Let $R$ be a commutative ring and $M$ a not necessary finitely presented $R$-module. I am looking for a prove or a counterexample to the following statement: $M$ is flat as an $R$-module if and only if $\mathrm{Tor}^R_1(R/I,M)=0$ for all radical ideals $I$.
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If you're not assuming $R$ is Noetherian, this is false.
Namely, let $R$ be a valuation domain with value group $\mathbb Q$, let $m$ be its maximal ideal, and let $M=k=R/m$. Then the only radical ideals of $R$ are $0$ and $m$, and one can show that $\operatorname{Tor}_1^R(R/I,k) = 0$ for $I=0, m$, but it is also clear that $k$ is not flat. A full argument, along with a way to construct additional such counterexamples, is given in Remark 3.5 of my joint work with Jay Shapiro, ``Strong Krull primes and flat modules'', available at http://arxiv.org/abs/1303.7458.
(Note that as you suspected, in the above situation $k$ is not finitely presented, since $m$ is not finitely generated.)
However, if $R$ is Noetherian, then your assertion is true, as noted in the comment by @user76758 after my original, ill-considered answer to your question. Namely, it is enough to consider finitely generated modules $N$ in the first coordinate of the $\operatorname{Tor}$, and then by taking a prime filtration of such $N$, one may merely check that $\operatorname{Tor}_1^R(R/P, M) = 0$ for any prime ideal (by induction on the [necessarily finite] length of said filtration).
Further results in this vein are explored in our article. For instance, if $R$ is a reduced but not necessarily Noetherian ring, flatness of $M$ is equivalent to the vanishing (or even the torsion-freeness) of $\operatorname{Tor}_1^R(R/I, M)$ for all finitely generated ideals $I$. (See Theorem 3.3 in the paper.)
For the Noetherian case, see my paper with Yongwei Yao, ``Criteria for flatness and injectivity'', published in Math. Z. (2011), also available at http://arxiv.org/abs/1103.4726.
answered Nov 28, 2013 at 19:12
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4$\begingroup$ For any noetherian ring $R$ it suffices to test against prime ideals $I$. Indeed, by direct limit considerations we just need ${\rm{Tor}}_1(N,M)=0$ for finitely generated $N$, and any such $N$ has a finite composition series whose successive quotients are $R/P$ for prime ideals $P$ since $R$ is noetherian. $\endgroup$ Commented Nov 28, 2013 at 23:06
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$\begingroup$ Ah; good point. We even mention that in passing in our paper (me & Yongwei Yao). Sorry to be absent-minded. Should I edit my answer to that effect? $\endgroup$ Commented Nov 29, 2013 at 5:00
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$\begingroup$ Do whatever you wish. If you alert me via comment that you make such an edit then I will delete my (then moot) comments. $\endgroup$ Commented Nov 29, 2013 at 5:32
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$\begingroup$ Thank you for the answer. I was aware of the noetherian case. Note that the ring in the counterexample is not Jacobson. I thought the statement is maybe true in case the ring is assumed to be Jacobson. $\endgroup$ Commented Dec 1, 2013 at 9:17
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1$\begingroup$ @user7d229955: Interesting question. I don't know. You could try $R=k[\{x^q \mid q \in {\mathbb Q}_{\geq 0}\}]$ (which is probably a Jacobson ring), where $k$ is a field, and see what happens with $M=k=R/m$, where $m$ is the maximal ideal generated by all the positive powers of $x$. My guess is that this will give you a Jacobson counterexample. $\endgroup$ Commented Dec 1, 2013 at 19:24