most recent 30 from http://mathoverflow.net 2013-05-23T14:19:55Z http://mathoverflow.net/feeds/question/11020 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11020/tensor-product-is-to-flat-as-hom-is-to Zev Chonoles 2010-01-07T07:23:15Z 2010-01-07T07:44:49Z

<p>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?</p>

http://mathoverflow.net/questions/11020/tensor-product-is-to-flat-as-hom-is-to/11021#11021 zeb 2010-01-07T07:27:32Z 2010-01-07T07:27:32Z

<p>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.</p>

http://mathoverflow.net/questions/11020/tensor-product-is-to-flat-as-hom-is-to/11022#11022 Charles Siegel 2010-01-07T07:28:38Z 2010-01-07T07:28:38Z

<p>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 <a href="http://en.wikipedia.org/wiki/Exact%5Ffunctor" rel="nofollow">wikipedia</a>.</p>

http://mathoverflow.net/questions/11020/tensor-product-is-to-flat-as-hom-is-to/11023#11023 Emerton 2010-01-07T07:44:49Z 2010-01-07T07:44:49Z

<p>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.</p>