6
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Projective module? $\endgroup$ Feb 29, 2016 at 5:30
  • 1
    $\begingroup$ 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. $\endgroup$
    – Stefan Kohl
    Feb 29, 2016 at 10:48

3 Answers 3

14
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks for the reference. I'm still learning about tensor products from Atiyah-Macdonald, and I was just wondering what the analogous concept was :) $\endgroup$ Jan 7, 2010 at 8:19
  • $\begingroup$ It's all there in Lang, I suppose. An advantage with a bigger algebra book is that you have everything in one place. $\endgroup$
    – Anweshi
    Jan 7, 2010 at 16:34
12
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.