I found a condition for flat module is projective
" Let $R\subset S$ be an extension of rings. Let $M$ be a flat left $R$-module. For $M$ be projective, it is necessary and sufficient that there exists an exact sequence
$$0\to K\to P\to M\to 0$$
of left $R$-module with $P$ projective and the scalar extension $S\otimes_R K$ a finitely generated left $S$-module."
Proof
The necessity is obvious. We prove sufficiency of the conditions stated. Let $Q$ be a projective $R$-module such that $P\oplus Q$ is free. Then with obivious maps, $0\to K\to P\oplus Q\to M\oplus Q\to 0$ is still an exact sequence; also $M\oplus Q$ is a flat module. If we can show that $M\oplus Q$ is projective, it means in particular that $M$ itself is projective. Hence without loss of generality, we can assume that there exists an exact sequence $0\to K\to F\to M\to 0$ of lefft $R$-modules with $F$ free and $S\otimes_R K$, a finitely generated $S$-module. We can also assume that $K$ is a submodule of $F$ and $M$ is the quotient module $F/K$.
Let $\alpha_1,\alpha_2,\dots,\alpha_n$ be a (finite) set of generators of the left $S$-module $S\otimes_R K$. The $\alpha_1$ has the form
$$\alpha_1=a_{i_1}\otimes x_{i_1}+ a_{i_2}\otimes x_{i_2}+\dots+a_{i_p}\otimes x_{i_p} $$ where $a_{i_j}\in S, x_{i_j}\in K$ forall $i,j$. Now $\{x_{i_j}\}, i=\overline{1,n},j=\overline{1,p}$ being a finite subset of $K$, the flatness of $M$ ensure the existence of a $R$-linear homomorphism $u:F\to K$ such that $u(x_{i_j})=x_{i_j}$ forall $i,j$. We claim that $u(x)=x$ forall $x\in K$; this mean $u$ splits the inclusions $K\to F$ and hence $M$ will be projective.
By the finiteness of $S\otimes_R K$ as $S$-module, there are elements $\lambda_1,\lambda_2,\dots,\lambda_n$ in $S$ such that $1\otimes x=\lambda_1\theta_1+\lambda_2\theta_2+\dots+\lambda_n\theta_n$. Let $I_s$ denote the identity endomorphism of $S$ onto itself. Then
$$ (I_s\otimes u)(\theta_i)=\displaystyle \sum\limits_{1 \le j \le p} {{a_{ij}} \otimes u\left( {{x_{ij}}} \right)} =\sum\limits_{1 \le j \le p} {{a_{ij}} \otimes x_{ij} }=\theta_i$$
This holds for $1\le i\le n$. Hence,
$$ \displaystyle \begin{array}{*{20}{l}}
{1 \otimes u\left( x \right)}&{ = \left( {{I_s} \otimes u} \right)\left( {1 \otimes x} \right)}&{ = \left( {{I_s} \otimes u} \right)\left( {{\lambda _1}{\theta _1} + {\lambda _2}{\theta _2} + \ldots + {\lambda _n}{\theta _n}} \right)}\\
{}&{}&{ = \displaystyle\sum\limits_{1 \le i \le n} {{\lambda _i}\left( {{I_s} \otimes u} \right)\left( {{\theta _i}} \right)} }\\
{}&{}&{ =\displaystyle \sum\limits_{1 \le i \le n} {{\lambda _i}{\theta _i} = 1 \otimes x} }
\end{array}$$
by what precedes. From this $u(x)=x$ will follow if we can show that the canonical homomorphism $K\to S\otimes_R K$ is injective. But this is clear since the canonical homomorphism $F\to S\otimes_R F$ is injective (F being free) and the square
$\require{AMScd}$
\begin{CD}
0@> >>K @> >> F\\
@. @VVV @VVV \\
@. S\otimes_RK @>>>S\otimes_R F\\
\end{CD} is commutative. This completes the proof.
Why "$u$ splits the inclusions $K\to F$ and hence $M$ will be projective" and What is the purpose of the proof after "$M$ wil be projective"? Thanks alots!
Best Answer
"$u$ splits the inclusion $K\to F$ and hence $M$ will be projective" means that $u$ will realise $K$ as a quotient of $F$, and the kernel of $u$ will be isomorphic to $M$. Indeed, $u(x)=x$ for all $x\in K$ implies that the kernel of $u$ intersects $K$ trivially, hence injects into $M$ under the projection map $F\to M$; and given any $m\in M$, take any preimage $f$ in $F$, and note that $f-u(f)$ is in the kernel of $u$ (again using that $u(x)=x$ for all $x\in K$) and maps to $m$, so the kernel of $u$ also surjects on $M$. Thus, $M$ would be realised as a submodule of a free module, and it would follow that $M$ is projective.
The rest of the proof is devoted to actually proving the claim that $u(x)=x$ for all $x\in K$.
Your second question is a perfect example of why proofs should be written in linear order; then these kinds of confusions would not arise. Even in that last part of the proof, the author manages to turn the logical order upside down one more time.