Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've heard that the idea of coordinate rings creates a language for talking about affine varieties without embedding them into affine space. But I don't really understand/agree. The definition of the coordinate ring of an affine variety $Y$ is $k[x_1,...,x_n]/I(Y)$. But if we know what $k$ and $n$ are, aren't we (at least implicitly) embedding $Y$ into some affine space?

Is there a way of interpreting this statement sensibly?

share|cite|improve this question
Probably some dependency on $k$ should be kept -- the old language is not so nice if you start changing base fields. But what if you just look at the category of finitely generated $k$-algebras that are also integral domains? Sure, the ones you'll write down will come with some prejudice in the form of the $x_i$, but they don't have to. – Dylan Moreland Feb 8 '12 at 1:36
up vote 6 down vote accepted

You can abstractly characterize the coordinate rings $A$ of (irreducible) affine varieties over a field $k$ as finitely-generated integral domains over $k$. This definition doesn't require that you pick a choice of generators, only that some finite generating set exists; such a choice of generators is equivalent to the choice of a surjection $$k[x_1, ... x_n] \to A$$

which in fact is equivalent to the choice of an embedding of $\text{Spec } A$ into affine space $\mathbb{A}^n$.

This flexibility is convenient when you want to construct new varieties out of old varieties. For example, let $A$ be the coordinate ring of some affine variety and $G$ a finite group which acts on the variety by algebraic maps. Then it acts on $A$ by algebra homomorphisms. The invariant subalgebra $$A^G = \{ a \in A : \forall g \in G, ga = a \}$$

is known to be finitely-generated, but it doesn't come with a distinguished set of generators, so it defines an affine variety $\text{Spec } A^G$ which models the quotient $(\text{Spec } A)/G$ and which doesn't come with a preferred embedding into affine space even if $A$ does.

share|cite|improve this answer
+1: This is eerily similar to the answer I wanted to type out but couldn't spare the time/effort for. In particular I wanted to give the example of taking invariants (maybe just of the full polynomial ring) under a finite group as an example where you have a theorem which tells you that this is isomorphic to some quotient $k[t_1,\ldots,t_n]/I$ but doesn't supply a canonical $n$ and $I$. Great minds, I suppose... – Pete L. Clark Feb 8 '12 at 4:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.