Exercise 2.4 Fulton's Algebraic Curves 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


*

*$X$ is a point

*$\Gamma(X)=k$

*$dim_k\Gamma(X)<\infty$


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
 A: 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.
A: 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.
