# Let $P$ and $Q$ be prime ideals of a ring $R$ such that every element of $R\setminus (P\cup Q)$ is a unit. Prove that either $P$ or $Q$ is maximal.

Let $$R$$ be a commutative ring with a unit. Let $$P$$ and $$Q$$ be prime ideals of $$R$$ such that every element of $$R\setminus( P\cup Q)$$ is a unit. We want to show that either $$P$$ or $$Q$$ is maximal.

I am not really sure what to try here. I can see that if $$a$$ is an element of a maximal ideal which contains $$P$$, but $$a\notin P$$, then $$a\in Q$$, and vice versa, but I am not sure how that helps me. I have mostly tried to come up with contradictions by supposing that neither $$P$$ nor $$Q$$ is maximal, then using things like $$R/P$$ and $$R/Q$$ are both integral domains with noninvertible elements, but I haven't had any luck. Any hints on how to proceed would be appreciated (for reference, I haven't taken a full commutative algebra course yet).

• I found an answer but did not use being "prime". Commented Jan 5, 2021 at 6:06

By Krull's theorem, there exists $$M$$ a maximal ideal of $$R$$. Since $$M$$ is proper and hence no element of $$M$$ is a unit, we must have $$M\subseteq P\cup Q$$. Now, suppose for contradiction that $$M\nsubseteq P$$ and $$M\nsubseteq Q$$. Then there is some $$m_p\in M\setminus P$$ and $$m_q\in M\setminus Q$$, and we have $$m:=m_p+m_q\in M$$. Since $$M\subseteq P\cup Q$$, this means $$m\in P\cup Q$$, and so either $$m\in P$$ or $$m\in Q$$; without loss of generality suppose $$m\in P$$. Then, since $$m_q\in M\subseteq P\cup Q$$, and $$m_q\notin Q$$, we must have $$m_q\in P$$. But this means that $$m_p=m-m_q\in P,$$ a contradiction. Thus we must have $$M\subseteq P$$ or $$M\subseteq Q$$, and so – by maximality of $$M$$, and since $$P$$ and $$Q$$ are prime and hence proper – either $$M=P$$ or $$M=Q$$, as desired.
You do not need $$P$$ and $$Q$$ to be prime here, just proper, and the argument above shows a more general result: if $$R$$ is any ring, and $$I$$, $$J_1$$, and $$J_2$$ are any ideals of $$R$$ with $$I\subseteq J_1\cup J_2$$, then either $$I\subseteq J_1$$ or $$I\subseteq J_2$$. We can extend this result even further by reintroducing the primality hypothesis; if $$P_1,\dots,P_n$$ are prime ideals of $$R$$, and $$I\subseteq P_1\cup\dots\cup P_n$$, then there is some $$i$$ such that $$I\subseteq P_i$$. Exercise: try to prove this by induction! (Note, however, that the primality hypothesis really is necessary. For instance, $$(\overline{x},\overline{y})=(\overline{x})\cup (\overline{x+y})\cup (\overline{y})$$ in $$\mathbb{F}_2[x,y]\big/(x^2,xy,y^2)$$, but $$(\overline{x},\overline{y})$$ is not contained in any of the ideals on the right.)
Firsly I supposed that the ideals are nontrivial. Let $$P' \supset P$$ and $$Q' \supset Q$$ be ideals. Obviously, if $$P'$$ or $$Q'$$ contains an element in $$R \setminus P \cup Q$$, then one of them must be $$R$$. Assume the other case, which is $$P' \subset P \cup Q$$ and $$Q' \subset P \cup Q$$. $$P'$$ contains $$q \in Q \setminus P$$ and $$Q'$$ contains $$p \in P \setminus Q$$. Note that $$P'$$ and $$Q'$$ contain both of $$p$$ and $$q$$, so their all combinations. Look at the element $$p-q$$. This is not element of $$P$$ and $$Q$$, by using the fact that ideals are additive subgroups. So $$p-q \notin P \cup Q$$, implies that $$p-q$$ is unit. Both of $$P'$$ and $$Q'$$ become $$R$$. This shows $$P$$ and $$Q$$ are maximals. I wonder that I did not use being prime.