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To prove that a given set is an affine variety, I want to first relate it to an ideal. If there exists an ideal that is finitely generated, is that enough to prove something is an affine variety (for example, every singleton set)?

Also, how would one rigorously prove that for a finite field k, every subset of the affine space $A_k^n$ is an affine variety V. Could one just say that if finite, then $k=\{a_1,...a_n\}$ ideal is given by $I(V)=(x_1-a_1,...,x_n-a_n)$, or would that need to be proved. Once that is established, is that it?

math1234567
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  • I think you mean V finite and F any field. – user42912 Mar 26 '14 at 14:34
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    I suppose your varieties are not necessarily irreducible. Proceed in steps: show that there is a topology on affine space such that the closed subsets are precisely the subsets of the form $V (I)$ for some ideal; then show that points are closed subsets. Note that every ideal of $k [x_1, \ldots, x_n]$ is finitely generated. – Zhen Lin Mar 26 '14 at 14:36

1 Answers1

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The idea is sound.

If you want to show that a subset $V$ of $A^n_k$ is an affine variety, it is enough to show that $V = V(I)$ for some ideal $I$ of $k[X_1,\dots,X_n]$. Note that finitely generatedness of $I$ you get for free. (As an aside, note that it is $V = V(I)$ you have to show, not $I = I(V)$).

For the second part, you seem to be mixing up the field $k$ and the subset $V$ of $A^n_k$ that you want to show is affine.

In general, a finite subset $V$ of $A^n_k$ is affine (whether or not $k$ itself is finite; but of course if $k$ is finite, every subset of $A^n_k$ is finite).

To avoid a notational mess, first show this for a subset $V$ consisting of a single point and then argue that a finite union of affine varieties is an affine variety again.

So, for $V = \{ (a_1,\dots,a_n) \}$, you need to find an ideal $I \subseteq k[X_1,\dots,X_n]$ such that $V(I) = V$. The ideal $I = (X_1 - a_1, \dots, X_n - a_n)$ works.

Magdiragdag
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