Sheaf morphism from closed subscheme is a closed immersion

Question

For $K=\bar{K}$ a field consider $X=\mathbb P^1_K$, $Z=\{P_1,\dots,P_n\}\subseteq X$ closed points. Give $Z$ the reduced induced closed subscheme structure and write $\iota:Z\to X$ the closed immersion. I now want to show that $\iota^{\#}:\mathcal O_X\to i_{*}\mathcal O_Z$ is indeed surjective.

To do so we can show surjectivity on stalks, which seems the easiest to me. Now we now what stalks on $X$ look like. Using exercise 1.19 (extending a sheaf by zero) we get that if $p\notin Z$ $(i_{*}\mathcal O_Z)_p=0$ if I'm correct.

Now the idea is to see what the others stalks look like, but I don't understand how sections look like with this sheaf. For example what is $\mathcal O_Z(Z)$ and for a more general case what are the global sections of the reduced induced closed subscheme structure for $Y\subseteq X$ ? And what does $\iota^{\#}$ really do, because since I don't understand the rings of $\mathcal O_Z$ I don't see how it acts.

Answer 1

Conceptually, we can think of $\mathcal{O}_Z$ as the sheaf of regular functions on a discrete set of points. Now, on any individual point, the only regular functions are constant, and our space is totally disconnected, so we would expect $\mathcal{O}_Z$ to be a direct sum of skyscraper sheaves, valued at $K$. Fortunately, our expectation is correct.

There is a couple of ways to see this. Personally, I like to reduce to the affine case, since this is often the strategy in this part of Hartshorne. To do this, we use the following lemma, which is not difficult to prove.

Let $f: X \to Y$ be a morphism of schemes so that $f(X) \subset U$ for an open subset $U \subset Y$. Then $f$ factors $X \to U \hookrightarrow Y$.

Now, assume $\infty \notin \{P_1, \dots, P_n\}$, so that the closed embedding $i: Z \hookrightarrow \mathbb{P}^1$ factors as $Z \hookrightarrow \mathbb{A}^1 \hookrightarrow \mathbb{P}^1$ where the first map is a closed embedding. In particular, we see that $Z$ is a closed subset of affine space, so that its cut out by an ideal $I(Z)$. If $P_i$ correspond to the maximal ideals $(x - a_i) \subset K[x]$, then we see that $$I(Z) = \prod_{i = 1}^n (x - a_i)$$

and that $Z \cong \operatorname{Spec} K[x]/I(Z)$. As such, the closed embedding $Z \hookrightarrow \mathbb{A}^1$ is exactly the morphism induced by the quotient $$K[x] \to K[x]/I(Z) \cong \prod_{i = 1}^n K[x]/(x - a_i) \cong K^n$$ where the first isomorphism is given by the Chinese remainder theorem.

This then completely describes $\mathcal{O}_Z$, and we see that the map $i^\#$ is something very concrete. For a regular function $\varphi \in \mathcal{O}_\mathbb{P}^1(U)$, it is exactly the map $$\varphi \mapsto (\varphi(P_{i_1}), \dots, \varphi(P_{i_j}))$$ where $\{P_{i_1}, \dots, P_{i_j}\} = Z \cap U$, which fits in exactly with our expectation.