I am looking at exercise 2.4 in William Fulton's "Algebraic Curves". It asks to prove that if $X\subset \mathbb{A}^n$ is nonempty affine variety, then the following are equivalent

  1. $X$ is a point
  2. $\Gamma(X)=k$
  3. $dim_k\Gamma(X)<\infty$

I have a problem with 3 implying 1. Suppose $X$ is the union of two points, say $1,2\in\mathbb{A}(\mathbb{C})$. Then $I(\{1,2\})=((x-1)(x-2))$ so

$\Gamma(X)=\mathbb{C}[x]/I(X)=\mathbb{C}[x]/((x-1)(x-2))$, and I am pretty sure that this has finite dimension as a $\mathbb{C}$ vector space. But yet our original variety is not a point. Does the question mean to have $X$ is a finite union of points instead? Though that you make number 2 not quite correct.

Thanks for the help

  • $\begingroup$ An affine variety must be cut out by a prime ideal, and the ideal you give is not prime. $\endgroup$ – Potato May 5 '13 at 5:13
  • $\begingroup$ What is the prime ideal that gives the variety $\{1,2\}$? I suppose it would be the radical of $((x-1)(x-2))$? $\endgroup$ – Moss May 5 '13 at 5:15
  • $\begingroup$ Ok, ill give it a try. $\endgroup$ – Moss May 5 '13 at 5:18
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    $\begingroup$ The radical of $I$ is $I$ itself. But it appears that Fulton (my copy is my office so cannot check) requires an algebraic set to be irreducible in order to be called a variety (or some such game with definitions is taking place). Or (as Potato said) the defining ideal should be prime. $I$ is clearly not prime and the set $\{1,2\}$ is not irreducible. $\endgroup$ – Jyrki Lahtonen May 5 '13 at 5:31

If you look at the beginning of section 2, Fulton defines an affine variety to be an irreducible algebraic variety, i.e. a closed irreducible subset of $\Bbb{A}^n$ with the Zariski topology.

Now in the case of $\Bbb{A}^1$, the union of two points is not an affine variety. Finite point sets are closed yes, but then $\{1,2\} = \{1\} \cup \{2\}$ and thus is not an affine variety.

Here's how I would do $(3) \implies (1)$. Let $X = V(I)$ for some ideal $I \subseteq k[X_1,\ldots,X_n]$. Suppose we know that $\dim_k \Gamma(X) < \infty$. Then by Corollary 4 of the Nullstellensatz we get that

$$\dim_k \Gamma(X) = \dim_k k[X_1,\ldots,X_n]/\sqrt{ I} < \infty \implies |V\left(\sqrt{ I}\right)| < \infty.$$

But now $V\left(\sqrt{I}\right) = V(I)$ which means that $|X| = |V(I)| < \infty$. A finite set of points is irreducible iff it consists of a point, so that $X$ is a point.


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    $\begingroup$ Thanks Ben, I forgot that Fulton defines varieties to be irreducible. $\endgroup$ – Moss May 5 '13 at 5:50
  • $\begingroup$ @Sebastian Dear Sebastian, no problems! I should say in your question above you are spot on in saying that $\Bbb{C}[x]/((x-1)(x-2))$ is finite dimensional over $\Bbb{C}$: By the Chinese remainder theorem, your ring is isomorphic to $\Bbb{C}[x]/(x-1) \times \Bbb{C}[x]/(x-2) \cong \Bbb{C} \times \Bbb{C}$ that is two dimensional over $\Bbb{C}$. Also you are right when you write that $I(\{1,2\}) = (x-1)(x-2)$. We have that $I(\{1,2\}) = I(\{1\}) \cap I(\{2\}) = (x-1) \cap (x-2) = (x-1)(x-2)$. The last equality comes from the fact that $(x-1)$ and $(x-2)$ are coprime. Regards, $\endgroup$ – user38268 May 5 '13 at 5:53
  • $\begingroup$ Is it always the case that for any affine varieties $X$ and $Y$: $I(X\cup Y)=I(X)\cap I(Y)$? $\endgroup$ – Moss May 5 '13 at 6:27
  • $\begingroup$ @Sebastian Dear Sebastian, your claim is certainly true. In fact it's true not just for affine varieties but for more general algebraic subsets of $\Bbb{A}^n$. $\endgroup$ – user38268 May 5 '13 at 6:31
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    $\begingroup$ Ah, thanks for the explanation, Benja: for some reason I had excluded the possibility that it was a photo of yourself :-) $\endgroup$ – Georges Elencwajg May 5 '13 at 7:40

By corollary 4 to Hilbert's Nullstellensatz, $V(I(V))=V$ is a finite set, say $\{P_1,\dots,P_r\}$. Hence $V=\{P_1\}\cup \dots \cup \{P_r\}$. Since $V$ is an affine variety, $r=1$, and thus $V$ is a point.


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