I'm in trouble showing
when $R$ is a noetherian commutative ring with unity, an $R$-module $M$ is flat if $\text{Tor}_1 (M, R/P) =0$ for any prime ideal $P$ of $R$.
First thought was to use localization, but since $M$ is not finitely generated, couldn't do any. Any hint for this?
It is sufficient to show $\operatorname{Tor_1}(M,N)=0$ for all finitely generated modules $N$. Since $R$ is noetherian there is a filtration of submodules: $$0=N_0\subset N_1\subset\dots\subset N_n=N$$ with $N_{i+1}/N_i\cong R/P_i$ for prime ideals $P_i$. By induction on the length $n$ we may assume $\operatorname{Tor_1}(M,N_{n-1})=0$. Then by considering the short exact sequence $$0\to N_{n-1}\to N\to R/P_{n-1}\to0$$ we get the long exact Tor sequence: $$\dots\to\underbrace{\operatorname{Tor_1}(M,N_{n-1})}_{=0}\to\operatorname{Tor_1}(M,N)\to \underbrace{\operatorname{Tor_1}(M,R/P_{n-1})}_{=0}\to\dots$$ i.e. $\operatorname{Tor_1}(M,N)=0$.
If we drop the noetherian assumption and replace the primes $P_i$ with just ideals $I_i$ the same proof works for the statement that $M$ is flat iff $\operatorname{Tor_1}(M,R/I)=0$ for all ideals $I$.