A module is flat if and only if all prime localizations are flat, equivalently all maximal localizations are flat

If $M$ is flat, then every localization $M_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ by [L1]. Therefore 1 implies 2, and 2 implies 3 trivially.

Assume 3. By [L2] it is enough to prove that for every finitely generated ideal $I\subseteq R$, the map $I{\otimes }_{R}M\to M$ is injective. Let $K$ be its kernel. If $K\mathrm{\ne }0$, then some maximal ideal $\mathfrak{m}$ contains $\mathrm{Ann}(k)$ for a nonzero $k\in K$, so $K_{\mathfrak{m}}\mathrm{\ne }0$.

But localization commutes with tensor products, so localizing the map of algebra at $\mathfrak{m}$ gives $I_{\mathfrak{m}}{\otimes }_{R_{\mathfrak{m}}}{M}_{\mathfrak{m}}\to M_{\mathfrak{m}}\mathrm{.}$ By assumption 3 and criterion [L2], this map is injective. Hence its kernel $K_{\mathfrak{m}}$ is zero, contradicting algebra. Therefore $K=0$.

So the ideal-injection criterion [L2] holds globally, and therefore $M$ is flat. Thus 3 implies 1.

The three conditions are equivalent.