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December 20th, 2012, 05:12 AM   #1
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Wedderburn - a simple ring that's not semisimple

Hi all.. I'm trying to prove the following but having no luck:

Claim:
Let F be a vector space of countably infinite dimension over a field k and let . Then R has a unique proper non-zero 2-sided ideal I. Put . Then S is not semisimple. The ring S is an example of a ring that is simple but not semisimple.

Attempted thoughts..
So since F is an infinite dimensional vector space over k, I wrote F as . But then I don't think we can pull the direct sum out of the endomorphism ring? I wasn't sure where to go looking for a unique non-zero 2-sided ideal.. Is this ring local? Is the unique ideal actually maximal (hence S would be simple)? Would it be best to just prove R is local? I wasn't sure how best to go about this.

Thank you!
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December 21st, 2012, 10:36 PM   #2
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Re: Wedderburn - a simple ring that's not semisimple

You cannot pull the sum out of the endomorphism ring. Endormorphism rings of vector spaces are much larger than the original vector spaces, but so your claim would imply . Even in the finite dimensional case this fails, as is isomorphic to the matrices over .

Part of this task is to really understand what endomorphisms of infinite dimensional vector spaces are. But consider this: the most important information about an endomorphism of a finite dimensional vector space is recorded in its rank.
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December 23rd, 2012, 04:19 PM   #3
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Re: Wedderburn - a simple ring that's not semisimple

also, endomorphisms can only (at best) preserve rank, or make it smaller, they cannot "enlarge" a subspace of V.

the fact that dim_k(V) is countable means we can index any basis by N, the natural numbers. this might be useful.
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December 25th, 2012, 10:32 AM   #4
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Re: Wedderburn - a simple ring that's not semisimple

Ok.. so I'm still confused how I would go about showing the existence of a unique proper non-zero 2-sided ideal of End_k(V), where V is the countably infinite dimensional vector space over a field k.
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December 26th, 2012, 08:02 PM   #5
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Re: Wedderburn - a simple ring that's not semisimple

have you tried:

{T in End(V): rank(T) < infinity} ?
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December 27th, 2012, 11:21 AM   #6
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Re: Wedderburn - a simple ring that's not semisimple

Have you considered the finite dimensional problem? What are the two sided ideals of ? Or, more concretely, ?

Once you understand how things work in the more simple situation, it should be more clear how to proceed in the larger problem. Viewed a certain way, the answers are the same; one must just see the appropriate generalization.

But more generally, don't get stuck trying to show there is just one ideal, try to find all two-sided ideals. When all is said and done, you should have concluded that there are not very many options.
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