Hartshorne's Chapter 1, exercise 1.9 asks us to show that irred. components of $Z(\mathfrak a)$ have dimension $\geq n-r$ if $\mathfrak a$ is an ideal generated by $r$ elements. I think I've reduced this to a problem in commutative algebra, but I'm not sure how to tackle it.
My start: for a variety $Y\subset\mathbb A^n$ we have $$\dim Y = n-\operatorname{height}I(Y),$$ thus we need to show that $\operatorname{height}I(Y)\leq r$ if $Y$ is an irred. component of $Z(\mathfrak a)$. I tried to argue by contradiction, supposing we had a chain $$\mathfrak p_0\subset\mathfrak p_1\subset\dots\subset\mathfrak p_{r+1}=I(Y),$$ but I'm not sure how to bring $\mathfrak a$ into play here as we have $I(Y)\supset\mathfrak a$, not the other way around. How would I approach this?
Best Answer
I want to point out that Hartshorne has given you all of the commutative algebra you really need. Let's say $\mathfrak{a}$ can be generated by $f_1, \dots, f_r$. We'll induct on $r$. The only observation is that any irreducible component $Y$ of $Z(\mathfrak{a})$ is contained in an irreducible component $Y'$ of $Z(f_1, \dots, f_{r-1})$, and in fact $Y$ is an irreducible component of $Y' \cap Z(f_r)$.
If you want to finish this off, you would have to justify a few things from the last paragraph and then use (Ex. 1.8) and the theorem (1.8Ab).