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.


1 Answer 1


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.

  • $\begingroup$ 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) ? $\endgroup$
    – GreginGre
    Jun 9, 2020 at 14:02
  • 1
    $\begingroup$ @GreginGre: You should just accept this answer and ask a followup question. $\endgroup$
    – tomasz
    Jun 9, 2020 at 14:19
  • 3
    $\begingroup$ @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. $\endgroup$
    – tomasz
    Jun 9, 2020 at 14:26

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