Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 21F (Algebraic Geometry)

For affine varieties $X\subset {\mathbb{A}}^m$ and $Y\subset {\mathbb{A}}^n$, a morphism of affine varieties is a map whose $n$ coordinate functions are restrictions to $X$ of polynomials in its $m$ ambient coordinates. Intrinsically, its coordinates belong to the coordinate ring $A(X)$; they must satisfy the defining relations of $Y$.

A morphism induces the pullback $f^{\ast }:A(Y)\to A(X)$, $g\mapsto g\circ f$, a unital $k$-algebra homomorphism. Conversely, given such a homomorphism $\phi$, take polynomial representatives $g_j$ on $X$ of the images of the coordinate classes $y_j$. Define $f(x)=(g_1(x),\dots ,g_n(x))$. If $P\in I(Y)$, then $\phi (\bar{P})=0$, so $P(g_1(x),\dots ,g_n(x))=0$ on $X$. Therefore the values lie in $Y$. The resulting map is a morphism and has pullback $\phi$ because coordinate classes generate $A(Y)$. Conversely recovering coordinate functions from $f^{\ast }$ returns $f$. This proves the bijection between morphisms and contravariant coordinate-ring homomorphisms.

Write the target coordinates as $x_1,x_2,x_3,x_4$. Every image point satisfies $x_3=x_2^2$ and $x_4=x_1{x}_2$. Conversely any point satisfying these equations is obtained by taking $t=x_2$, $u=x_1$. Hence the image is the closed affine varietyThe substitution homomorphism $k[x_1,x_2,x_3,x_4]\to k[u,t]$ has kernel exactly the ideal $J=(x_3-x_2^2,x_4-x_1{x}_2)$: reduce any polynomial by these two relations to a polynomial in $x_1,x_2$, which vanishes after substitution only if it is zero. Equivalently the quotient by $J$ is $k[x_1,x_2]$, so $J$ is prime. Since $k$ is algebraically closed, or directly by the same polynomial-vanishing argument, $I(X)=J$.