An elementary fact about tensor product of modules

Question

Let $A$ be an $R$-right module, $N$ be a submodule of $R$-left module $M$, $\pi:M\rightarrow M/N $ is the natural epimorphism. What is $\ker\left(\pi\otimes1_{A}\right) $?

I appreciate your help.

Answer 1

I guess you are wondering about the map $ M \otimes A \rightarrow M/N \otimes A$. $A$ is free and so flat as $A$-module which means $\ker \pi \otimes A$ is a submodule of $M \otimes A$. It is clearly in the kernel of $\pi \otimes 1$. Now think of a an element of $\pi \otimes 1$. You can see it's of the form $u \otimes 1$ for $u \in M$. Then again by flatness, argue that $u \in \ker \pi$.

Answer 2

We have the short exact sequence $$ 0 \to N \xrightarrow[]{i} M \xrightarrow[]{\pi} M/N \to 0$$

Tensor on the right by $A$ and get a right exact sequence

$$N \otimes A \xrightarrow[]{i\otimes 1_A} M \otimes A \xrightarrow[]{\pi\otimes 1_A} (M/N)\otimes A \to 0$$

Therefore, the kernel is the submodule of $M\otimes A$ generated by $i(n)\otimes a$ with $n\in N$ and $a \in A$.

Note that $i \otimes 1_A$ may not be injective.

Answer 3

$\DeclareMathOperator{\Tor}{Tor}$ Let $N\subseteq M$ and $A$ be as in the question. In general, there is not a simple description of the kernel of the map $M\otimes A\to M/N\otimes A$.

Given modules $X$ and $Y$ (on the right and on the left, respectively) there is a standard way to construct an abelian group $\Tor(X,Y)$; this can be done in several ways, one of which is very similar to the construction of the tensor product itself and which you can find explained in MacLane's book Homology. Along with the construction of this abelian group one shows that whenever we have a short exact sequence $$0\to X'\to X\xrightarrow{f} X''\to 0$$ of right modules, we get an exact sequence $$\Tor(X'',Y)\xrightarrow{\partial} X'\otimes Y\to X\otimes Y\xrightarrow{f\otimes1_Y} X''\otimes Y\to 0$$ for a certain map $\partial$. In this way we get a description of the kernel of $f\otimes1_Y$ as the quotient $(X'\otimes Y)/\partial(\Tor(X'',Y))$.

In your specific case, what we get from the short exact sequence $0\to N\to M\xrightarrow{f} M/N\to 0$ is the exact sequence $$\Tor(M/N,A)\to N\otimes A\to M\otimes A\xrightarrow{f\otimes1_A} M/N\otimes A\to 0$$ and the isomorphism $\ker(f\otimes1_A)\cong (N\otimes A)/\partial(\Tor(M/N,A))$.

One can find examples showing that pretty much anything can happen, so this description is as good as it gets.

For example, if $M/N$ is free or flat, or if $A$ is flat, then $\Tor(M/N,A)$ is zero and the kernel is just $X'\otimes Y$. It may happen also that the $Tor$ is nonzero and the map $\partial$ is zero. It may happen that the map $\partial$ is an isomorphism, and then your kernel is zero. It may well happen that neither $M/N$ nor $A$ are flat, yet $\Tor(M/N,A)$ is nevertheless zero. Etc.

In general, the kernel depends on the actual structures of the modules $M$, $N$, and $A$, and on the way $N$ sits inside $M$ (so that one can pick a different submodule $N'$ in $M$ which is isomorphic to $N$ but for which the kernel in question is different).

Answer 4

Let $K\stackrel f\to M\stackrel g\to L\to 0$ and $K'\stackrel {f'}\to M'\stackrel {g'}\to L'\to 0$ be two short exact sequences. Then $$\ker (g\otimes g')=\operatorname{im}(f\otimes1_{M'})+\operatorname{im}(1_M\otimes f').$$

In your case $\ker(\pi\otimes 1_A)=\operatorname{im}(\iota\otimes1_{A})$, where $\iota:N\to M$ is the inclusion. One can also write $$\ker(\pi\otimes 1_A)=\langle \{x\otimes a:x\in N, a\in A\}\rangle.$$ (One can prove directly the last claim: denote the right hand side by $D$, and show that there is a well-defined map $q:M/N\otimes A\to (M\otimes A)/D$ such that $q(\pi(x)\otimes a)=x\otimes a\bmod D$, so $\ker(\pi\otimes 1_A)\subseteq D$. The converse is obvious.)