ideals are no rings (in the sense of algebraic geometry any ring has a $1$), but the restriction map on the ideal sheaf is the restriction of the restriction map of the structure sheaf. And those are ring homomorphisms, hence map $1$ to $1$. $\endgroup$
    – MooS
    Commented Mar 13, 2016 at 18:33
  • 1
    $\begingroup$ So $\operatorname{res}^{\mathcal{I}}_{X,U} = \operatorname{res}^{\mathcal{O}_X}_{X,U}|_{\mathcal{I}(U)}$, and since $\operatorname{res}^{\mathcal{O}_X}_{X,U}(1) = 1$, we have $\operatorname{res}^{\mathcal{I}}_{X,U}(1) = 1$ as $1 \in \mathcal{I}(X)$? $\endgroup$
    – walkar
    Commented Mar 13, 2016 at 18:38
  • 1
    $\begingroup$ Yes, this is the correct reasoning. $\endgroup$
    – MooS
    Commented Mar 14, 2016 at 11:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .