2
$\begingroup$

Let $X$ be a scheme over a field $k$. Let $R$ be the $k$-algebra of global sections of $X$'s structure sheaf.

Is there any way to view scheme morphisms $X\rightarrow \mathrm{Spec}(k[t])$ as elements of $R$? Or are these two notions unrelated?

(Apologies if this question is ill-formed; I have little background in schemes.)

$\endgroup$

2 Answers 2

4
$\begingroup$

If you stipulate that $X\to \operatorname{Spec}(k[t])$ is a morphism of $k$-schemes, then yes! In this case, $\pi:X\to \operatorname{Spec}(k[t])$ corresponds to a $k$-algebra homomorphism of global sections $\pi^\#:k[t]\to R$. In turn, this corresponds to $\pi^\#(t)\in R$ because a $k$-algebra homomorphism from $k[t]$ is uniquely determined by the image of $t$.

$\endgroup$
3
$\begingroup$

You absolutely can. For any ring $A$ and scheme $X$, we have a natural bijection $$Hom( X, Spec(A) ) \to Hom( A, \Gamma( X, O_X) ), $$ where the left side is scheme morphisms and the right side is ring morphisms. See Hartshorne exercise II.2.4. This bijection can be upgraded to that between $k$-schemes and $k$-algebra morphisms in which case, as user816709 points out, your result follows.

$\endgroup$

You must log in to answer this question.

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