# Unique prime ideal factorization in noetherian domains?

[I changed the title and the body of the question. Below I explain why I did so, and paste the previous version.]

Let (UPIF) (for "Unique Prime Ideal Factorization") be the following condition on a noetherian domain $$A$$:

If $$\mathfrak p_1,\dots,\mathfrak p_k$$ are distinct nonzero prime ideals of $$A$$, and if $$m$$ and $$n$$ are distinct elements of $$\mathbb N^k$$, then we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$

The main question is

Do all noetherian domains satisfy (UPIF)?

Of course Dedekind domains satisfy (UPIF), but other noetherian domains $$A$$ also do. Indeed, as noted by user26857, if each nonzero prime ideal of $$A$$ is invertible or maximal, then $$A$$ satisfies (UPIF). To see this, assume by contradiction $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$ We can also assume that all the $$\mathfrak p_i$$ are maximal, and that $$m_1>n_1$$. Then $$\mathfrak p_1^{m_1}$$ contains $$\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}$$ but doesn't contain $$\mathfrak p_1^{n_1}$$. As $$\mathfrak p_1^{m_1}$$ is primary, this implies that the radical $$\mathfrak p_1$$ of $$\mathfrak p_1^{m_1}$$ contains $$\mathfrak p_2^{n_2}\cdots\mathfrak p_k^{n_k}$$, and thus $$\mathfrak p_1$$ contains one of the other $$\mathfrak p_i$$, contradiction. In particular, one dimensional noetherian domains and domains of the form $$B[X]$$, $$B$$ principal ideal domain, $$X$$ an indeterminate, satisfy (UPIF).

Here are the reasons why I changed the title and the body of the question (and added the "noetherian" tag): user26857 answered the original question in a comment, but didn't want to upgrade his comment to an answer. If they had, I would have accepted the answer and asked a follow-up question, but I thought it would be better, under the circumstances, to avoid creating a new question.

Here is the previous version of the question:

Previous title: Unique non-idempotent prime ideal factorization in domains?

Previous question:

Let $$A$$ be a domain; let $$\mathfrak p_1,\dots,\mathfrak p_k$$ be distinct non-idempotent prime ideals of $$A$$; and let $$m$$ and $$n$$ be elements of $$\mathbb N^k$$ such that $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$ Does it follow that $$m=n\ ?$$

[Recall that a domain is a commutative ring with one in which $$0\ne1$$ and $$a\ne0\ne b$$ implies $$ab\ne0$$.]

I suspect the answer is No, but I haven't been able to find a counterexample.

Edit

(1) If $$A$$ is a noetherian domain, then $$(0)$$ is the only idempotent prime ideal of $$A$$.

(2) Say that a domain satisfies Condition (D) (for "Dedekind") if the multiplicative monoid generated by the non-idempotent prime ideals is free (over the obvious basis).

Then the above question can be stated as: "do all domains satisfy (D)?"

Of course Dedekind domains satisfy (D), I but I know no non-Dedekind domain satisfying (D). (And, as indicated, I know no domain not satisfying (D).) For instance I'd be happy to know if $$K[X,Y]$$ satisfies (D). (Here $$K$$ is a field and $$X$$ and $$Y$$ are indeterminates.)

• About your 1st comment: Thanks for your comment! Unfortunately I don't understand it. Could you give some more details? Not sure if it helps to consider the following particular case: Suppose $\mathfrak p_1^2\mathfrak p_2=\mathfrak p_1\mathfrak p_2^2$ for $\mathfrak p_i$ distinct nonzero primes of $K[X,Y]$. How do you show this is impossible? I agree that the $\mathfrak p_i^2$ are primary, but I don't see how to use it. Perhaps you use primary decompositions, but in any case I need more help. - About your 2nd comment: Awesome! Can you say more? – Pierre-Yves Gaillard Dec 27 '18 at 20:20
• $p_1^2\supseteq p_1p_2^2$; $p_1\nsubseteq p_1^2$, then $p_2^2\subseteq p_1$, so $p_2\subseteq p_1$. If both have the same height we are done. Suppose $p_1$ is maximal and $p_2$ is not. Repeat the reasoning with $p_2$ instead of $p_1$. – user26857 Dec 27 '18 at 20:29
• If $V$ is a rank one valuation ring with $m=m^2$, then set $R=V[x]$, where $x^2\in V$. Then $M=m+(x)$ satisfies the requirement. (Sorry, I was too optimistic claiming that $R$ is local.) I'd consider $V$ as the valuation ring of the unique non-discrete valuation $v:K(X,Y)\to\mathbb R$ which is trivial on $K$, $v(X)=1$ and $v(Y)=\sqrt 2$, and $x=\sqrt X$. – user26857 Dec 27 '18 at 20:31
• I'm sorry but I don't see how you get $p_2^2\subseteq p_1$. – Pierre-Yves Gaillard Dec 27 '18 at 21:05
• @user26857 - At any rate your 2nd comment answers the question, so I must ask you if you'd consider turning your comment into an answer (even if I suspect you won't want to do this). – Pierre-Yves Gaillard Dec 27 '18 at 21:34

As user26857 answered the question in a comment, and prefers not to post it as an answer, I'll try to write the answer myself. I think I've understood user26857's argument, but I may be wrong. So, in the lines below, everything that's true is due to user26857, and everything that's false is due to me.

More precisely:

If $$A$$ is a noetherian integral domain, if $$\mathfrak p_1,\dots,\mathfrak p_k$$ are distinct nonzero prime ideals of $$A$$, and if $$m$$ and $$n$$ are distinct elements of $$\mathbb N^k$$, then we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$

Proof. In the setting of the question, suppose by contradiction that we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}$$ with $$m\ne n$$.

Enumerate the $$\mathfrak p_i$$ in such a way that each $$\mathfrak p_i$$ is a minimal element of the set $$\{\mathfrak p_i,\dots,\mathfrak p_k\}$$, and write $$\mathfrak p_{ij}$$ for the localization of $$\mathfrak p_i$$ at $$\mathfrak p_j$$.

For all $$i$$ we get $$(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{n_1}\cdots(\mathfrak p_{ii})^{n_i}.\quad(1)$$ Note the following consequence of the determinant trick, or Nakayama's Lemma:

$$(2)$$ If $$\mathfrak a$$ and $$\mathfrak b$$ are ideals of $$A$$, then the equality $$\mathfrak a\mathfrak b=\mathfrak b$$ holds only if $$\mathfrak a=(1)$$ or $$\mathfrak b=(0)$$.

Let's prove $$m_i=n_i$$ by induction on $$i$$:

Case $$i=1$$: We have $$(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{n_1}$$ by $$(1)$$. If we had $$m_1\ne n_1$$ we could assume $$m_1, and would get $$(\mathfrak p_{11})^{n_1-m_1}(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{m_1},$$ contradicting $$(2)$$.

From $$i-1$$ to $$i$$: We have $$(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{n_i}.\quad(3)$$ If we had $$m_i\ne n_i$$ we could assume $$m_i and we could write $$(3)$$ as $$(\mathfrak p_{ii})^{n_i-m_i}\mathfrak b=\mathfrak b$$ with $$(\mathfrak p_{1i})^{n_i-m_i}\ne(1)$$ and $$\mathfrak b\ne(0)$$, contradicting $$(2)$$. (Here $$\mathfrak b$$ is the left side of $$(3)$$, and we assume $$2\le i\le k$$.) $$\square$$

Note that the argument still works if $$A$$ is not noetherian, but the $$\mathfrak p_i$$ are finitely generated.