Why is it that $ V_{+}(I) \cup V(J) $ = $ V_{+}(I \cap J), $ where $ I $ and $ J $ are homogeneous ideals of a graded ring $ S$?

Question

Define $ V_{+}(I) = \lbrace \mathfrak{p} \in \text{Proj}(S) \;|\; \mathfrak{p} \supset I \rbrace. $

I can see that $ V_{+}(I) \cup V_{+}(J) \subset V_{+}(I \cap J) $ since if $ \mathfrak{p} \in V_{+}(I) \cup V_{+}(J), $ then $ \mathfrak{p} $ contains $ I $ or $ J $ and therefore contains $ I \cap J $ either way.

But I am struggling to show that $ V_{+}(I \cap J) \subset V_{+}(I) \cup V_{+}(J). $

Answer 1

Let $\mathfrak{p} \in V_+(I \cap J).$ Suppose for a contradiction that $\mathfrak{p} \notin V_+(I)$ and $\mathfrak{p} \notin V_+(J).$ Then there exists $i \in I$ and there exists $j \in J$ for which we have $i \notin \mathfrak{p}$ and $j \notin \mathfrak{p}.$ However, we have $ij \in (I \cap J) \subseteq \mathfrak{p}.$ Since $\mathfrak{p}$ is prime, it follows $i \in \mathfrak{p}$ or $j \in \mathfrak{p}.$ This is a contradiction.