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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asked Jan 7, 2010 at 7:23
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$\begingroup$ Projective module? $\endgroup$– Alex DegtyarevCommented Feb 29, 2016 at 5:30
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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 ♦Commented Feb 29, 2016 at 10:48
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3 Answers
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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.
answered Jan 7, 2010 at 7:28
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$\begingroup$ Thanks for the reference. I'm still learning about tensor products from Atiyah-Macdonald, and I was just wondering what the analogous concept was :) $\endgroup$ Commented Jan 7, 2010 at 8:19
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$\begingroup$ It's all there in Lang, I suppose. An advantage with a bigger algebra book is that you have everything in one place. $\endgroup$– AnweshiCommented Jan 7, 2010 at 16:34
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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.
