Prime submodule equivalence statement

Question

By a prime submodule $P$ of an $R$-module $M$ we mean, $P\neq M $ and for every ideal $I$ of $R$ and every submodule $N$ of $M$, if $IN\subseteq P$ , then either $N\subseteq P$ or $IM\subseteq P$.

Let $P$ be a proper submodule of $M$ with $P:M=\operatorname{Ann}(M/P)=N$. Then $P$ is a prime submodule of $M$ if and only if $M/P$ is a torsion-free $R/N$-module.

My questions: $1.$ Can anyone help me to prove the previous equivalences?

$2.$ Is it true for any ring or for commutative rings only?

Thanks in advance

Answer 1

$\DeclareMathOperator\Tor{Tor}\DeclareMathOperator\Ann{Ann}$ I think your statement is false, but it can be corrected as follows:

Let $R$ be a commutative ring with unit, $M$ be an $R$-module, $P\subset M$ be an $R$-submodule and $\mathfrak p=\Ann_R(M/P)$. Then $P$ is a prime submodule of $M$ if and only if $\mathfrak p$ is a prime ideal and $\Tor_{R/\mathfrak p}(M/P)=\{0\}$.

Assume $P$ to be a prime $R$-submodule. Then $\mathfrak p=\Ann_R(M/P)$ is a prime ideal in $R$. For let $a,b\in R$ such that $ab\in\mathfrak p$ so that $Mab\subseteq P$. Then $Ma\subseteq M$, $bR\subseteq R$ and $(Ma)(bR)\subseteq Mab\subseteq P$. Since $P$ is prime, we have $Ma\subseteq P$ or $M(bR)=Mb\subseteq P$ that's $a\in\mathfrak p$ or $b\in\mathfrak p$.

Let $P$ be a prime submodule of $M$ and $x\in M$ such that $x+P\in\Tor_{R/\mathfrak p}(M/P)$. Then there exists $r$ such that $(x+P)(r+\mathfrak p)=0\in M/P$ and $r+\mathfrak p\neq 0$ in $R/\mathfrak p$. We have $xr\in P$. Let $N=rR$ and $\mathfrak a=rR$. Then $N\mathfrak a=xrR\subseteq P$ hence $N\subseteq P$ or $M\mathfrak a\subseteq P$. In the first case, $x\in P$ that's $x+P=0\in M/P$. On the other hand, if $Mr\subseteq P$, hence $r\in\Ann_R(M/P)=\mathfrak p$ so that $r+\mathfrak p=0\in R/\mathfrak p$ which is a contradiction. This proves $\Tor_{R/\mathfrak p}(M/P)=\{0\}$.

Conversely, let $\mathfrak p=\Ann_R(M/P)$ be a prime ideal in $R$ and $\Tor_{R/\mathfrak p}(M/P)=\{0\}$. Let $N\subseteq M$ be a submodule and $\mathfrak a\subseteq R$ be an ideal such that $N\mathfrak a\subseteq P$ and $N\not\subseteq P$; we have to show $M\mathfrak a\subseteq P$. Let $x\in M$ and $a\in\mathfrak a$. We have $Na\subseteq N\mathfrak a\subseteq P$, hence if $y\in N-P$ we have $ya\in P$, so that $(y+P)(a+\mathfrak p)=0$ in $M/P$. Since $\Tor_{R/\mathfrak p}(M/P)=\{0\}$ and $y+P\neq 0$ in $M/P$ we must have $a+\mathfrak p=0$ in $R/\mathfrak p$, that's $a\in\mathfrak p$. Consequently, $xa\in M\mathfrak p\subseteq P$.

If $R$ is not commutative, $\Ann_R(M/P)$ can be no a two-sided ideal of $R$, hence the assumption $R$-commutative is necessary.