Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 1 / d / Solution

Here subvariety means a closed affine subvariety. Suppose first that $F^{\#}$ is onto, and write $X\subseteq {\mathbb{A}}^n$, $Y\subseteq {\mathbb{A}}^m$. Put $J=\ker F^{\#}$ and $Z=V_Y(J)$. Choose $g_i\in A(Y)$ with $F^{\#}{g}_i=x_i{|}_X$ and let $G:Z\to {\mathbb{A}}^n$ have components $g_i{|}_Z$.

If $h\in I(X)$, then $F^{\#}(h(g_1,\dots ,g_n))=h{|}_X=0$, so $h(g_1,\dots ,g_n)\in J$. It vanishes on $Z$, proving $G(Z)\subseteq X$. The chosen lifts give $G\circ F={\mathrm{id}}_X$, and $F(X)\subseteq Z$ by the definition of $J$.

Let the components of $F$ be polynomial functions $f_1,\dots ,f_m$ on $X$. For each target coordinate $y_j$, the element $y_j-f_j(g_1,\dots ,g_n)$ has zero pullback, hence belongs to $J$. It vanishes on $Z$, giving $F\circ G={\mathrm{id}}_Z$. Both maps have polynomial components, so $F$ is an isomorphism onto the closed subvariety $Z$.

Conversely, if $F:X\to Z\subseteq Y$ is an isomorphism onto a closed subvariety, restriction $A(Y)\to A(Z)$ is surjective, and the pullback $A(Z)\to A(X)$ of that isomorphism is an isomorphism. Their composition is $F^{\#}$, provingThe word closed matters: an isomorphism onto a merely locally closed subset need not induce a surjective coordinate-ring map. For example ${\mathbb{G}}_m\hookrightarrow {\mathbb{A}}^1$ induces $k[t]\hookrightarrow k[t,t^{-1}]$.