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\title{Math 631, Lecture 4}
\author{Notes by Bryden Cais}
\date{Sept. 13, 2002}
\begin{document}
\maketitle

In this lecture we will define and study maps between varieties.

\begin{defn}
    A \textbf{regular map} or \textbf{morphism} between two affine $k$ varieties $X,Y$ is a collection of maps $\{f_K\}$ for all
    $k$ algebras $K$ such that for any homomorphism of $k$ algebras $\varphi:K\longrightarrow K^{'}$ the following
    diagram is commutative:
    \begin{eqnarray*}
         \begin{CD}
          X(K)   @>f_K>>  Y(K)\\
        @V\varphi^{\bigoplus n} VV      @VV \varphi^{\bigoplus n} V\\
          X(K^{'})   @>f_{K^{'}}>>     Y(K^{'})
            \end{CD}
    \end{eqnarray*}
    where $\varphi^{\bigoplus n}:K^n\longrightarrow K^{{'}n}$ is the natural map induced by $\varphi$ and $X(K)\subset K^n, Y(K)\subset K^m$.
    We will denote the set of morphisms from $X$ to $Y$ by $\mor(X,Y).$
\end{defn}


\begin{defn}
    The \textbf{coordinate ring} of a $k$ variety $X$ is the ring $$\O(X):=k[T]/I(X).$$
\end{defn}

Recall that we have a bijection $X(K)\leftrightarrow \hom{\O(X)}{K}$.  Let us now look at maps between varieties in these terms.
Given a homomorphism $\varphi:\O(Y)\rightarrow \O(X),$ we have a natural map of sets
$$\hom{\O(X)}{K}\longrightarrow \hom{\O(Y)}{K}$$
given by
$$\psi\mapsto \psi\circ\varphi.$$  Therefore, any homomorphism (of $k$ algebras) $\varphi\O(Y)\rightarrow\O(Y)$
defines a morphism $f:X\rightarrow Y$ by setting, for each $K$ and any $\alpha\in\hom{\O(X)}{K}=X(K)$
$$f_K=\alpha\circ\varphi.$$
Observe that $f_K:\O(Y)\rightarrow K$ is in $Y(K)$ (as it ought to be).
We thus have a natural map
$$\xi:\hom{\O(Y)}{\O(X)}\longrightarrow \mor(X,Y)$$ given by
$$\xi(\psi)(\alpha):=\alpha\circ\psi\in\hom{\O(Y)}{K}$$
for any $\alpha\in \hom{\O(X)}{K}$ (i.e. $\xi(\psi):X(K)\rightarrow Y(K)$).

\begin{thm}
The map $\xi$ defined in this way is in fact a bijection.
\end{thm}

\begin{proof}
Let $f:X\rightarrow Y$ be a morphism.  Then
$$f_{\O(X)}:\hom{\O(X)}{\O(X)}\longrightarrow \hom{\O(Y}{\O(X)}.$$
The image of the identity homomorphism $\id_{\O(X)}\in\hom{\O(X)}{\O(X)}$ under $f_{\O(X)}$
is some homomorphism $\psi\in\hom{\O(Y)}{\O(X)}.$  If we can show that $\xi(\psi)=f$ then
we will have shown that $\xi$ is surjective.  To do this, we must show that for any $k$ algebra $K$ and any $\alpha\in\hom{\O(X)}{K}$
we have $\xi(\psi)(\alpha)=f_K(\alpha)$.  This follows from the definition of a morphism: the diagram

\begin{eqnarray*}
         \begin{CD}
         X(\O(X))=\hom{\O(X)}{\O(X)}   @>f_{\O(X)}>>     \hom{\O(Y)}{\O(X)}=Y(\O(X))\\
        @V\alpha\circ VV                                    @VV \alpha\circ V\\
         X(K)=\hom{\O(X)}{K}               @>f_K>>         \hom{\O(Y)}{K}=Y(K)\\
         \end{CD}
\end{eqnarray*}
is commutative.  We have $\id_{\O(X)}\in \hom{\O(X)}{\O(X)}$ mapping to $\alpha\circ\id_{\O(X)}=\alpha:\O(X)\rightarrow K$.
The image of $\alpha$ in $\hom{\O(Y)}{K}=Y(K)$ under $f_K$ is just $f_K(\alpha)$.  By commutativity of the diagram, this must
coincide with the image of $\psi\in\hom{\O(Y)}{\O(X)}$ under $\alpha\circ$ in $\hom{\O(Y)}{K}$, which is just $\alpha\circ\psi=\xi(\psi)(\alpha).$
We therefore have
$$\xi(\psi)(\alpha)=f_K(\alpha)$$ for any $\alpha\in\hom{\O(X)}{K}$, and therefore that $\xi(\psi)=f$ and we are done with surjectivity.
To prove that $\xi$ is injective, let $\psi,\phi\in\hom{\O(Y)}{\O(X)}$ be distinct and suppose that
$\xi(\psi)=\xi(\phi).$  Then for all $k$ algebras $K$ and any $\alpha\in\hom{\O(X)}{K}$ we have
$$\alpha\circ\psi=\alpha\circ\phi.$$  However, if we choose $K=\O(X)$ and $\alpha=\id_{\O(X)}\in\hom{\O(X)}{\O(X)}$ then
we must have $\psi=\phi,$ which shows that $\xi$ is injective.  Therefore, $\xi$ is a bijection, as claimed.
\end{proof}

This bijection allows us to think of morphisms between varieties $X,Y$ as homomorphisms of the corresponding coordinate rings, going in the
reverse dirction: $\O(Y)\rightarrow \O(X)$ corresponds to a morphism $X\rightarrow Y$.
\begin{defn}
    An \textit{isomorphism} of $k$ varieties $X,Y$ is an invertible morphism.
\end{defn}

Using this definition and our interpretation of morphisms $X\rightarrow Y$ as homomorphisms $\O(Y)\rightarrow \O(X)$,
we see that

\begin{prop}
    Two affine algebraic $k$ varieties $X,Y$ are isomorphic if and only if their coordinate $k$ algebras $\O(X)$ and $\O(Y)$
    are isomorphic.
\end{prop}

So we have this fine identification of $\mor(X,Y)$ with $\hom{\O(Y)}{\O(X)},$ \textit{so what}?  The point is that this enables us
to prove something really profound:

\begin{thm}
    \textit{Every} morphism $f:X\rightarrow Y$ of affine $k$ varieties is given by a collection of polynomials.
\end{thm}

\begin{proof}
    Any morphism $f:X\rightarrow Y$ corresponds to a homomorphism $\psi:\O(Y)\rightarrow \O(X)$ with
    $f_K(\alpha)=\alpha\circ\psi:\O(Y)\rightarrow K$ for any $\alpha\in \hom{\O(X)}{K}$.  Now $\O(Y),\O(X)$
    are quotients of the polynomial rings $k[U_1,\ldots,U_r],k[T_1,\ldots,T_s]$ respectively, so $\psi$
    is determined by where it sends the generators $U_i$.  Moreover, these generators must be cosets of
    $I(X)$ in $k[T_1,\ldots,T_s],$ that is
    $$\psi(U_i)=P_i(T_1,\ldots,T_s)+I(X)$$ for $P_i\in k[T_1,\ldots,T_s]$.
    Now since $\psi$ as a homomorphism $k[U_1,\ldots,U_r]\rightarrow k[T_1,\ldots,T_s]/I(X)$
    factors through $I(Y)$, we have, for any $F\in I(Y)$,
    $$F(P_1(T_1,\ldots,T_s),\ldots,P_r(T_1,\ldots,T_s))\in I(X).$$
    Our morphism $f:X\rightarrow Y$ is given by $f_K(\alpha)=\alpha\circ\psi$.  Recall, however, that
    $\alpha\in\hom{\O(X)}{K}$ is given by sending $T_i\mapsto a_i\in K$ for $1\leq i\leq s $.  Therefore,
    we have
    $$\alpha\circ\psi (U_i)=\alpha\circ (P_i(T_1,\ldots,T_s)+I(X))=P_i(a_1,\ldots,a_s)$$
    since $I(X)$ is contained in the kernel of the map $\alpha:k[T_1,\ldots,T_s]\rightarrow K$.
    Therefore, we have
    $$f_K(a_1,\ldots,a_s)=(P_1(a_1,\ldots,a_s),\ldots,P_r(a_1,\ldots,a_s))$$
    for any $(a_1,\ldots,a_s)\in X(K)$.  In short, $f$ is given by a collection of polynomials.
\end{proof}

Recall that we can interpret points $a\in \sol(S,K)$ for any system of algebraic equations $S$ defining a variety $X$ as homomorphisms
$a:\O(X)\rightarrow K$ by defining $a(p(T)):=P(a)$ for any $P\in k[T]$.  This kind of \textbf{duality} is ubiquitous in algebraic
geometry.  In this case, we move from considering elements of $k[T]$ as functions on $K^n$ to considering elements of $K^n$
as functions on $k[T]$.  Another example is the following: any morphism
$$X\longrightarrow \A^1_k$$ may be thought of as a homomorphism $$k[T_1]\longrightarrow \O(X).$$  Any such homomorphism
is determined by its value at $T_1$.  We can therefore think of such homomorphisms as \textit{elements} of $\O(X)$.
In summary,
\begin{prop}
    There is a bijection between elements of $\O(X)$ and morphisms $f:X\longrightarrow \A_k^1.$
\end{prop}

This enables us to make sense out of the following: let $f\in\mor(X,Y)$.  Then we have a homomorphism
$$f^*:\O(Y)\longrightarrow\O(X).$$  Now let $\varphi\in\mor(Y,\A_k^1)=\O(Y)$.  Then the composition
of morphisms $\varphi\circ f$ is well defined and is a map $X\longrightarrow \A^1_k$, i.e. an element
of $\O(X)$.  By our correspondence, we therefore have
$$f^*(\varphi)=\varphi\circ f.$$
\end{document}
