A question about one of Hartshorne's propositions

Question

Hartshorne says that for $S_1,S_2\in A[x_1,x_2,\dots,x_n]$, where $A$ is a commutative ring, $Z(S_1)\cup Z(S_2)=Z(S_1S_2)$.

Shouldn't it be $Z(S_1)\cup Z(S_2)=Z(S_1\cap S_2)$?

We know that $S_1S_2\subseteq S_1\cap S_2$.

Answer 1

No. For example take $S_1 = \left\{x_1\right\}$ and $S_2 = \left\{x_2\right\}$. Then $S_1 \cap S_2 = \emptyset$ and $Z(S_1 \cap S_2) = \mathbb{A}^n \neq Z(x_1) \cup Z(x_2)$.

Answer 2

• If $I$ and $I'$ are ideals of some commutative ring, then clearly $(I\cap I')^2 \subseteq I I' \subseteq I\cap I'$. Therefore also $Z((I\cap I')^2) \supseteq Z(I I') \supseteq Z(I\cap I')$.

• If $J$ is an ideal, then $Z(J^2) = Z(J)$. Well, $Z$ can mean several slightly different things and you didn't exactly state the context so I have to guess, but if $Z(J)$ stands for the set of prime ideals containing $J$, then the key point is that $J^2 \subseteq \mathfrak{p}$ implies $J \subseteq \mathfrak{p}$ for $\mathfrak{p}$ a prime ideal, which is clear; and if $J$ is an ideal of $k[x_1,\ldots,x_n]$ with $k$ a field, and $Z(J)$ stands for the set of $\underline{x}$ such that $f(\underline{x})=0$ for all $f\in J$, then the key point is that if $f g$ vanishes at some $\underline{x}$ then already one of $f$ or $g$ has to vanish.

• Putting these two remarks together, $Z(I\cap I') = Z(I I')$.