TITLE: What is the geometric object corresponding to a subalgebra in a polynomial ring
QUESTION [19 upvotes]: Many introductory texts on algebraic geometry set up some sort of algebra-geometry dictionary in which radical ideals correspond to varieties, and so on. I am wondering if there is a geometric way to think about a subalgebra in a polynomial ring?  For instance for invariant rings one may develop some intuition, but what about infinitely generated subalgebras?  A simple example would be the infinitely generated monomial algebra
$k[x,xy, xy^2, xy^3, \dots]$ as a subalgebra in $k[x,y]$ for some field $k$.

REPLY [5 votes]: One should note that there is a formal analogue of this for arbitrary affine schemes. If $A,B$ are commutative rings, then a homomorphism $f:A \to B$ induces a continuous map $f^*:\mathrm{Spec}(B) \to \mathrm{Spec}(A)$.
One can show (see e.g. Exercise 1.21(v), Introduction to Commutative Algebra, Atiyah and Macdonald), the map $f^*$ is dominant, meaning that the image is dense (which, in topology, is good enough as being surjective) iff the kernel of $f$ is contained in the nilradical of $A$ (which is simply the radical of the ideal $0$).
Just as radical ideals rather than arbitrary ideals correspond to algebraic subsets, for any ring $A$, the natural map $\mathrm{Spec}(A/\mathfrak{a}) \to \mathrm{Spec}(A)$ is a homeomorphism whenever $\mathfrak{a}$ is an ideal consisting entirely of nilpotents.
Therefore, the map on $\mathrm{Spec}$'s, i.e. the geometric map associated to a map of rings, is dominant iff the map $f$ realizes $A$ modulo some nilpotents as a subalgebra of $B$. In particular, if $A$ is reduced, the map $f^*$ is dominant iff $f$ is injective.