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 $I$.



Checking flatness using radical ideals
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>.


