Simple ring has a unique simple module

[Guest dazai]

I get stuck on the question " for any simple ring (no 2-sided ideal except 0 and itself) has a unique faithful irreducible module upto isomorphism". can u pls help me to work it out?

 

many thanks

[Johnny5]

I get stuck on the question " for any simple ring (no 2-sided ideal except 0 and itself) has a unique faithful irreducible module upto isomorphism". can u pls help me to work it out?

 

many thanks

 

I would be willing to, but it's been a long time since i've solved these kinds of problems.

 

You will have to tell me what the definitions are.

 

What is a ring?

What is a simple ring?

What is an ideal?

I don't remember what a module is, or a submodule.

 

I wouldn't mind at least trying to solve the problem, or watching others help you solve it.

[Lyssia]

My guess would be to assume that there are two distinct such modules, do a bit of wiggling about, and then be forced (by reductio ad absurdum) to concede that your assumption that those two modules are distinct was wrong, and thus that they are identical.

 

My ring theory's horrible rusty so I'd need to dig out some books to go into any detail, though. Hope this helps a little.

[matt grime]

Here are the tools you need, I think, It may be overkill, but I don't think I've left anything out.

 

If f is a map from R to Hom(M,M), then Im(f) is iso to R/Ann(M) where Ann(m) is the set of all x in M such that rx=0 for all r in R.

 

If f is faithful, then Im(f)=R

 

If M is simple then M=Rm for any m in M.

 

Any simple module is iso to R/L for some maximal left ideal L.

 

note, I'm assuming you mean left module. Is R unital? Commutative? Shuold we reallyl be talking about right modules? Answers to any of these?

[Guest dazai]

R is neither commutative or unital, and i'm talking about left module. I found on a text that if R is a simple module then R is isomorphic to End_D(I) for some left ideal I and D = End_R(I). Does it mean that R has a unit (R is a simple ring then R is unital).