# Isomorphism of tensor product involving a principal ideal

This question arose when dealing with a long exact sequence of Tor.

Let $R$ be a (not necessarily commutative) ring, $g$ a central element of $R$ and $M$ a right $R$-module. We have an exact sequence of left $R$-modules $$0 \rightarrow Rg \rightarrow R \rightarrow R/Rg \rightarrow 0.$$ Tensoring with a right $R$-module $M$ and using that $\text{Tor}_R^1(M,R)=0$, we obtain an exact sequence $$0 \rightarrow \text{Tor}_R^1(M,R/Rg) \rightarrow M \otimes_R Rg \rightarrow M \otimes_R R \rightarrow M \otimes_R R/Rg \rightarrow 0.$$ I want to identify $\text{Tor}_R^1(M,R/Rg)$. I'm under the impression that $M \otimes_R R \cong M$ and $M \otimes_R R/Rg \cong M/Mg$ (although now I'm doubting these facts too).

My question is: is $M \otimes_R Rg \cong Mg$?

I've done the following (with most of my working omitted): we can define an $R$-biadditive map $\overline{\phi}:M \times Rg \rightarrow Mg$ by $(m,rg) \mapsto mrg$, and so this factors through to give a map $\phi: M \otimes_R Rg \rightarrow Mg$, $m \otimes rg \mapsto mrg$. The map $\psi : Mg \rightarrow M \otimes_R Rg, mg \mapsto m\otimes g$ is then inverse to $\phi$.

However, I'm told that $\text{Tor}_R^1(M,R/Rg) \neq 0$, and if $M \otimes_R Rg \cong Mg$, then the form of the exact sequence involving Tor implies that $\text{Tor}_R^1(M,R/Rg) = 0$.

• You should probably explain the context... – Mariano Suárez-Álvarez Apr 14 '15 at 8:49
• I'll edit my post then. – user216089 Apr 14 '15 at 8:50

You have perfectly defined the map $\phi$ however, the map $\psi$ is not well defined. What happen if $mg=m'g$, or equivalently, if $(m-m')g=0$ ? do we necessarily have $(m-m')\otimes g=0$ in $M\otimes_R Rg$ ?
Take $R=\mathbb{Z}, g=2$ and $M=\mathbb{Z}/2\mathbb{Z}$. Then $2\mathbb{Z}\simeq\mathbb{Z}$ so that $M\otimes 2\mathbb{Z}\simeq M$. However, $M.2=0$. The thing is $1.g=0$ in $M.2$ but $1\otimes 2\neq 0$ in $M\otimes 2\mathbb{Z}$.
The $Tor$ is precisely the difference between $Mg=\mathop{Im}\phi$ and $M\otimes_R Rg$.
• Great, that makes perfect sense. I forgot that I even had to check that $\psi$ was well-defined. Thanks! – user216089 Apr 14 '15 at 9:24