= Solution
Suppose first that $\overline{f(X)}=Y$. If $g\in\ker f^*$, 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 $f^*$ 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 $f^*(g)=0$, contradicting injectivity. Therefore
$$
\boxed{\ \overline{f(X)}=Y\iff f^*:k[Y]\to k[X]\text{ is injective}.\ }
$$
This is the coordinate-ring criterion for a <dominant morphism of affine varieties>.
For an irreducible affine variety, define dimension by
$$
\dim X=\operatorname{trdeg}_k k(X),
\qquad k(X)=\operatorname{Frac}(k[X]).
$$
Irreducibility makes both coordinate rings <integral domains>. The injective map $f^*:k[Y]\hookrightarrow k[X]$ extends to an embedding of function fields
$$
k(Y)\hookrightarrow k(X).
$$
An algebraically independent family in $k(Y)$ remains algebraically independent in $k(X)$, so the <dimension of an irreducible affine variety by transcendence degree> gives
$$
\boxed{\ \dim X\geq\dim Y.\ }
$$
Back to article page
