$X$ is a scheme over $k$. Considering the map $f:X\to k$

Question

This is a really basic question sorry.

Say $X$ is a scheme over $k$. Then this means that we have a morphism of schemes $X\to \text{Spec}(k)$ right. If $k$ is a field, then $\text{Spec}(k)=\{(0)\}$, and this has sheaf sending $\{(0)\}$ to $k$. So we have that $f$ sends everything to $(0)$.

Now we also need a morphism of sheaves $f^\# :O_k \to f_*O_X$, but there are no restriction maps to consider a commuting square with for $O_k$.

So I suppose I just need $f^\#: O_k((0))\to f_* O_X((0))$ $$f^\#:k\to O_X(X),$$

so this just requires a homomorphism from $k$ into $O_X(X)$?

Am I misunderstanding anything? What is the significance of requiring this morphism. Thank you

Answer 1

Thanks to Roland for the comment above.

Let $U\subset X$, then $O_X(U)$ has $k$-algebra structure induced from the composition: $$k\to O_X(X)\to O_X(U).$$