algebraic variety of dimension 0 How to prove that if algebraic variety has zero dimension it must be finite set of points.
In other words: how to prove that it can't be infinite set ???
Definitions
algebraic variety V - if V is the common zero set of a polynomial system over Field K (polynomes from $K[x_1,...,x_n]$)
dimension of V is $ trdeg(K(V) : K)$, $K$ - is assumed to be algebraically closed

Let's suppose that we've broken $V$ down into irreducible components, and therefore without loss of generality assume $V$ to be irreducible. I'll show that $V$ must be a single point.
Since the transcendence degree of $K(V)$ over $K$ is $0$, $K(V)$ must be an algebraic field extension of $K$. Since $K$ is algebraically closed, $K(V)$ must equal $K$. In particular, this means that the only regular functions on $V$ are constant polynomials; algebraically speaking, if $I \subset K[x_1,\dots, x_n]$ corresponds to $V$, then $K[x_1,\dots,x_n]/I= K$. This implies that $I$ is a maximal ideal; i.e. $V$ is a point.

$V$ must be a finite number of points because $k[x_1,\dots, x_n]$ is a Noetherian ring (see Hilbert Basis Theorem). This means there are no infinitely ascending chains $I_1 \subsetneq I_2 \subsetneq \cdots$. On the geometric side this means there are no infinite descending chains of subsets $X_1\supsetneq X_2\supsetneq\cdots$. So, if $V$ had an infinite number of points, we could form an infinite descending chain by peeling off one point at a time, a contradiction.
