2

One condition, (2) of Definition 25.4.1, for a morphism of ringed spaces $i:Z\rightarrow X$ to be a closed immersion is that $$O_X \rightarrow i_*O_Z$$

is surjective.

I have two confusions

(a) $i^*O_X \rightarrow O_Z$ is surjective does this show $O_X \rightarrow i_*O_Z$ is surjective? This doesn't seem to be the case for me. The counit of $i^*,i_*$ doens't seem to be special.

(b) I could prove $(i^*O_X)_x=O_{X,i(x)}$ but is it the case $(i_*O_Z)_{i(z)}=(O_Z)_{z}$?

I do not see why both cases have to be true, it would be nice if a counter example is provided too.

Bryan Shih
  • 9,934
  • b) No, the second result is just false in general, imagine for instance that $i$ isn’t injective: in $(i_*O_Z)_{i(z)}$, you would get contributions from the whole fiber above $i(z)$. – Aphelli Oct 18 '19 at 22:50
  • @Mindlack I think in this case the assumption is still that $i$ is a closed immersion so it's injective and you'd need different reasoning there. For the OP: you look like you have a couple typos here - one of those $i_$ should be a $i^$ in (a), and one of those arrows is going the wrong way. – KReiser Oct 18 '19 at 23:26
  • Terribly sorry, don't know how that happened... – Bryan Shih Oct 19 '19 at 02:43

1 Answers1

1

First, as the pullback of the structure sheaf is the structure sheaf, the answer to (a) is that the natural map $i^*\mathcal{O}_X\to \mathcal{O}_Z$ is always an isomorphism for any morphism of schemes. So you're right to be suspicious here.

Second, $(i_*\mathcal{O}_Z)_{i(z)} = \mathcal{O}_{Z,z}$ is true. You can find this result on wikipedia or in most introductory algebraic geometry books. The point is that we can combine the definition $\Gamma(U,i_*\mathcal{F})=\Gamma(U\cap Z,\mathcal{F})$ for any open $U\subset X$ and any sheaf $\mathcal{F}$ on $Z$ with the fact that $Z$ has the subspace topology and every open set in $Z$ comes from the intersection of some open set in $X$ with $Z$.

KReiser
  • 74,746