Properties of Morphism of a scheme which is both reduced and irreducible.

Question

Let, $f:X \to Y$ be a morphism of schemes with $X$ being both reduced and irreducible.

1. If the morphism $f|_{U}$ is constant, $\forall$ affine open $U \subset X$,. Show that $f(x)=f(y), \forall x,y \in X$

2. Let $Y=\operatorname{Spec}(A)$ . For every affine open $U=\operatorname{Spec}(B)$ of $X$, the homomorphism $A \to B$ associated to $f|_{U} : U \to Y$ maps inside the subring $\Gamma(X,O_X)$

For Q.1, $X$ is irreducible, so $X$ is connected. So basically have to show that given fixed $x_0 \in X$, $\{x \in X:f(x)=f(x_0)\}$ is both open and closed. Now how to use the affine subsets to glue them together? And also by using the fact that $X$ doesn't have any zero divisors. I'm stuck here.

For Q.2, having no intuition regarding this one!

Thank you for help

Answer 1

For 1., let $x_0 \in X$ and consider the inverse image of the closure of $f(x_0)$: it is a closed subset of $X$ and contains some affine open subset (dense) so is closed. So if $x \in X$, $f(x)$ is a specialization of $f(x_0)$.

As schemes are $T_0$, this shows that $f$ is constant.

For 2, note that if $f$ induces morphisms $f^{\sharp}_{U,V}: \mathcal{O}_Y(V) \rightarrow \mathcal{O}_X(U)$ compatible with restriction for $f(U) \subset V$.

This means that $f^{\sharp}_{Y,U}$ is the composition of the restriction and $f^{\sharp}_{Y,X}$, which makes the claim easy.

Answer 2

Part 1 follows from the fact that each fiber of $f$ is an open set and $X$ is a disjoint union of the fibers of $f$

For part 2 you have the morphism of schemes $$U\overset{j}\hookrightarrow X\xrightarrow{f}Y$$.

Observe that $f|_U=f\circ j$

It then follows that $f|_U^\#=j^\#\circ f^\#$

Thus $$f|_U^\#(A)=j^\#\circ f^\#(A)\subset j^\#(\Gamma(X,\mathcal O_X))$$

Thus we only need to show $j^\#$ is injective so u can identify $\Gamma(X,\mathcal O_X)$ as a subring of $B$. Suppose $f\in \mathcal O(X)$ such that $f|_U=0$. We want to show $f=0$ or $f_x=0 \ \forall \ x\in X$. Choose a point $x$ and an affine open set $V=\operatorname{Spec} R$ containing $x$. Then $R$ is an integral domain as $X$ is reduced and irreducible. Also $V\cap U\neq \phi$ by irreducibility. Let $p\in V \cap U$ be a prime ideal of $R$. Since we have $f=0$ in $R_p$, $f=0$ in $R$. Thus $f_x=0$ and this shows $f=0$.

Now you can conclude.