Solution to exercise 1.3H of Vakil's 'Foundations of Algebraic Geometry' I have been trying to solve exercise 1.3H of Ravi Vakil's AG notes for quite a while now, and I just cannot seem to do it. The question asks me to prove (from the concrete definition, presumably without much heavy categorical machinery) that the tensor functor is right exact. 

This question has been asked before (see Proving that the tensor product is right exact) but the answers given only contain hints to help with finding the solution, and unfortunately I cannot seem to get the answer even with these hints, no matter how hard I try. 

I feel as if I am not really grasping what equality means for tensor products, and I really feel I need to see a (mostly completed) proof to rectify this. 

Edit: Rings are assumed commutative.
Let $A\to B\to 0$ be an exact sequence of modules; we must show $A\otimes M\to B\otimes M\to 0$ is exact. An element of $B\otimes M$ is of the form $\sum b_i\otimes m_i$, where the representation is unique up to bilinear rewriting. But this is the image of $\sum a_i\otimes m_i$, where each $a_i$ maps to $b_i$.

That's pretty much it. To make sure this wasn't too facile, let's check whether such a proof would incorrectly prove left exactness, supposing now $0\to A \stackrel{f}{\to} B$ were exact; if $\sum f(a_i)\otimes m_i=0$, there's no reason we should have each $f(a_i)=0$. So there's no problem. 

Here's a less hands on proof that still isn't categorically very sophisticated: in the situation of the first paragraph we want to prove $A\otimes M\to B\otimes M$ is an epimorphism, so that any two maps $g,h:B\otimes M\to C$ which are equal on $A\otimes M$ are equal. But maps out of the tensor product are equal if and only if the represent the same bilinear map out of $B\times M$; since $A\times M\to B\times M$ is a surjection of sets, the claim follows.
Let $A\to B\to 0$ be an exact sequence of modules; we must show $A\otimes M\to B\otimes M\to 0$ is exact. An element of $B\otimes M$ is of the form $\sum b_i\otimes m_i$, where the representation is unique up to bilinear rewriting. But this is the image of $\sum a_i\otimes m_i$, where each $a_i$ maps to $b_i$.

That's pretty much it. To make sure this wasn't too facile, let's check whether such a proof would incorrectly prove left exactness, supposing now $0\to A \stackrel{f}{\to} B$ were exact; if $\sum f(a_i)\otimes m_i=0$, there's no reason we should have each $f(a_i)=0$. So there's no problem. 

Here's a less hands on proof that still isn't categorically very sophisticated: in the situation of the first paragraph we want to prove $A\otimes M\to B\otimes M$ is an epimorphism, so that any two maps $g,h:B\otimes M\to C$ which are equal on $A\otimes M$ are equal. But maps out of the tensor product are equal if and only if the represent the same bilinear map out of $B\times M$; since $A\times M\to B\times M$ is a surjection of sets, the claim follows.