# The condition for flat module is projective.

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!

1. "$$u$$ 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.
2. The rest of the proof is devoted to actually proving the claim that $$u(x)=x$$ for all $$x\in K$$.