Vakil's The Rising Sea Exercise 3.7.H (Version 2022)

Question

The original exercise is on the page 127:

In $\mathbb{A}_n = \text{Spec}\ k[x_1,\dots,x_n]$, the subset cut out by $f(x_1,\dots,x_n)\in k[x_1,\dots,x_n]$ should certainly have irreducible components corresponding to the distinct irreducible factors of $f$.

My question is:

We can conclude that every prime ideal containing $f(x)$ must contain one of its irreducible factors, say $p_i(x)$.

But why the principal ideal $(p_i(x))$ generated by $p_i(x)$ is a minimal prime ideal (in $\text{Spec}\ k[x_1,\dots,x_n]/(f)$)?

Answer 1

Because $(p(x))$ is a prime ideal of height 1 for irreducible $p(x)$. Namely, if $\mathfrak{p}$ is a nonzero prime ideal contained in $(p(x))$, we can find an irreducible polynomial $q(x)\in\mathfrak{p}$. So $p(x)\mid q(x)$ and hence $(p(x))=(q(x))$ since both of them are irreducible. Thus we also have $(p(x))\subseteq\mathfrak{p}$ and hence they are equal.