Timeline for Why are injective modules more complicated than projective modules?
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20 events
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| Jan 29, 2021 at 18:37 | comment | added | Qi Zhu | @LeonidPositselski Thank you for this thorough answer! A lot of what you say seems to point towards that projectives and injectives (covers/envelopes) behave quite differently which is one reason why they maybe should not be considered duals. I'm not sure if that's a valid reason, being dual does not mean being similar in my eyes. E.g. also limits behave quite differently than colimits which are (arguably) dual notions. | |
| Jan 28, 2021 at 13:04 | comment | added | Leonid Positselski | [cont'd] ... Projective modules over commutative Noetherian rings behave quite differently. (E.g., even the classification of finitely generated projective modules, even over Dedekind domains, involves the Picard group, and generally the group $K_0$ etc.) | |
| Jan 28, 2021 at 13:01 | comment | added | Leonid Positselski | In fact, these duality-analogies are a bit tricky. One can argue that the proper analogues of injective modules are not the projective modules, but the flat cotorsion modules. Injective modules over commutative Noetherian rings were classified by Matlis; they are direct sums of the injective envelopes of the residue fields, taken over the spectrum points of the ring. There is a very similar classification, due to Enochs, of flat cotorsion modules over commutative Noetherian rings, as products of certain modules sitting at the prime ideals. | |
| Jan 28, 2021 at 12:58 | comment | added | Leonid Positselski | @QiZhu Flat covers always exist, to begin with (for all modules over an arbitrary ring, or for quasi-coherent sheaves over a scheme, under mild assumptions on the scheme, etc.). Projective covers usually don't exist (in fact, projective covers of arbitrary modules only exist over perfect rings, which are quite special -- a Noetherian commutative ring is only perfect when it is Artinian). | |
| Jan 28, 2021 at 12:22 | comment | added | Qi Zhu | @LeonidPositselski Could you elaborate on why flat covers are the "right" dual analogues of injective envelopes and not the projective ones? (Sorry for reviving such an old comment but I found your remark quite interesting.) | |
| Jul 27, 2012 at 18:22 | comment | added | Leonid Positselski | @Fernando: concerning covers, it appears that the "right" dual analogues of the injective envelopes of modules are the flat covers, not the projective ones. Flat covers of modules (and also of sheaves of modules, etc.) exist quite generally. | |
| Jul 27, 2012 at 17:32 | comment | added | Leonid Positselski | Given such an abelian group homomorphism $f$, consider the corresponding element of $Hom_R(P,Hom_{\mathbb Z}(R,\mathbb Q/\mathbb Z)) = Hom_{\mathbb Z}(R\otimes_R P, \mathbb Q/\mathbb Z) = Hom_{\mathbb Z}(P,\mathbb Q/\mathbb Z)$; so we get an $R$-module homomorphism $g\colon P \to Hom_{\mathbb Z}(R,\mathbb Q/\mathbb Z)$. Make the component of the desired embedding $M\to J(M)$ corresponding to the direct factor indexed by $f$ equal to the map $g$. | |
| Jul 27, 2012 at 17:22 | comment | added | Leonid Positselski | Here is the construction of a functorial injection of an arbitrary $R$-module $M$ to an injective $R$-module $J(M)$ dual-analogous to the above-mentioned construction of the "free $R$-module on the elements of $M$". Set J(M) to be the product of copies of $Hom_{\mathbb Z}(R,\mathbb Q/\mathbb Z)$ indexed by all the abelian group homomorphisms $M\to \mathbb Q/\mathbb Z$. | |
| Jul 27, 2012 at 16:07 | comment | added | Charles Staats | Note: My last comment is related to Qiaochu's answer, which--among other things--offers a construction for a "better" natural injective in the case $R=\mathbb Z$, but also gives some reasons why this may not work in greater generality. | |
| Jul 27, 2012 at 16:03 | comment | added | Charles Staats | On the other hand, the "injective envelope" $I(M)$ is not a functor, whereas the "free abelian group on the elements of $M$" $F(M)$ is functorial. So perhaps this qualifies as a "deep" reason: if $M$ is an $R$-module, the "natural way to produce a projective that surjects onto $M$" is $F(M) \to M$, which is functorial. The "natural way to produce an injective module into which $M$ injects" is $M \hookrightarrow I(M)$, which is not functorial (and hence, not really "natural" at all, in the technical sense). However, I don't know about the existence of a "better" natural injective. | |
| Jul 27, 2012 at 15:57 | comment | added | Charles Staats | ...the indecomposable modules $I(R/\mathfrak p)$, where $\mathfrak p$ ranges over the primes of $R$. | |
| Jul 27, 2012 at 15:54 | comment | added | Charles Staats | There is, in fact, a classification of sorts for injective modules over a commutative noetherian ring; I believe this is discussed in Maclane's book on homological algebra (but probably in many others as well). The basic idea is this: every $R$-module $M$ has an "injective envelope"; this is an injection of modules $M \hookrightarrow I(M)$ such that every nonzero submodule of $I(M)$ intersects $M$. One can show that $I(M)$ is injective, and that every injective module containing $M$ also contains a (non-unique) copy of $I(M)$. Then an injective module is precisely a direct sum of... | |
| Jul 27, 2012 at 8:49 | answer | added | Liviu Nicolaescu | timeline score: 12 | |
| Jul 27, 2012 at 1:53 | comment | added | paul garrett | I myself think $\mathbb Q/\mathbb Z$ is more complicated than $\mathbb Z$. | |
| Jul 27, 2012 at 1:36 | comment | added | name | @roy Do you mean why is Q/Z more complicated than Z? I imagine you are thinking that every abelian group has a projective resolution by direct sums of Z, and similarly, every abelian group has an injective resolution by products of Q/Z? | |
| Jul 27, 2012 at 1:20 | history | edited | Karl Schwede |
Added a tag since I think some people in commutative algebra might have some comments on this.
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| Jul 27, 2012 at 1:05 | answer | added | Qiaochu Yuan | timeline score: 26 | |
| Jul 27, 2012 at 1:05 | comment | added | roy smith | why is Q more complicated than Z? | |
| Jul 27, 2012 at 0:55 | comment | added | Fernando Muro | I don't quite agree. Injective envelopes occur more easily than projective covers, and Grothendieck abelian categories have injectives but need not have projectives. Some people find easier to think of projectives because they think of free objects, which behave like vector spaces to some extent. But once you're into the abelian category world, injectives are even easier. | |
| Jul 27, 2012 at 0:39 | history | asked | temp | CC BY-SA 3.0 |