# $Specm(A)$ affine variety $\Leftrightarrow$ A is finitely generated $k$-algebra

I want to prove that "the spectrum of maximal ideals of a ring $A$ is a variety of $\mathbb{A}^n_k$ for some $n$ if and only if $A$ is a finitely generated $k$-algebra". I assume that $k$ is algebraically closed.

Any hints on how to make a start for each direction?

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What is the definition of an affine variety over $k$? –  Makoto Kato Oct 4 '12 at 1:33
What do you mean by "the spectrum is an affine variety"? A priori the spectrum is a set (maybe a topological space) and it doesn't make sense to ask whether a set (or a topological space) is a variety. –  Qiaochu Yuan Oct 4 '12 at 1:36
@QiaochuYuan: Why not? A variety is both a set and a topological space. –  Manos Oct 4 '12 at 1:42
@MakotoKato: I suppose an affine variety of $\mathbb{A}^n_k$ for some $n$. –  Manos Oct 4 '12 at 1:44
You probably mean to prove something like: $A$ is a finitely generated $k$-algebra iff there exists a subvariety of $\Bbb A^n_k$ whose coordinate ring is isomorphic to $A$, and under this isomorphism one may identify points on the subvariety with maximal ideals of $A$. –  Andrew Oct 4 '12 at 2:20

The formulation of the question given by Andrew in the comments is meaningful but false (take $A = k[x]/x^2$). The correct statement comes from replacing "finitely generated $k$-algebra" with "finitely generated integral domain over $k$" and follows from the Nullstellensatz.
Just to make sure i understand what "finitely generated ring over $k$" is: a ring $A$ that is a $k$-module and is finitely generated as a $k$-module? –  Manos Oct 4 '12 at 15:37
@Manos: no, it means finitely generated as a $k$-algebra. People use "finite" to mean finitely generated as a $k$-module. –  Qiaochu Yuan Oct 4 '12 at 16:10
So finitely generated integral domain over $k$ is just a finitely generated $k$-algebra with no zero divisors...Could you please also give me some insight into your counterexample? I can see that $k[x]/x^2$ is not an integral domain. How do you see that its spectrum of maximal ideals is not isomorphic to a variety of $\mathbb{A}^n_k$? –  Manos Oct 4 '12 at 16:15
@Manos: again, that question doesn't make sense. What I'm actually claiming is that $k[x]/x^2$ is not isomorphic to the ring of functions on any variety, and this follows because the ring of functions on any variety is an integral domain. –  Qiaochu Yuan Oct 4 '12 at 16:16