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 December 20th, 2012, 05:12 AM #1 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 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 $R=End_k(V)$. Then R has a unique proper non-zero 2-sided ideal I. Put $S=R/I$. 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 $F=\bigoplus_1^\infty k$. 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!
 December 21st, 2012, 10:36 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 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 $End_k(k)= k$ so your claim would imply $V= End_k(V)$. Even in the finite dimensional case this fails, as $End_k(k^n)$ is isomorphic to the $n \times n$ matrices over $k$. 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.
 December 23rd, 2012, 04:19 PM #3 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 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.
 December 25th, 2012, 10:32 AM #4 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 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.
 December 26th, 2012, 08:02 PM #5 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Wedderburn - a simple ring that's not semisimple have you tried: {T in End(V): rank(T) < infinity} ?
 December 27th, 2012, 11:21 AM #6 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Wedderburn - a simple ring that's not semisimple Have you considered the finite dimensional problem? What are the two sided ideals of $End_k(k^n)$? Or, more concretely, $M_{n \times n}(k)$? 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.

 Tags ring, semisimple, simple, wedderburn

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