ac.commutative algebra - Tensor product is to flat as Hom is to ?

It might also be helpful to know that projective is equivalent to being a summand of a free module (apply $Hom(M,text{--})$ to a presentation of $M$), and hence projectives are flat.
The converse is not true in general (e.g. $mathbb Q$ is flat as a $mathbb Z$-module, but not projective), but for finitely presented modules over commutative rings, flat and projective are equivalent.