Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 25I / c / Solution

Suppose first that $\overline{f(X)}=Y$. If $g\in \ker f^{\ast }$, then $g\circ f=0$, so the zero set of the regular function $g$ contains $f(X)$ and hence its closure $Y$. Thus $g=0$ in the coordinate ring $k[Y]$, and $f^{\ast }$ is injective.

Conversely, if $\overline{f(X)}$ were a proper closed subset of $Y$, its vanishing ideal in $k[Y]$ would contain a nonzero regular function $g$. Then $g$ vanishes on $f(X)$, so $f^{\ast }(g)=0$, contradicting injectivity. ThereforeThis is the coordinate-ring criterion for a dominant morphism of affine varieties.

For an irreducible affine variety, define dimension byIrreducibility makes both coordinate rings integral domains. The injective map $f^{\ast }:k[Y]\hookrightarrow k[X]$ extends to an embedding of function fieldsAn algebraically independent family in $k(Y)$ remains algebraically independent in $k(X)$, so the dimension of an irreducible affine variety by transcendence degree gives