# Does a prime ideal contains an irreducible element?

The context. Let $$R$$ be an integral domain. It is known that a domain $$R$$ is a UFD if and only if any nonzero prime ideal contains a prime element.

It is also known that $$R$$ is a UFD if and only if any non zero element has a decomposition as a product of a unit and irreducible elements (which is automatic if $$R$$ is Noetherian, for example) and any irreducible element is prime.

Thius, if $$R$$ is a Noetherian domain which is NOT a UFD, we known that there exists a nonzero prime ideal which does NOT contain a prime element.

The natural question coming into my mind is now:

Question 1. Let $$R$$ be a Noetherian integral domain which is not a field. Does any nonzero prime ideal of $$R$$ contains an irreducible element ?

Question 2. If answer to Q1 is NO, can we find sufficient conditions for which the answer to Q1 becomes YES ?

The answer is YES for $$A[X]$$ where $$A$$ is a PID (a full description of prime ideals are known, and they all contain an irreducible element)

I think I have proved it is also true for $$\mathbb{Z}[\sqrt{-d}], d>0$$ squarefree such that $$d\not\equiv -1 \mod 4$$ (I have not checked the details), but I have no clue how to prove it in general or how to find a counterexample (if there is any).

Edit In fact, Q1 is trivial. Any nonzero non unit $$a\in \mathfrak{p}$$ (prime ideal) maybe written as a product of irreducible elements. Since $$\mathfrak{p}$$ is prime, one of these irreducible elements belong to $$\mathfrak{p}$$.

So the real question is:

Real question. Let $$R$$ be an integral domain which has irreducible elements. Does any nonzero prime ideal contains an irreducible element ?

If there are counter examples, they are non noetherian.

Let $$a$$ be a nonzero element of your prime ideal $$P$$. If it's not irreducible, it has a proper factorisation $$a=a_1b_1$$ where, one may assume that $$a_1\in P$$ and $$b_1$$ is not a unit. Again if $$a_1$$ is not irreducible, then $$a_1=a_2b_2$$ where $$a_2\in P$$ and $$b_2$$ is not a unit. If we keep going, we get a strictly increasing chain of principal ideals $$(a_1)\subset (a_2)\subset\cdots$$ contradicting the Noetherian condition.
• Thanks ! In fact, I came with a proof (which is more or less yours) . I leave my post anyway. The next question would be: can we find a non noetherian counterexample, provided that $R$ is supposed to have irreducible elements (to avoid counterexamples like the ring of all algebraic integers) ? Commented Jun 9, 2020 at 14:02
• @GreginGre: Anyway, I think the answer is no. For instance, if $S$ is a valuation ring which is not DVR, then $R=S[x]$ is a domain and it has irreducibles (e.g. $x$ is irreducible) and for the maximal ideal $\mathfrak m$ in $S$, the ideal $\mathfrak m[x]$ is prime in $S[x]$ and has no irreducibles. Commented Jun 9, 2020 at 14:26