Chapter 1, Exercise 1.9: A lower bound on the dimension of irreducible components

Let $\mathfrak a\subseteq A=k[x_1,\dots,x_n]$ be an ideal which can be generated by $r$ elements. Then every irreducible component of $Z(\mathfrak a)$ has dimension $\geq n-r$.

This is an exercise that looks somewhat nontrivial, but ultimately reduces to a geometric realization of a lemma from commutative algebra generalizing Krull's Hauptidealsatz. For this reduction, we use the following consequence of Theorem 1.8Ab): if $Y$ is any affine variety in $\mathbb A^n$, then $$\dim Y = \dim A/I(Y) = n - \operatorname{height}I(Y).$$ In our case, if $Z(\mathfrak a)=\bigcup G_i$, then we need only prove that $\operatorname{height}I(G_i)\leq r$. In particular, the ideals $I(G_i)$ are minimal prime ideals over $\mathfrak a=I(Y)$ (where $Z(\mathfrak a)=Y$), so that the following lemma from commutative algebra (Krull's height lemma) will suffice.

Lemma 1. If $A$ is a noetherian ring, $\mathfrak a$ is a prime ideal generated by $r$ elements, and $\mathfrak p$ is a minimal prime ideal over $\mathfrak a$, then $\operatorname{height}\mathfrak p\leq r$.

This is true, and is sufficient.