# Quotients of direct sums of modules

1. Jan 21, 2012

### gauss mouse

Hi, I keep seeing indirect uses of a result which I think would be stated as follows:

If a module $M$ over the unital associative algebra $A$ is written
$M\cong S_1\oplus\cdots\oplus S_r$ (where the $S_i$ are simple modules), then in any comosition series of $M$, the composition factors are, up to order and isomorphism, $S_1,\ldots,S_r$. Perhaps this statement is not correct, but it's the best I can do when I can't find an explicit statement anywhere.

I think I would use the Jordan Holder theorem to prove this. I would argue as follows: a composition series for $S_1\oplus\cdots\oplus S_r$ is given by
$S_1\oplus\cdots\oplus S_r>S_1\oplus\cdots\oplus S_{r-1}>S_1\oplus\cdots\oplus S_{r-2}>\cdots>S_1>\{0\}.$
Now, $(S_1/\{0\})\cong S_1,\ (S_1\oplus S_2)/S_1\cong S_2,\ (S_1\oplus S_2\oplus S_3)/(S_1\oplus S_2)\cong S_3,\ldots, (S_1\oplus\cdots\oplus S_r)/(S_1\oplus\cdots\oplus S_{r-1})\cong S_r$ and so by the Jordan Holder theorem, in any comosition series of $M$, the composition factors are, up to order and isomorphism, $S_1,\ldots,S_r$.

I have a feeling that I am going wrong somewhere, perhaps in my cancelling when I do things like $(S_1\oplus S_2)/S_1\cong S_2$. The terminology of "direct sum" would suggest that this is not allowed.

Last edited: Jan 21, 2012
2. Jan 21, 2012

### morphism

Assuming the S_i's are simple modules, what you've written is correct.

To justify canceling off a factor in the direct sums, just use the first isomorphism theorem: e.g., the map $S_1 \oplus S_2 \to S_2$ defined by $(s_1,s_2) \mapsto s_2$ is surjective and has kernel equal to $S_1\oplus 0$.

3. Jan 21, 2012

### gauss mouse

Yes, they are simple, I've edited that in. Thanks a million for your help.

4. Jan 21, 2012

### mathwonk

this looks fishy.

5. Jan 21, 2012

### morphism

Is there anything specific that's making you skeptical?