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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?

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    $\begingroup$ 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. $\endgroup$ – Dylan Moreland Feb 8 '12 at 1:36
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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.

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    $\begingroup$ +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... $\endgroup$ – Pete L. Clark Feb 8 '12 at 4:17

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