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

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 $$f(x)=f(y), \forall x,y \in X$$
1. 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

## 2 Answers

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.

• I've upgraded your post to use \operatorname{Spec} to format $\operatorname{Spec}$, which produces proper spacing. Jun 21, 2020 at 23:38

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.

• Thanks a lot. But I am completely new to Algebraic Geoemtry, can you be a bit more in details for Q 2? Are you saying that just look at the Global section of the ringed map?
– user802191
Jun 21, 2020 at 20:33
• What I’m saying is that Q2 is one of the numerous “abstract nonsense” questions on schemes that just make you want to pull your hair out because you just aren’t sure of what’s to prove anymore. But yes, the think is to wrote down as properly as you can the action on global sections of the maps $U \rightarrow X$ and $f: X \rightarrow Y$, since their composition is the spectrum of the homomorphism $A \rightarrow B$. Jun 21, 2020 at 20:38
• Okay, but sorry to say that I am still not getting it. Can you kindly write a bit more regarding Q 2?
– user802191
Jun 21, 2020 at 20:41
• The homomorphism $A \rightarrow B$ is the global section of the map of schemes $U \rightarrow Y$, which is $U \rightarrow X$ followed by $f$. So, taking the actions on the global sections, $A \rightarrow B$ is the composition of $A \rightarrow \mathcal{O}_X(X)$ (coming from $f$) and the natural map $\mathcal{O}_X(X) \rightarrow B$ (the restriction). Jun 21, 2020 at 20:48