# Semisimple modules

1. Aug 15, 2008

### peteryellow

Can somebody help me with the following proof:

Let M be a semisimple module, say M = +_IS_i, where + denotes direct sum and S_i is a simple module.
Then the number of summands is finite if and only of M is finitely generated.

I have problem with understanding the proof of the following in my notes:

if M is finitely generated then the number of summands is finite

Can somebody help me in this argument.

2. Aug 15, 2008

### morphism

This follows immediately from the definition of a "finitely generated" module.

3. Aug 15, 2008

### peteryellow

I know that it is quite clear but still there is an argument which I dont understand and I somebody can help me with this, I will be greatful.

4. Aug 15, 2008

### morphism

What argument, exactly?

5. Aug 15, 2008

### mathwonk

try proving the contrapositive, that if M is any non zero module, that an infinite direct sum of copies of M cannot be finitely generated.

recall the definition of direct sum, and in particular that only a finite number of summands can occur in each element of a direct sum.

6. Aug 15, 2008

### peteryellow

Ok the argument for this theorem in my notes which I dont understand is:

Let M be finitely generated by u_1,...,u_r say. For each u_j we can find finitely many terms S_i whose sum contains u_j. Hence all the u_j are contained in the sum of a finite subfamily of the S_i and this family generates M so that I must be finite.

I dont understand details of this so it will be good if you can help me with the details. Thanks.

7. Aug 15, 2008

### peteryellow

mathwonk, I have a proof of this which I dont understand.

8. Aug 15, 2008

### morphism

If you understand the appropriate definitions ("direct sum" and "finitely generated"), then the details will be crystal clear.

9. Aug 15, 2008

### peteryellow

Nut why is it true that this family generates M

10. Aug 15, 2008

### morphism

It generates it in the sense that its sum is M. (And this is true because M is generated, as a module, by u_1, ..., u_r.)

11. Aug 15, 2008

Thanks alot