# Hartshorne Proof Question

• Aug 3rd 2010, 02:22 PM
xboxlive89128
Hartshorne Proof Question
I have a question from Algebraic geometry by Robin Hartshorne. My question is regarding a proof in this book. The proof is the following:

Lemma. Let $X$ be a variety and $Y \subseteq \mathbb{A}^n$ be an affine variety. Let $x_1, \ldots, x_n$ be the coordinate functions on $\mathbb{A}^n$. A map of sets $\psi : X \rightarrow Y$ is a morphism if and only if $x_i \circ \psi$ is a regular function on $X$ for each $i=1, \ldots, n$.

Theorem. Let $X$ be a variety and $Y$ be an affine variety. Then there is a natural bijective mapping of sets
$\alpha : \text{Hom}(X, Y) \overset{\sim}{\rightarrow} \text{Hom}(A(Y), \mathcal{O}(X))$. Here the leftside indicates morphisms of varieties and the rightside indicates homomorphisms of $k$-algebras.

Proof. Let $\phi: X \rightarrow Y$ be a morphism. Then $\phi$ takes regular functions on $Y$ to regular functions on $X$. So $\phi$ induces a map $\mathcal{O}(Y)$ to $\mathcal{O}(X)$ . This is clearly a homomorphism of $k$-algebras. However, by a result in Hartshorne, $\mathcal{O}(Y) \cong A(Y)$. So we get a homomorphism $A(Y) \rightarrow \mathcal{O}(X)$. This defines $\alpha$.

Conversely, let $h : A(Y) \rightarrow \mathcal{O}(X)$ be a homomorphism of $k$-algebras. Suppose $Y$ is a closed subset of $\mathbb{A}^n$ such that $A(Y)=\frac{k[x_1, \ldots, x_n]}{I(Y)}$. Let $\overline{x_i}$ be the image of $x_i$ in $A(Y)$. Now consider the elements $\sigma_i=h(\overline{x_i}) \in \mathcal{O}(X)$. The $\sigma_i$ are global functions on $X$. So we can define a map $\psi : X \rightarrow \mathbb{A}^n$ by $\psi(P)=(\sigma_1(P), \ldots, \sigma_n(P))$ where $P \in X$.

Now we need to show that the image of $\psi$ is contained in $Y$ . Because $Y=V(I(Y))$ , it suffices to show that for any $P \in X$ and any $f \in I(Y)$, $f(\psi(P))=0$. However, $f(\psi(P))=f(\sigma_1(P), \ldots, \sigma_n(P))$. Note that $f$ is a polynomial and $h$ is a homomorphism of $k$-algebras, so we have $f(\sigma_1(P), \ldots, \sigma_n(P))=h(f(\overline{x_1}, \ldots , \overline{x_n} ))(P)=0$ because $f \in I(Y)$. So $\psi$ defines a map from $X$ to $Y$ that induces the homomorphism $h$. It now remains to show that $\psi$ is a morphism. This is a consequence of the lemma.

I don't understand this part of the proof:

It now remains to show that $\psi$ is a morphism. This is a consequence of the lemma.

Why is $\psi$ a morphism? How is this a consequence of the lemma? I can't figure this out. Thank you.