0
$\begingroup$

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?

  • 1
    What is the definition of an affine variety over $k$?2012-10-04
  • 1
    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.2012-10-04
  • 0
    @QiaochuYuan: Why not? A variety is both a set and a topological space.2012-10-04
  • 0
    @MakotoKato: I suppose an affine variety of $\mathbb{A}^n_k$ for some $n$.2012-10-04
  • 1
    @Manos: being a variety is not a _property_ of a set or topological space, it is an extra _structure_ attached to a set or topological space (exactly what that structure is depends on what formalism you're working in). You can't ask whether a topological space is a variety any more than you can ask whether a set is a group. It's a type error (http://en.wikipedia.org/wiki/Type_safety).2012-10-04
  • 0
    @QiaochuYuan: I see your point. I changed the statement of the question. Is it meaningful now?2012-10-04
  • 0
    What do you mean by a variety of $\mathbb{A}_{k}^{n}$?2012-10-04
  • 0
    @MakotoKato: I consider $\mathbb{A}^n_k$ the affine space with the Zariski topology. So an affine variety should be an irreducible algebraic subset of $\mathbb{A}^n_k$.2012-10-04
  • 0
    Is $k$ algebraically closed?2012-10-04
  • 0
    @MakotoKato: Yes, i will add that to the question.2012-10-04
  • 0
    $Specm(\mathbb{Z}_p)$ is a point which is isomorphic to a one point variety as a toplogical space, where $\mathbb{Z}_p$ is a finite prime field. Since $\mathbb{Z}_p$ is not algebraically closed, your assertion does not hold.2012-10-04
  • 0
    @MakotoKato: I don't see your point...I assume that $k$ is algebraically closed.2012-10-04
  • 2
    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 $A$.2012-10-04
  • 0
    @Andrew: This seems to be the right formulation, thanks!2012-10-04
  • 0
    I assume $A$ is a commutative ring with a unity. $\mathbb{Z}_p$ is such a ring.2012-10-04

1 Answers 1

3

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.

  • 0
    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?2012-10-04
  • 0
    @Manos: no, it means finitely generated as a $k$-algebra. People use "finite" to mean finitely generated as a $k$-module.2012-10-04
  • 0
    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 $\mathbb{A}^n_k$?2012-10-04
  • 0
    @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.2012-10-04
  • 0
    Got it. Thanks a lot.2012-10-04
  • 0
    So if $Specm(A)$ is an affine variety, then its sheaf of regular functions $\mathfrak{O}(Specm(A))$ is isomorphic to say $k[x_1,\cdots,x_n]/\mathcal{I}$, where $\mathcal{I}$ is some prime ideal of $k[x_1,\cdots,x_n]$. So i need to show that $\mathfrak{O}(Specm(A))$ is isomorphic to $A$, right?2012-10-05
  • 0
    @Manos: again, that question doesn't make sense. The maximal spectrum of an arbitrary finitely generated algebra $A$ is _a priori_ just a set or a topological space. One way to fix this problem is to choose an embedding into affine space, but until you do something like that you can't tell me what a regular function on the spectrum is without assuming something like what ypu want to prove.2012-10-05
  • 0
    I see your point. So let's assume that we have such an embedding $\sigma: Specm(A) \hookrightarrow \mathbb{A}^n_k$ such that $Y=\sigma(Specm(A))$ is an affine variety. Let $\mathfrak{O}(Y)$ be the ring of regular functions of $Y$, which we know that it is a finitely generated integral domain over $k$. We know that we have $1-1$ correspondence between $Y$ and $Specm(\mathfrak{O}(Y))$, which gives us a $1-1$ correspondence between $Specm(A)$ and $Specm(\mathfrak{O}(Y))$. Can we deduce that $A$ is a finitely generated integral domain over $k$?2012-10-05
  • 0
    @Manos: no. Without further hypotheses, $A = k[x]/x^2$ is still a counterexample. The correct statement of the question is the following modification to Andrew's statement in the comments: _if_ $A$ is a finitely generated integral domain over $k$, then there is an affine variety with ring of regular functions $A$, and moreover the points of this affine variety can be identified with the maximal spectrum of $A$. A choice of embedding into affine space is one way to construct the desired variety (and depending on your definition of affine variety it may be the only way).2012-10-05