Let $(R, P)$ be a local ring. If $M$ is a finitely presented module, then $M$ is flat iff $\mathrm{Tor}_{1}^R(M,R/P)=0$.
First, choose a minimal set of generators for $M$, and use these we can define a map from a free module $F$ onto $M$, let $K$ be the kernel, since $M$ is finitely presented, $K$ is finitely generated. Now, tensor with $R/P$, if $\mathrm{Tor}_{1}^R(M,R/P)=0$, I want to show $K=0$, then we are done. If $\mathrm{Tor}_{1}^R(M,R/P)=0$, we have a short exact sequence
$0\longrightarrow K\otimes R/P\longrightarrow F\otimes R/P\longrightarrow M\otimes R/P\longrightarrow0$
Is it true by counting the numbers of generators with Nakayama's lemma to deduce that $K=0$ ?
Since we can view all those tensor products as vector spaces over the field $R/P$, the short exact sequence above splits, therefore the consequence follows directly.
Best Answer
Yes, you can deduce that $K = 0$.
As you propose, chose a minimal set $\{ m_1, \dots, m_k \}$ of generators. Then, by Nayakama's lemma you obtain a basis for $M \otimes R/P \simeq M/PM$. (It is a classical result that minimal generating sets of finitely generated module $M$ over a local ring $(R,P)$ are in bijection with basis of the $R/P$-vector space $M/PM$, see here for example in Consequences, Local rings ; it is one of the many forms (or corollaries actually) of Nayakama's lemma).
Thus, $\dim_{R/P} (M \otimes R/P) = k$
Since $F = R^k$, we have $F \otimes R/P \simeq (R/P)^k$ and $\dim_{R/P} (F \otimes R/P) = k = \dim_{R/P} (M \otimes R/P)$.
Seeing all morphisms from the exact sequence you mentioned as linear maps between $R/P$-vector spaces, by equality of dimension, $F \otimes R/P \longrightarrow M \otimes R/P$ is an isomorphism, thus injective and $K \otimes R/P = 0$.
Using Nayakama's lemma again, $K = 0$ and $M$ is free thus flat.