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Proposition Claim: Let $A$ be a finite-dimensional algebra over a field $k$. Suppose $T,T'$ are simple $A$-modules and that there is a non-split extension $M$ of $T$ by $T'$. Then $T$ and $T'$ are composition factors of the same projective indecomposable module.

Proof. We have a short exact sequence $$0\to T'\xrightarrow{\iota_1} M\xrightarrow{\pi_1} T\to 0.$$

Let $P_T$ be the projective indecomposable corresponding to $T$, so $P_T/\mathrm{rad}(P_T)\cong T$, i.e. we have another short exact sequence $$ 0\to \mathrm{rad}(P_T)\xrightarrow{\iota_2}P_T\xrightarrow{\pi_2}T\to 0. $$

By the universal property of projective modules, using the surjective maps $\pi_1$ and $\pi_2$, there is a unique map $\varphi:P_T\to M$ satisfying $\pi_1\circ \varphi=\pi_2$.

Since the second sequence is exact, we know $\pi_2\circ\iota_2=0$ and so $\pi_2\circ\varphi\circ\iota_1=0$. Notice that $\ker\pi_2=T'$.

Hence by the universal property of $\ker\pi_2$, there exists a unique nonzero map $\psi:\mathrm{rad}(P_T)\to T'$. Since $T'$ is simple, $\psi$ is surjective and by the first isomorphism theorem $$ \mathrm{rad}(P_T)/\ker\psi\cong T'. $$ Hence $P_T$ has a filtration $$ P_T\supset \mathrm{rad}(P_T)\supset \ker\psi\supset 0 $$ with composition factors $T$ and $T'$ as required.

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Consider writing the projective cover of M as a direct sum of PIMs. –  Jack Schmidt Sep 2 '11 at 3:44
    
rad(PT′) need not be isomorphic to T, it will just have T as a top composition factor. –  Jack Schmidt Sep 2 '11 at 3:45
    
Please do consider examples! Path algebras and their quotients provide an immense wealth of them ---if you are not familiar with them, do yourself a favor and read the first chapter of the book [Simson, Daniel; Skowronski, Andrzej; Assem, Ibrahim (2007), Elements of the Representation Theory of Associative Algebras, Cambridge University Press], for example. Most humans learn from examples (I am told by a few of his former students that Maurice Auslander was an exception to this, though...) –  Mariano Suárez-Alvarez Sep 2 '11 at 3:48
    
You cannot conclude that $\psi$ is surjective using solely the fact that $T'$ is simple: $\psi$ could very well be zero... (Also: one usually only uses the term composition series when the subquotients of the sequence of submodules are simple: this is not true in the series $P_T\supset \mathrm{rad}(P_T)\supset \ker\psi\supset 0$ at the end of your edit) (Please, when you edit a question, do not remove old text, because otherwise you more or less render old comments and answers unintelligible: add text at the end instead) –  Mariano Suárez-Alvarez Sep 2 '11 at 6:10
    
$\psi$ is nonzero because it fits into the commutative diagram no? (I have said that $\psi$ is nonzero in my proof). Changing to the word "filtration" is enough to complete the proof, since we may just refine this filtration to a composition series with the right top two factors. –  Clinton Boys Sep 2 '11 at 6:24

1 Answer 1

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Suppose

0 ---> T' ---> M ---> T ---> 0 

is a non-split extension with $T$ and $T'$ simple. Let $P\to T$ be the projective cover of $T$, whose kernel is $\mathrm{rad}(P)$. You should show that there is a commutative diagram with exact rows

0 ---> rad(P) ---> P ---> T ---> 0 
         |         |      |
         |         |      |
         V         V      V
0 --->   T'   ---> M ---> T ---> 0 

in which the arrow $T\to T$ is the identity and the map $\mathrm{rad}(P)\to T'$ is not zero.

Can you conclude from this that $T'$ is a composition factor in $P$?

For an example where your argument fails, consider the path algebra of the quiver $$\overset1\bullet\to\overset2\bullet\to\overset3\bullet$$ There is an extension of the simple $S_1$ by the simple $S_2$, but the radical of the projective cover $P_1$ of $S_1$ has length $2$, so it cannot be isomorphic to $S_2$ (depending on your conventions on modules, you might need to interchange $1$s and $3$s in what I wrote....)

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I see where your map $P\to M$ comes from (basically the def. of projective). Where does your map $\mathrm{rad}(P)\to T'$ come from? –  Clinton Boys Sep 2 '11 at 4:35
    
@user1375: once you have constructed the map $P\to M$ so that the rightmost square commutes, you can see that the composition $\mathrm{rad}(P)\to P\to M\to T$ is zero: it follows from this that the image of the composition $\mathrm{rad}(P)\to P\to M$ is contained in the kernel of the map $M\to T$. &c. (BTW: Please consider changing your user name to something less generic, as it is much comfortable to communicate with someone with a more identifiable name!) –  Mariano Suárez-Alvarez Sep 2 '11 at 4:38
    
Sorry, still not following. Are you using some universal property to construct this map? –  Clinton Boys Sep 2 '11 at 4:45
    
If a map $f:X\to Y$ has its image in the kernel $K$ of $g:Y\to Z$, and we write $\iota:K\to Y$ the inclusion of thay kernel into $Y$, then there exists a unique map $\bar f:X\to K$ such that $\iota\circ\bar f=f$. That is the characteristic property of the kernel of a map that I am using. –  Mariano Suárez-Alvarez Sep 2 '11 at 5:12
    
Right, the universal property of the kernel. Not sure how I missed that. OK so now we need some module $X$ such that $P/X\cong T'$? –  Clinton Boys Sep 2 '11 at 5:32

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