5

This is a follow-up to this question.

Let $A$ be a domain; let $\mathfrak p_1,\dots,\mathfrak p_k$ be distinct prime ideals of $A$ such that $\mathfrak p_i^{j+1}\ne\mathfrak p_i^j$ for all $1\le i\le k$, $j\ge1$; and let $m$ and $n$ be elements of $\mathbb N^k$ such that $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?

user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $\mathfrak p_i^{j+1}\ne\mathfrak p_i^j$ for all $j\ge1$ is equivalent to $\mathfrak p_i\ne(0)$.)

Pierre-Yves Gaillard
  • 19,712

1 Answers1

4

The answer is no.

Let $G$ be the abelian group $\mathbb{Z}\times \mathbb{Z}$ with the lexicographic order, and let $M\subset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:=\{(m,n):m\geq 1\text{ or }n\geq 1\}\subset M$ and $I_2:=\{(m,n):m\geq 1\}\subset M$. For $j\in\mathbb{Z}_{\geq 0}$ and $I\subset M$ an ideal, write $jI:=\{i_1+\ldots+i_j:i_1,\ldots,i_j\in I\}\subset M$. Check that for all $j$, we have $jI_1\neq (j+1)I_1$ and $jI_2\neq (j+1)I_2$, but $I_1+I_2=I_2$.

We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $\mathfrak{p}_1=k[I_1]$, and $\mathfrak{p}_2=k[I_2]$. Then $\mathfrak{p}_1^j\neq \mathfrak{p}_1^{j+1}$ and $\mathfrak{p}_2^j\neq \mathfrak{p}_2^{j+1}$, but $\mathfrak{p}_1\mathfrak{p}_2=\mathfrak{p}_2$.

An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},\ldots]$, $\mathfrak{p}_1=(x)$, $\mathfrak{p}_2=(y,yx^{-1},yx^{-2},\ldots)$.

Julian Rosen
  • 16,142