Quotients of direct sums of modules

[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.

 

[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.

 

[gauss mouse]

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

 

[mathwonk]

this looks fishy.

 

[morphism]

this looks fishy.

Is there anything specific that's making you skeptical?