Counterexample to $F\otimes_R k(P)\ne 0$ iff $IF\ne F$ for $F$ flat

Question

Let $F$ be a flat $R$-module. By this, TFAE:

1. $F$ is faithfully flat, i.e., $L\otimes_R F\ne0$ for any $R$-module $L \ne 0$;
2. $IF\ne F$ for all proper (in particular, prime) ideals $I\subset R$.

I was wondering whether the two conditions above are equivalent to

3. $F\otimes_R k(P)\ne 0$ for all prime ideals $P\subset R$.

Yesterday I had the following argument in mind:

Condition 2 holds iff $F\otimes R/P\ne 0$. Since $F$ is faithfully flat (due to the fact that $(2)\iff (1)$), this takes place iff $R/P\ne0$. Condition 3 holds iff $F\otimes (R/P)_P\ne 0$ iff $(R/P)_P\ne 0$ (again by the faithful flatness of $F$). So proving $(2)\iff (3)$ is the same as proving $R/P=0$ iff $(R/P)_P=0$. We know that for any $R$-module $N$, $N=0$ iff $N_Q=0$ for all primes $Q$. So apply this for $N=R/P$.

Today I realized that this argument is of course false, for this $N=R/P$ varies as $P$ varies, and one cannot apply that criterion. So I believe the third condition is not equivalent to either of the first two ones.

Is there a counterexample to the equivalence of $(3)$ with one of the first two conditions?

(Note that they are also equivalent to condition 3 with "prime" replaced by "maximal".)

Answer 1

Your logic seems rather confused. You can't get equivalent versions of (2) or (3) using the assumption that $F$ is faithfully flat, because $F$ is only assumed to be flat, not faithfully flat. For any prime ideal $P$, $R/P$ and $(R/P)_P$ are both nonzero ($R/P$ is nonzero since $P$ is a proper ideal, and $(R/P)_P$ is just the field of fractions of the domain $R/P$ since the elements of $R\setminus P$ just map to all the nonzero elements of $R/P$).

To prove (3) is equivalent to (1) and (2), note that (1) implies (3) since $k(P)\neq 0$ for any $P$. Conversely, if (2) fails so $IF=F$ for some proper ideal $I$, let $P$ be a maximal ideal containing $I$. Then $PF=F$, so $R/P\otimes_R F=F/PF=0$. But $R/P=k(P)$ (since $P$ is maximal), so this means (3) fails.

Answer 2

2. (and 1. also) obviously implies 3.

And 3. with $P$ running through prime ideals obviously implies the same statement with $P$ running through maximal ideals, hence 3. implies 2. by your note in your very last sentence.