Timeline for Indecomposable projectives correspond to irreducibles - reference
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 10, 2012 at 9:18 | vote | accept | Sasha | ||
| Feb 10, 2012 at 9:18 | |||||
| Dec 21, 2011 at 21:47 | answer | added | Jim Humphreys | timeline score: 4 | |
| Dec 18, 2011 at 18:19 | history | edited | Sasha | CC BY-SA 3.0 |
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| Dec 18, 2011 at 18:17 | answer | added | Sasha | timeline score: 3 | |
| Dec 18, 2011 at 17:39 | comment | added | Florian Eisele | I was just referring to the second assertion: "Given an indecomposable projective of finite length, how to show that it has a unique irreducible quoteint?". That could be true. I agree that the first claim (projective indecomposables <-> simples) is wrong. I don't think even the Krull-Schmidt property can fix this. Take for instance $R=\mathbb Z_{(5)}[x]/(x^2+1)$ ($\mathbb Z_{(5)}$ the localization of $\mathbb Z$ at $(5)$). $R$ is indecomposable but has two simple quoutients because $\mathbb F_5[x]/(x^2+1) \cong \mathbb F_5 \oplus \mathbb F_5$. | |
| Dec 18, 2011 at 17:03 | comment | added | Ralph | Florian. the OP states that "In an abelian category, indecomposable projectives of finite length correspond bijectively to irreducibles". What are the indecomposable projective resp. irreducible objects in the cat. of f.g. ab. groups ? (cf. my 2nd comment). // The situation may improve, if all(!) projectives have finite length. But this condition isn't required by the OP. Still I'm pretty sure that one needs a progenerator (corresponds to the rank-one free module in the module category) and the unique decomposition into irreducibles. | |
| Dec 18, 2011 at 16:40 | comment | added | Florian Eisele | He says projectives "of finite length". If that means what I think it does (i.e., the same it means in the case of a module category), the category of abelian groups is not a counter-example since $\mathbb Z$ is not of finite length. I assume when a projective has finite length the situation is the same as with modules over artinian rings, where indecomposable projectives have simple top. | |
| Dec 18, 2011 at 13:32 | comment | added | Ralph | That doesn't suffice. The category of finitely generated abelian groups has enough projectives and is a counter-example. I suggest you have a look at the proof in the module category (over an appropriate ring) and see the properties used there, before you try to generalize. | |
| Dec 18, 2011 at 11:41 | history | edited | Sasha | CC BY-SA 3.0 |
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| Dec 18, 2011 at 11:40 | comment | added | Sasha | Sorry, I forgot to add the condition that the category has enough projectives. | |
| Dec 18, 2011 at 9:24 | comment | added | Ralph | I think further restrictions on the categrory are needed (e.g. Krull-Schmidt property should hold). Take for example the category of finitely generated abelian groups. The only indecomposable projective of finite length is 0 while the irreducibles are 0 and the cyclic groups of prime order. | |
| Dec 18, 2011 at 9:05 | comment | added | Ralph | Don't you have to assume that the category has a progenerator ? | |
| Dec 18, 2011 at 7:50 | history | asked | Sasha | CC BY-SA 3.0 |