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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

Moss
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  • An affine variety must be cut out by a prime ideal, and the ideal you give is not prime. – Potato May 05 '13 at 05:13
  • What is the prime ideal that gives the variety ${1,2}$? I suppose it would be the radical of $((x-1)(x-2))$? – Moss May 05 '13 at 05:15
  • Ok, ill give it a try. – Moss May 05 '13 at 05:18
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    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. – Jyrki Lahtonen May 05 '13 at 05:31

3 Answers3

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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.

Q.E.D.

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    Thanks Ben, I forgot that Fulton defines varieties to be irreducible. – Moss May 05 '13 at 05:50
  • @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, –  May 05 '13 at 05:53
  • Is it always the case that for any affine varieties $X$ and $Y$: $I(X\cup Y)=I(X)\cap I(Y)$? – Moss May 05 '13 at 06:27
  • @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 $\Bbb{A}^n$. –  May 05 '13 at 06:31
  • Yes I see, it follows immediately by applying "V" to both sides, ie looking at the vanishing loci of both sides – Moss May 05 '13 at 06:34
  • Dear @Sebastian, taking $V$ on both sides confuses the issue. Indeed, the result is purely set-theoretic and has nothing to do with algebraic geometry: the equality $I(X\cup Y)=I(X)\cap I(Y)$ just says that a function is zero on $X\cup Y$ if and only it is zero on $X$ and is zero on $Y$. – Georges Elencwajg May 05 '13 at 07:25
  • Very clear explanation, Benja: +1. May I ask who the gentleman on the photo is? – Georges Elencwajg May 05 '13 at 07:30
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    Ah, thanks for the explanation, Benja: for some reason I had excluded the possibility that it was a photo of yourself :-) – Georges Elencwajg May 05 '13 at 07:40
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Let me give an algebraic proof for $(3)\Rightarrow (1)$. We can always embed $k$ as a subring of $\Gamma(V)$ by regarding them as constant functions. Then condition (3) implies that $\Gamma(V)$ is a finite $k$-module, which tells you that every element of $\Gamma(V)$ is integral over $k$, and hence in $k$ because $k$ is algebraically closed. Therefore, we have $\Gamma(V)=k$, and $V$ is forced to be a point.

In your example, your $X=\{1,2\}$ is not irreducible, and hence not an affine variety.

Chuhao HUANG
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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.

Martin
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khalil alazzawe
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