# Every prime ideal is either zero or maximal in a PID.

$(1)$ Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, show that every prime ideal is either zero or maximal.

In many books I have found the proof of the above statement where they show that

(2)Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, then every nonzero prime ideal is maximal.

I have revised the question now, how can prove that $(1)$ is true using $(2)$. I am of the opinion that both questions are similar and that's why I am asking this question if I am mistaken then please explain why these two are different. Thanks.

• I'm not sure I understand your question. If the prime ideal is $0$, then it generally is not maximal. Jul 8 '12 at 2:41
• In a PID, $0$ is a maximal ideal if and only if the PID is actually a field. Jul 8 '12 at 2:49
• To the revised question: You are correct in that they are very close statements. The zero ideal is prime in an integral domain, and any other prime ideal is, well, not the zero ideal. Jul 8 '12 at 2:56
• @POTUS: You are essentially asking whether $\neg A\implies B$ is equivalent to $A\lor B$. But recall that $P\implies Q$ is logically equivalent to $\neg P\lor Q$, so $\neg A\implies B$ ("nonzero implies maximal") is equivalent to $\neg\neg A\lor B$ ("either not-nonzero or maximal"), and $\neg\neg A$ is the same as $A$ ("not-nonzero" is the same as "zero"). If you are not clear on the logical equivalence of $P\implies Q$ with $\neg P\lor Q$, then that's where you need to start. Jul 8 '12 at 3:12
• @ArturoMagidin Thanks, that could be my problem. Jul 8 '12 at 3:20

In your revised question, you ask us to show that Every nonzero prime ideal is maximal in a nontrivial PID $\implies$ Every prime ideal is either zero or maximal in a nontrivial PID.

This is trivial. Take a prime ideal. If it's zero, that's fine. If it's not, then the assumed statement says exactly that it's maximal. There's nothing left to show.

Showing the other direction is as trivial, too.

EDIT

The OP asks in a comment:

Sometimes the trivial things are difficult to understand! You said, "If it's zero, that's fine." and my question is that: why is it fine? Everyone says it is trivial but I cannot wrap my brain around it. Can you elaborate and explain it with a microscopic lens?

The statement we want to show is Every prime ideal is either zero or maximal in a nontrivial PID. In other words, if we take a prime ideal, we have to show that it is either zero, or that it is maximal.

1. If it is not zero, then by assumption we know that it is maximal (this is the statement from (2) ).

2. If it is zero, than we don't care. Why don't we care? Because we wanted to show that prime ideals are either zero or maximal. So we have shown that all nonzero prime ideals are maximal, and the zero ideal is in fact the zero ideal. That's why I can say that "The zero ideal is zero, and that's fine." There is nothing to prove about that case.

• Sometimes the trivial things are difficult to understand! You said, "If it's zero, that's fine." and my question is that: why is it fine? Everyone says it is trivial but I cannot wrap my brain around it. Can you elaborate and explain it with a microscopic lens? Jul 8 '12 at 3:08
• @POTUS: Yes, let me expand. Give me a few minutes as I edit my answer. Jul 8 '12 at 3:09

Hint $\$ By definition $\rm\ [P\ne 0\:\Rightarrow\: P\ max]\iff [P=0\ \ or\ \ P\ max]$

Recall that, by definition: $\rm\ [A\Rightarrow B]\iff [\lnot A\ \ or\ \ B]$

• I don't think this is really helpful to anyone who would ask this question. Jul 8 '12 at 3:42
• @Alex Out of curiousity, why do you think that? Jul 8 '12 at 3:51
• The user is asking if "every nonzero prime is maximal" implies "every prime is either zero or maximal". I think that anyone familiar enough with formal logic to understand what you've written, would find the question trivial. Jul 8 '12 at 3:58
• @Alex That may be true. Alternatively, some students do know the formal definition and have either forgotten it, or are not well-practiced, in which case this answer may prove helpful. Jul 8 '12 at 4:02