$\forall\, 0\to N\to M\to F\to 0 $ exact, $\forall R$-module $E$, $F$ is flat $\implies 0 \to N\otimes E\to M\otimes E\to F\otimes E \to0$ is exact. In $R$-mod category, $\forall\, 0\to N\to M\to F\to 0 $ is short exact sequence, $\forall R$-module $E$.
$\forall\, 0\to N\to M\to F\to 0 $ exact, $\forall R$-module $E$, $F$ is flat $\implies 0 \to N\otimes E\to M\otimes E\to F\otimes E \to0$ is exact.
How to prove $F$ is flat $\implies$$\quad0 \to N\otimes E\to M\otimes E\to F\otimes E \to0$ is exact?
My thought:
I gave an answer. I wonder if there's an alternative and more direct proof, just using definition of flat module instead of introducing another short exact sequence, since it's not easy for beginners to think of that method. Thanks in advance :)
 A: Let me try and make the question more precise. You are given a short exact sequence $0\to M\to N\to F\to 0$, with $F$ flat and you want to prove that, for every $E$, the sequence $0\to M\otimes E\to N\otimes E\to F\otimes E\to 0$ is exact.
Take an exact sequence $0\to X\to Y\to E\to 0$ with $Y$ projective and build the diagram with exact rows and columns
$$\require{AMScd}
\begin{CD}
{} @. {} @. {} @. 0 \\
 @. @. @. @VVV \\
{} @. M\otimes X @>a>> N\otimes X @>b>> F\otimes X @>>> 0 \\
@. @VcVV @VdVV @VeVV @. \\
0 @>>> M\otimes Y @>f>> N\otimes Y @>g>> F\otimes Y @>>> 0 \\
@. @VhVV @ViVV @VjVV @. \\
{} @. M\otimes E @>k>> N\otimes E @>l>> F\otimes E @>>> 0 \\
@. @VVV @VVV @VVV @. \\
{} @. 0 @. 0 @. 0
\end{CD}
$$
where we have used the fact that $F$ and $Y$ are flat. 
Here's the diagram chasing. Let $x\in\ker k$; then $x=h(y)$ and $if(y)=kh(y)=0$, so $f(y)\in\ker i$; hence $f(y)=d(z)$. Since $eb(z)=gd(z)=gf(z)=0$, we conclude $b(z)\in\ker e$, so $b(z)=0$. This implies $z=a(t)$.
Thus $fc(t)=da(t)=d(z)=f(y)$; since $f$ is injective, we have $y=c(t)$ and finally
$$
x=h(y)=hc(t)=0
$$
A: We can choose a short exact sequence $0\to K\to L\to E\to 0 $ s.t. $L$ is free module, and use properties: $(1)\,\otimes $ is right exact $(2)L$ is free ($\implies$flat) $(3) F$ is flat for diagram
$$N\otimes K\to M\otimes K\to F\otimes K$$
$$\downarrow \quad\quad\quad\quad\downarrow \quad\quad\quad\quad\downarrow$$
$$N\otimes L\to M\otimes L\to F\otimes L$$
$$\downarrow \quad\quad\quad\quad\downarrow \quad\quad\quad\quad\downarrow$$
$$N\otimes E\to M\otimes E\to F\otimes E$$
Snake Lemma then yields $\quad0 \to N\otimes E\to M\otimes E\to F\otimes E \to0$.
Using the same idea, we can show this:
$0 \to F' \to F \to F''$ is exact and $F''$ is flat, then $F'$ is flat $\iff F$ is flat.
Especially, for exact sequence $0 \to F^0\to F^1 \to \cdots \to F^n$, if $F^1, \cdots, F^n$ are flat, then $F^0$ is flat.
