Prove Endomorphism Ring of Simple Module is a Field

[Don Aman]

I spent like 6 hours trying to prove that the endomorphism ring of a simple module is a field, helping my friend with his homework. Now I'm convinced that the result is not true. By Schur's lemma, all endomorphisms of a simple module (except 0) are isomorphisms, so it's a division ring, but I don't think it will be abelian. Only, so far I haven't found a counterexample. Can someone help me find one?

-Don

 

[matt grime]

Module for what? Perhaps with more information we could deduce it is a 1 dimensional division algebra, thus a field. Perhaps it is finite, when we know that all finite division rings are fields. IN general though I believe it is "just a division algebra"

 

[Don Aman]

Module for what? Perhaps with more information we could deduce it is a 1 dimensional division algebra, thus a field. Perhaps it is finite, when we know that all finite division rings are fields. IN general though I believe it is "just a division algebra"

So every simple module that I've looked at so far has an endomorphism ring that is actually a field. What I'm looking for is a simple module whose endomorphism ring is not a field. Based on what you said about finite modules always having fields, I suppose I will have to look for an infinite module. That's already a little helpful.

 

[matt grime]

How about this: suppose D is a division ring that is not commutative (eg the quaternions), then look at

http://www.maths.warwick.ac.uk/~rumynin/rings2002/notes/node29.html

 

[Don Aman]

How about this: suppose D is a division ring that is not commutative (eg the quaternions), then look at

http://www.maths.warwick.ac.uk/~rumynin/rings2002/notes/node29.html

the hint in that page made me wonder if there is an isomorphism $\operatorname{End}_R(R^R)\cong R$. But that seems unlikely to me. the cardinalities probably don't even match. I'm going to play with it some.

 

[mathwonk]

for one thing some definitions of "field" do not include commutativity so check you source. for another the map Hom(R,M)-->M taking f to f(1), certainly seems inverse to the map M-->Hom(R,M) taking x to the map t-->tx, at least if M is a left R module.

Hence Hom(R,R) seems to be isomorphic to R as left R module, whether R abelian or not. Now check the ring structure. I.e. does xy got to the composition?

well xy(t) = txy. while y(t) = ty, and x(y(t)) = tyx, oops.

bah, i never enjoy non commutativity.

OK trythe map R -->Hom(R,R) taking x to the map t-->x(t) = xt.

then it should take xy to (xy)(t) = x(y(t)). hence multiplication goes to composition.

well anyway, it is apparently not true that Hom(R,R) is commutative when R is not.

 

[Hurkyl]

Some books also say, way back in the beginning, that all rings will be commutative!

 

[Don Aman]

for one thing some definitions of "field" do not include commutativity

Some books also say, way back in the beginning, that all rings will be commutative!

I want fields commutative and rings noncommutative. So I'm looking for a simple module whose endomorphism ring (which is necessarily a division ring) is not a field (so is not commutative).

I presume it will have to be a module over a non commutative ring, though I'm not positive.

I didn't follow everything else you wrote yet, mathwonk, I'm still working through it, but I'm going home now, and I wanted to say that I've checked it now, and in the case that $R=\mathbb{Z}_2$, $\operatorname{End}_R(R^R)$ has order 16, so it is definitely not isomorphic to R

 

[mathwonk]

you have the object in question written wrong. the point is that End(R) is isomorphic to R, not End(R^R).

in the case of R = Z/2Z there are two endomorphisms, zero and the identity, hence End(R) certainly IS isomorphic to R.

 

[Don Aman]

you have the object in question written wrong. the point is that End(R) is isomorphic to R, not End(R^R).

Well, I didn't get that from your post, I got the idea that R was isomorphic to End(R^R) from as an (incorrect) possibility from the question in the page that matt grime linked above. It says that R^R simple implies R is a division ring, and I know that the endomorphism ring of a simple module is a division ring, so I just guessed that maybe the endomorphism ring of R^R would be R. But yes, it's simply not true.

I was still digesting your suggestion when I left last night. Let me consider it again now.

Hm... I wasn't clear from what you said whether you thought this map was an isomorphism of rings, but it doesn't seem to be. For example, if r and s are in R, then rs goes to the map $t \mapsto rst$, whereas the product of the maps of r and s is $t\mapsto rtst\neq rst$. Unless I've made a mistake, this is not an isomorphism of rings, right?

 

[matt grime]

As a left module End(R) is isomorphic to R^{op} the opposite ring, as s right module End(R) is R.

 

[Don Aman]

As a left module End(R) is isomorphic to R^{op} the opposite ring, as s right module End(R) is R.

Oh, I totally see that now. It *is* an isomorphism of rings. My mistake earlier was that I was making the product of maps be pointwise multiplication, instead of composition. I see why it has to be R^op as well. And Rop is isomorphic to R, right? And R, viewed as a right R-module, is simple as a module when R is a division ring. So I'm done. Thanks for your help.

 

[Don Aman]

I spent like 6 hours trying to prove that the endomorphism ring of a simple module is a field, helping my friend with his homework.

So I showed the construction to my friend, and he showed me that his textbook wanted him to prove it. It was in Artin. Does that just mean that Artin made a mistake?

 

[mathwonk]

your friend wouldn't be one of those people who declines to read the chapter before trying the problems would he? tell him to read the first sentence (!) in the chapter, on page 450.

Once you told me it was Artin, I knew the choice between whether he or your friend made the mistake was pretty easy money.