For a flat module, faithful flatness is equivalent to detecting nonzero modules and residue fields

If $M$ is faithfully flat and $N\mathrm{\ne }0$, then the map $0\to N$ is nonzero. If $N{\otimes }_{R}M$ were zero, tensoring would turn the nonzero map $0\to N$ into the zero map, contradicting exactness reflection in [L1]. Thus 1 implies 2.

Condition 2 implies 3 by taking $N=\kappa (\mathfrak{p})$, and 3 implies 4 by restricting to maximal primes.

Assume 4. Let $N\mathrm{\ne }0$ and choose $0\mathrm{\ne }x\in N$. The cyclic submodule $Rx\cong R\mathrm{/}\mathrm{Ann}(x)$ injects into $N$. Choose a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(x)$. Then there is a surjection $R\mathrm{/}\mathrm{Ann}(x)\twoheadrightarrow R\mathrm{/}\mathfrak{m}\mathrm{.}$ Tensoring with $M$ preserves the injection by [L2], and the target tensor is nonzero by 4. Hence $N{\otimes }_{R}M\mathrm{\ne }0.$ So 4 implies 2.

Assume 2 and let $N_1\to N_2\to N_3$ be a complex whose tensor with $M$ is exact. Since $M$ is flat, it is enough to prove exactness at $N_2$. If $x\in \ker (N_2\to N_3)$ is not in the image of $N_1\to N_2$, then it defines a nonzero element of the quotient $Q:=\ker (N_2\to N_3)\mathrm{/}\mathrm{im}(N_1\to N_2)\mathrm{.}$ But tensoring with $M$ kills $Q$, because the tensor complex is exact. This contradicts 2. Hence $Q=0$, and the original complex is exact. Therefore $M$ is faithfully flat.

The four conditions are equivalent.