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In summary: I won't say.In summary, after spending 6 hours trying to prove that the endomorphism ring of a simple module is a field, the conversation ultimately concluded that the result is not true. The conversation explored different possibilities and potential counterexamples, including using Schur's lemma and looking at finite modules. However, it was ultimately found that the construction in question is not an isomorphism of rings and does not result in a field.

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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 ismatt grime said:

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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

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

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

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

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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.

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

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

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).Hurkyl said:Some books also say, way back in the beginning, that all rings will be 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 [itex]R=\mathbb{Z}_2[/itex], [itex]\operatorname{End}_R(R^R)[/itex] has order 16, so it is definitely not isomorphic to R

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

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

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 [itex]t \mapsto rst[/itex], whereas the product of the maps of r and s is [itex]t\mapsto rtst\neq rst[/itex]. Unless I've made a mistake, this is not an isomorphism of rings, right?

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

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Oh, I totally see that now. Itmatt grime said:As a left module End(R) is isomorphic to R^{op} the opposite ring, as s right module End(R) is R.

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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?Don Aman said: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.

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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.

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

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An endomorphism ring is a mathematical term used in abstract algebra to describe the set of all endomorphisms of a given mathematical object, such as a module or vector space. In simpler terms, it is the set of all possible transformations or mappings from the object to itself.

A simple module is a type of module in abstract algebra that has no proper submodules. In other words, it cannot be broken down into smaller modules. It is the most basic and irreducible type of module.

Proving that the endomorphism ring of a simple module is a field is important because it provides a deeper understanding of the structure and properties of the module. It also has many applications in other areas of mathematics, such as representation theory and algebraic geometry.

The endomorphism ring of a simple module is closely related to the field of scalars used to define the module. In fact, the endomorphism ring is isomorphic to the field of scalars. This means that they have the same structure and properties, and can be thought of as the same mathematical object in different forms.

There are several techniques used to prove that the endomorphism ring of a simple module is a field. These include using properties of simple modules, using properties of fields, and applying the first isomorphism theorem from abstract algebra. Other techniques may also be used depending on the specific module and field in question.

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