# Indecomposable projectives correspond to irreducibles - reference

Hello,

We have the following assertion:

In an abelian category that has enough projectives and in which every object has finite length, indecomposable projectives correspond bijectively to irreducibles (both up to isomorphism).

I see how to construct a well defined map from irreducibles (up to iso.) to indecomposable projectives, and to show that it is injective. What I do not know is how to show the following:

Given an indecomposable projective of finite length, how to show that it has a unique irreducible quoteint?

(I have edited the question; Sorry about the initial imprecision).

Thanks, Sasha

• Don't you have to assume that the category has a progenerator ? – Ralph Dec 18 '11 at 9:05
• 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. – Ralph Dec 18 '11 at 9:24
• Sorry, I forgot to add the condition that the category has enough projectives. – Sasha Dec 18 '11 at 11:40
• 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. – Ralph Dec 18 '11 at 13:32
• 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. – Florian Eisele Dec 18 '11 at 16:40

It's useful for questions like this to go back to the basic literature where some of these ideas are developed in context. (Serganova's lecture notes look helpful but if course rely on older sources.) The applications I'm familiar with arise in various areas of module theory including modular representations of finite groups, but the arguments can be imitated for abelian categories satisfying reasonable finiteness conditions; one could even resort to embedding such categories in module categories (by standard theorems going back to Freyd and Mitchell).

The two-part treatise Methods of Representation Theory (Wiley, 1981+) by Curtis and Reiner provides a lot of general background beyond what is used traditionally for finite groups. In particular, Section 6C of Part I has a clear discussion of projective covers. Most of the interesting results in this direction apply to modules over arbitrary artinian rings. Once one has enough projectives, it's straightforward to show that any finitely generated module has a projective cover (unique up to isomorphism).

These ideas for the BGG category $\mathcal{O}$ of a semisimple Lie algebra (which is artinian) come up in Section 3.9 of my 2008 AMS text but don't involve the special features of that module category. Not only do projective covers of simple modules in the category exist, but one sees immediately from the definition of "essential" map that any indecomposable projective in the category has a unique simple quotient. (Modules here actually have finite length.) To be explicit, the given projective $P$ has at least one simple quotient $L$, which in turn has a projective cover $P_L$. Now you get maps between $P$ and $P_L$, which from the definitions are isomorphisms in both directions.

My main point is that quite a bit of systematic work (for module categories) has been done on these questions, so it's a good idea to be aware of the relevant literature when it's needed for applications.

Hello,

I am sorry that I did not formulate the question precisely enough. The claim I wanted to be true is:

Let $\mathcal{A}$ be an abelian category, in which every object has finite length, and there are enough projectives. Then, there is a bijective correspondence between indecomposable projectives and irreducible objects.

The only compound which I did not know how to proof was:

An indecomposable projective object of finite length has an unique irreducible quotient.

Now, I saw elements which lead to a proof in those notes:

http://math.berkeley.edu/~serganov/math252/notes9.pdf

Here is a proof:

Lemma 1: Every endomorphism of an indecomposable object of finite length is either nilpotent, or isomorphism.

Lemma 2: If the sum of two endomorphisms of an indecomposable object of finite length is an isomorphism, one of them is also an isomorphism.

In the above notes there are the (short and easy) proves of these two lemmas. Now, let $P$ be an indecomposable projective of finite length. We need to show that it has a unique maximal proper sub-object (thus unique irreducible quotient). Suppose not: let $K_1, K_2 \subset P$ be two different maximal proper sub-objects. Then we have a surjection $K_1 \oplus K_2 \to P$, which splits since $P$ is projective. Hence we get two maps $P \to K_1, P \to K_2$, which when considered as maps $P \to P$, sum to the identity map. Thus, by lemma 2 above, one of them is an isomorphism, which is impossible by length consideration.