Pullback of global sections of a vector bundle over a rational point

118 Views Asked by At

Suppose I've got some projective curve $X$ over a field $K$ of characteristic $0$, and $z \in X(K)$ a rational point $z: \text{Spec}( K) \rightarrow X$. Now suppose, for simplicity, we consider the structure sheaf on $\mathcal{O}_X$.

I think that the way to think about $z^{*}\mathcal{O}_{X}$ is as evaluation of the sections of $g \in \mathcal{O}_X(X)$ at the point $Q=z(P) \in X$ where $P$ is the point in $\text{Spec}(K)$ corresponding to the zero ideal. Now, I know that given $z$ we have an obvious map $\mathcal{O}_X(X) \rightarrow K$, but I'd like to be clearer about what it is.

It's clear that $z^*\mathcal{O}_X= \mathcal{O}_{\text{Spec}(K)}$. This is because we have

$$ z^*\mathcal{O}_X = z^{-1} \mathcal{O}_X \otimes_{z^{-1} \mathcal{O}_X} \mathcal{O}_{\text{Spec}(K)}$$

Now

$$z^{-1} \mathcal{O}_X : U \mapsto \lim_{\substack{ \rightarrow \\ V \supseteq b(U)}} \mathcal{O}_X(V) $$

The only open sets on $\text{Spec}(K)$ are $\emptyset$ or the total space. So then we either find $z^{-1} \mathcal{O}_X(U)=0$ if $U= \emptyset$ or $z^{-1} \mathcal{O}_X = \mathcal{O}_{X,Q}$ otherwise. Hence, $z^*\mathcal{O}_X(U) = 0$ if $U = \emptyset$ and $z^*\mathcal{O}_X(U) = \mathcal{O}_{X,Q} \otimes_{\mathcal{O}_{X,Q}} K$ otherwise.

If we let $U$ be a small open affine $\text{Spec}(R)$ on $X$ containing $Q$, and let $g$ be some global section of $\mathcal{O}_X$, then $\mathcal{O}_{X,Q}=R_Q$. I believe then that evaluation of $g$ at $Q$ is viewed as the composition

$$ g \mapsto g' = \text{res}_{X,U}(g) \in R \mapsto g_Q \in R_Q \mapsto \overline{g_Q} \in k(Q) \hookrightarrow K$$

where the kernel of the third map is going to the be ideal $Q \cdot R_Q$. Let $\text{ev}_Q : \mathcal{O}_X(X) \rightarrow K$ be the evaluation map above.

So what I'm thinking then is something like this: if we have a global section $g$ of $\mathcal{O}_X$, then we find that in $z^*\mathcal{O}_X(U) = \mathcal{O}_{X,Q} \otimes_{\mathcal{O}_{X,Q}} K$, the element $g_Q \otimes 1 = 1 \otimes \text{ev}_Q(g) \mapsto \text{ev}_Q(g) \in K$. So then we can view the pullback above as being evaluation of the sections at $Q$.

Does this look vaguely right?