TITLE: Tensor product is to flat as Hom is to ?
QUESTION [6 upvotes]: Sorry if I'm missing something here, but what do we call $M$ if the functor $H_M:N\mapsto Hom(M,N)$ is exact? Is this in fact equivalent to being flat through some adjointness properties?

REPLY [14 votes]: We call such modules projective.  If you take $N\mapsto Hom(N,M)$ then you get injective modules.  This is fairly basic, and covered in any homological algebra book, and mentioned on wikipedia.

REPLY [12 votes]: 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.

REPLY [6 votes]: I'm pretty sure that $M$ is called projective in this case, and if $N \rightarrow Hom(N,M)$ is exact then $M$ is called injective. I might have it backwards, though.