# Tensor product is to flat as Hom is to ?

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?

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Projective module? – Alex Degtyarev Feb 29 at 5:30
Why the close votes? -- This is a 6-year-old question which has a highly-voted accepted answer. Of course it's just a question about definitions, which would likely be closed if asked nowadays, but anyway. – Stefan Kohl Feb 29 at 10:48

## 3 Answers

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.

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Thanks for the reference. I'm still learning about tensor products from Atiyah-Macdonald, and I was just wondering what the analogous concept was :) – Zev Chonoles Jan 7 '10 at 8:19
It's all there in Lang, I suppose. An advantage with a bigger algebra book is that you have everything in one place. – Anweshi Jan 7 '10 at 16:34

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.

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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.

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