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Two questions.

Q1. Let $A$ be a finite-dimensional algebra over a field with block decomposition $A=A_1\oplus A_2\oplus\cdots\oplus A_r$ and $M$ an $A$-module. If every composition factor of $M$ lies in the block $A_i$ for some $i$, why must $M$ lie in $A_i$?

Q2. I would like to understand the statement of the following proposition:

Two simple $A$-modules $S$ and $T$ lie in the same block of $A$ if and only i there are simple $A$-modules $S=S_1, S_2,\ldots,S_m=T$ such that $S_i,S_{i+1}$, $1\leq i<m$, are composition factors of an indecomposable projective $A$-module.

What does this mean? We need to find a ``sequence'' of simple modules going from $S$ to $T$ such that successive modules in the sequence are composition factors of the same projective indecomposable?

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Please do not use abbreviations like PIM (specially if you only use them once!) without explaining their meanings ---an easy corollary of this is that you should not use them if you only use them twice, for you do not gain anything :) – Mariano Suárez-Alvarez Sep 2 '11 at 0:45
On the other hand: I do not understand what you are asking in your Q2. The condition you state indeed means that... – Mariano Suárez-Alvarez Sep 2 '11 at 0:46
I suppose I'm using a different definition of lying in the same block, and this statement is claiming they are equivalent. My definition of $M$ lying in $A_i$ is that $A_iM=M$ and $A_jM=0$ for $j\neq i$. – Clinton Boys Sep 2 '11 at 1:13
But that is the same thing as $1_{A_i}m=m$ for all $m\in M$ and $1_{A_j}m=0$ for all $m\in M$ and all $j\neq i$; moreover The elements $e_i=1_{A_i}$ with $i\in\{1,\dots,k\}$, are central idempotents. – Mariano Suárez-Alvarez Sep 2 '11 at 1:18

(I am going to assume that your algebra $A$ is artinian over a field... If you intended this, please edit the question to add the hypothesis)

To answer your first question, it is enough to observe that an extension of modules in a block is still in the block. More precisely, that if $M'$ and $M''$ are in a block $B$ and $$0\to M'\to M\to M''\to 0$$ is a short exact sequence of left $A$-modules, then $M$ is also in $B$. Suppose then we are in such a situation, and that $B$ is the block corresponding to the central idempotent $e\in A$. Multiplication by $e$ is the identity map in $M'$ and in $M'$, so it is also the identity on $M$: if $m\in M$, then $em$ and $m$ have the same image in $M''$, so that $em-m$ is the image of some $m'\in M'$ under the map $M'\to M$; since $em'=m'$ and since this map is $A$-linear, $0=e(em-m)=em-m$. That multiplication by $e$ is the identity on $M$ means that $M$ is in $B$, as we wanted.

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