Q:

How to show $W$ is irreducible, $V \subseteq W$ and $\dim V = \dim W$ implies $V = W$ for algbraic sets

Suppose I have two algebraic sets $V, W$ in $\mathbb{A}^n_{\mathbb{C}}$. Suppose $W$ is irreducible, $V \subseteq W$ and $\dim V = \dim W$. I want to prove that $V = W$ but I seem to be missing something... I would appreciate any hints or answers!

A:

This follows almost by definition. If $V$ is irreducible and $n = {\rm dim} \ V$ , then $n$ is the length of the longest possible chain of proper inclusions of irreducible algebraic sets inside $V$:
$$ V_0 \subset V_1 \subset V_2 \dots \subset V_n = V.$$
Similarly, if  $W$ is irreducible and $m = {\rm dim} \ W$ , then $m$ is the length of the longest chain of proper inclusions of irreducible algebraic sets inside $W$:
$$ W_0 \subset W_1 \subset W_2 \dots \subset W_m = W.$$
But if $V \subseteq W$, we are guaranteed the existence of at least one such chain of inclusions within $W$:
$$ V_0 \subset V_1 \subset V_2 \dots \subset V_n = V \subseteq W.$$
Thus if ${\rm dim} \ W = {\rm dim} \ V = n$, this chain must be of length no longer than $n$, hence the final inclusion must be an equality. So we have $V = W$.
[Okay, in your question you didn't assume that $V$ is irreducible. If $V$ is reducible, you can define the dimension of $V$ to be the maximum of the dimensions of its irreducible components. Then apply the same argument to each irreducible component of $V$ in turn.]

