If $\mathfrak p$ is a prime ideal of an integral domain, is $\mathfrak pM$ a prime submodule of a torsion-free module $M$?

Question

Let $M$ be a torsion-free module of an integral domain $R$. A submodule $N$ of a module $M$ is said to be prime if for all $r\in R$ and $m\in M$, $rm\in N$ implies either $rM\subseteq N$ or $m\in M$. My question is, is it true that if $\mathfrak p$ is a prime ideal of $R$, then $\mathfrak pM$ is a prime submodule of $M$? I've tried to search for a counter-example on this over $\mathbb Z$, but I can't find any. Putting $rm$ as $rm=p_1m_1+\ldots+p_nm_n$ doesn't really help much for me either. Is there any idea on how to prove/unprove this? Or is there any additional condition that can lead to the objective? (Maybe finitely-generated modules?) Or maybe there is something I've overlooked? Thank you very much in advance.

Answer 1

The answer is no. For example, take $R=\mathbb{Q}[x,y]$ and $\mathfrak{p}=yR$ and $M=R\times R\big/\langle(x,-y)\rangle$, and let $r=x$ and $m=\overline{(1,0)}$. Certainly $\mathfrak{p}$ is prime, and we have $rm=\overline{(x,0)}=\overline{(0,y)}\in \mathfrak{p}M$.

Claim: $M$ is torsion-free.

Proof: Suppose $\alpha\cdot\overline{(f,g)}=0$ for some $\alpha\in R\setminus\{0\}$ and $\overline{(f,g)}\in M$. Then there exists $\beta\in R$ with $(\alpha f,\alpha g)=(\beta x,-\beta g)$. So $\alpha f/x=\beta=-\alpha g/y$, hence $\alpha yf=-\alpha xg$, hence, since $\alpha$ is non-zero and $R$ is a domain, $yf=-xg$. So, since $R$ is a UFD, $(f,g)=(xh,-yh)$ for some $h\in R$, as desired. $\blacksquare$

Claim: $m\notin\mathfrak{p}M$.

Proof: Otherwise $m=y\cdot\overline{(f,g)}$ for some $f,g\in R$, ie $(1-yf,-yg)=(\alpha x,-\alpha y)$ for some $\alpha\in R$. But this is impossible, since $1-yf$ is not divisible by $x$ in $R$ for any $f\in R$. $\blacksquare$

Claim: $rM\nsubseteq\mathfrak{p}M$.

Proof: It suffices to show $\overline{(0,x)}=r\cdot\overline{(0,1)}\notin\mathfrak{p}M$; otherwise $\overline{(0,x)}=y\cdot\overline{(f,g)}$ for some $f,g\in R$, ie $(-yf,x-yg)=(\alpha x,-\alpha y)$ for some $\alpha\in R$. But this is impossible, since $x-yg$ is not divisible by $y$ in $R$ for any $g\in R$. $\blacksquare$