TITLE: Exact sequence with flat module tensored by module stays exact
QUESTION [5 upvotes]: The following theorem is given in Liu's book Algebraic Geometry and Arithmetic Curves, proposition 1.2.6:
Let $A$ be a ring. Let $0\to M^\prime\to M\to M^{\prime\prime}\to 0$ be an exact sequence of $A$-modules. Let us suppose that $M^{\prime\prime}$ is flat. Then the sequence $0\to M^\prime \otimes N\to M\otimes N\to M^{\prime\prime} \otimes N\to 0$ is also exact for any $A$-module $N$.
Liu gives proof in http://176.58.104.245/ALGANT/TONG/Liu-1-4.pdf page 9.
I have problems to understand the proof. The mistake is easy to fix, Proposition 2.12 should be 1.12. But what does the "Let us write $N$ as the quotient of a free $A$-module $L$ by a submodule $K$" means? Is it $N=L/K$? And how one can do the diagram chasing part?

REPLY [5 votes]: Consider the diagram of the proof:
$$\require{AMScd}
\begin{CD}
{} @. M'\otimes K @>f>> M\otimes K @>g>> M''\otimes K \\
@. @VVhV @VViV @VV{\alpha}V \\
0 @>>> M'\otimes L @>j>> M\otimes L @>k>> M''\otimes L \\
@. @VV{\beta}V @VVlV \\
{} @. M'\otimes N @>m>> M\otimes N \\
@. @VVV @VVV \\
{} @. 0 @. 0
\end{CD}
$$
Let $x\in\ker m$; then $x=\beta(y)$ and $lj(y)=0$, so $j(y)=i(z)$.
Thus $\alpha g(z)=ki(z)=kj(y)=0$, so $g(z)=0$ and therefore $z=f(u)$.
Since $jh(u)=if(u)=i(z)=j(y)$, we have $y-h(u)\in\ker j=\{0\}$, so $y=h(u)$ and finally $x=\beta(y)=\beta h(u)=0$.