By Thomas Kahle


2012-07-02 20:14:36 8 Comments

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 $k$.

4 comments

@David Corwin 2012-08-06 18:15:50

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.

@Will Sawin 2012-07-02 22:52:32

Because geometry and algebra are connected by the contavariant $\textrm{Spec}$ functor, a subalgebra corresponds to the image of a dominant morphism. So subalgebras of $k[x_1,..,x_n]$ are affine schemes with dominant maps from $\mathbb A^n_k$.

Sometimes, like in your example, this is in addition a surjective morphism, so it looks like the quotient of $\mathbb A^n$ by some equivalence relation. But sometimes, the simplest example being $k[x,xy] \subset k[x,y]$, it fails to be surjective. Here there is a map from the plane to the plane whose image is the plane minus a line plus a point.

Since the image is always dense, we can geometrically think of it as the quotient of $\mathbb A^n_k$ by an equivalence relation followed by a completion or partial compactification.

@Martin Brandenburg 2012-07-02 20:33:15

Sometimes these algebras are fiber products along surjective homomorphisms, and Karl Schwede's paper on Gluing Schemes tells us how the spectrum looks like. For example:

1) The spectrum of $k[x,xy,xy^2,\dotsc] \cong k[x,y] \times_{k[y]} k$ is $\mathbb{A}^2_k$ with the line $x=0$ contracted to a point. This is example 3.5 in the cited paper.

2) The spectrum of $k[x^2,x^3] \cong k[x] \times_{k[x]/(x^2)} k$ is $\mathbb{A}^1_k$ with a rational point doubled, i.e. the cuspidal curve.

3) The spectrum of $k[x^2-1,x^3-x] \cong k[x] \times_{k[x]/((x-1)(x+1))} k$ is $\mathbb{A}^1_k$ with two rational points identified, i.e. the nodal curve.

4) The spectrum of $\{(a,b) \in \mathbb{Z}^2 : a \equiv b \text{ mod } 6\}$ is the gluing of two copies of $\mathrm{Spec}(\mathbb{Z})$ along the closed points $(2),(3)$. Interestingly, if you localize at all primes $\neq 2,3$, you will get an affine scheme which is weakly homotopy equivalent to $S^1$, the Pseudocircle.

@MTS 2012-07-02 21:20:48

I never saw the pseudocircle before. What a fascinating thing!

@Martin Brandenburg 2012-07-03 14:24:35

I think a combination of results by M. Hochster and E. Clader shows that every geometric simplicial complex is homotopy equivalent to an affine scheme. Is this known at all?!

@Paul Siegel 2012-07-03 14:37:37

This answer kind of rocks my world. Particularly the pseudocircle bit...

@Will Sawin 2012-07-03 19:35:50

@Martin: How would you get a triangle, say?

@Mariano Suárez-Álvarez 2012-07-04 03:05:01

Maybe it is only a weak homotopy equivalence?

@DES-SupportsMonicaAndTransfolk 2012-07-02 23:03:12

One should mention the special case where there is a group $G$ acting on $k[x_1, \ldots, x_n]$ and $A$ is the algebra of invariant functions. Then, by definition, $\mathrm{Spec} \ k[x_1, \dots, x_n]^G$ is the GIT quotient $\mathbb{A}^n//G$. The question of how this is related to the topological quotient $\mathbb{A}^n/G$ is subtle, but in many cases they are equal or extremely close to equal. Look up a textbook on Geometric Invariant Theory for details.

Related Questions

Sponsored Content

1 Answered Questions

9 Answered Questions

1 Answered Questions

1 Answered Questions

1 Answered Questions

[SOLVED] Rings with finitely generated nilradical

Sponsored Content