Proving that the tensor product is right exact

Let
AαBβC0 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 ARMαIdBRMβIdCRM0 is
exact.

The part I have trouble with is that kerβIdim αId.

If we had β(b)m=0 if and only if β(b)=0 or m=0,
we could easily conclude using the exactness of the original sequence. However, it is false, right ? (I think of C3Z/2Z, we have g21=g2=g0=0, where g is a generator of C3.)

I can’t see how to proceed then… When a tensor cm is zero, what can we say on c and m in general ?

Answer

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 αid. You get an induced map (BM)/DCM. Let’s try to define an inverse: if (c,m)C×M then choose a bB such that β(b)=c, and send (c,m) to bmmod. You can check that this is well defined using the exactness of the original sequence.

Attribution
Source : Link , Question Author : Klaus , Answer Author : Dylan Moreland

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