Proving that the tensor product is right exact

Question

Let $A\stackrel{\alpha}{\rightarrow}B\stackrel{\beta}{\rightarrow}C\rightarrow 0$ a exact sequence of left $R$-modules and $M$ a left $R$-module ($R$ any ring).

I am trying to prove that the induced sequence $$A\otimes M\stackrel{\alpha\otimes Id}{\rightarrow}B\otimes M\stackrel{\beta\otimes Id}{\rightarrow}C\otimes M\rightarrow 0$$ is exact.

The part I have trouble with is that $\ker{\beta\otimes Id}\subset\text{im }{\alpha\otimes Id}$.

If we had $$\beta(b)\otimes m=0 \text{ if and only if } \beta(b)=0\text{ or }m=0,$$ we could easily conclude using the exactness of the original sequence. However, it is false, right ? (I think of $C_3\otimes \mathbb{Z}/2\mathbb{Z}$, we have $g^2\otimes 1=g\otimes 2=g\otimes 0=0$, where $g$ is a generator of $C_3$.)

I can't see how to proceed then... When a tensor $c\otimes m$ is zero, what can we say on $c$ and $m$ in general ?

Answer 1

The proof mentioned by Frederik and Loronegro is great because it provides a first example of how it can be useful to know that two functors are adjoint: left adjoints are right exact. However, you can also argue as follows. Let $D$ be the image of $\alpha \otimes \operatorname{id}$. You get an induced map $(B \otimes M)/D \to C \otimes M$. Let's try to define an inverse: if $(c, m) \in C \times M$ then choose a $b \in B$ such that $\beta(b) = c$, and send $(c, m)$ to $b \otimes m \bmod D$. You can check that this is well defined using the exactness of the original sequence.

Answer 2

First of all, if you start with an exact sequence $A\to B\to C\to 0$ of left $R$-modules, then $M$ should be a right $R$-module, so that the tensor products $M\otimes A$, etc. are well defined.

Second, it happens that for the proof that I will explain, it is easier to consider the functor $M\otimes\underline{}$ which is applied to the exact sequence. Then we can use the isomorphism $M\otimes A\cong A\otimes M$ to prove the exactness of the sequence $A\otimes M\to B\otimes M\to C\otimes M\to 0$, in case that $A,B,C$ are right $R$-modules and $M$ is a left $R$-module.

I don´t know a direct proof of the proposition and I think it may be difficult. The proof I know uses indeed the natural isomorphism mentioned by @Frederik (I think that in his comment there is a misorder of the modules involved). With the notation used by @Klaus, the natural isomorphism that is convenient is $Hom(M\otimes A,Q)\cong Hom(A,Hom(M,Q))$, where $Q$ is an injective cogenerator right $R$-module (for example, the injective hull of the direct sum of a complete set of non-isomorphic simple modules). We can consider the functor $(\underline{})^*=Hom(\underline{},Q)$, so the latter natural isomorphism can be stated as $(M\otimes A)^*\cong Hom(A,M^*)$. This functor $(\underline{})^*$, which is contravariant, so that it reverses the direction of morphisms, has the following property:

For $R$-modules $K,N,L$, the sequence $K\to M\to N$ is exact if, and only if, the sequence $N^*\to M^*\to K^*$ is exact.

Therefore, the sequence $M\otimes A\to M\otimes B\to M\otimes C\to 0$ is exact if, and only if, $0\to (M\otimes C)^*\to (M\otimes B)^*\to (M\otimes A)^*$ is exact, if and only if, $0\to Hom(C,M^*)\to Hom(B,M^*)\to Hom(A,M^*)$ is exact.

But the contravariant functor $Hom(\underline{},M^*)$ is left exact, that is, if the sequence $A\to B\to C\to 0$ is exact, then the sequence $0\to Hom(C,M^*)\to Hom(B,M^*)\to Hom(A,M^*)$ is exact, and this is very much easier to prove directly, rather than the right exactness of the functor $M\otimes\underline{}$, which @Klaus was trying.