Let

Aα→Bβ→C→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⊗RMα⊗Id→B⊗RMβ⊗Id→C⊗RM→0 is

exact.The part I have trouble with is that kerβ⊗Id⊂im α⊗Id.

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 C3⊗Z/2Z, we have g2⊗1=g⊗2=g⊗0=0, where g is a generator of C3.)I can’t see how to proceed then… When a tensor c⊗m 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 α⊗id. You get an induced map (B⊗M)/D→C⊗M. Let’s try to define an inverse: if (c,m)∈C×M then choose a b∈B such that β(b)=c, and send (c,m) to b⊗mmod. 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*