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Prove that If $\text{dim}\ F (V)>1$ be finite, prove that $L(V)$ has no two sided ideals other than $(0)$ and $L(V)$.

Further

Prove that the conclusion above is false if $V$ is not finite dimensional over $F$.

I assumed that $I$ is an ideal properly contained in $L(V)$ and I attempted that any element in $I$ is $0$. Please help me with the problem.I tried too but wasn't sucessful.

I am unable to understand the approach they did using simple groups etc..We hadnt been taught simple groups in our course..I will be thankful if someone can give an elemetary proof of this result..

Morever that question was different(shown as possible duplicate).Because that was wrong

The Phenotype
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Bahubali
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  • What is $L(V)$ supposed to mean? The ring of linear transformations? – rschwieb Mar 08 '18 at 18:41
  • @rschwieb yes it means ring of linear transformations – Bahubali Mar 08 '18 at 18:51
  • I cannnot still get the answer! Can anyone give me a detailed answer here...For a finite dimensional case I am thinking to consider an ideal containing an non-zero element an I wish to prove that it contains an unit element(an invertible LT) – Bahubali Mar 09 '18 at 05:24
  • sure, call the nonzero element x. you can achieve that by fixing a basis, and then showing that you can find $r_i$ $s_i$ such that $r_ixs_i$ is the projection onto the $i$-th basis vector. When you sum all those, you get the identity matrix. That would demonstrate the ideal generated by a nonzero element contains the identity, and is the whole ring. – rschwieb Mar 09 '18 at 11:17

1 Answers1

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Laboring under the impression that $L(V)$ means the ring of linear endomorphisms of $V$.

If $\dim(V)<\infty$, it never has nontrivial two-sided ideals. This is proven lots of places where the ring of square matrices is identified with the full ring of matrices.

On the other hand, if $\dim(V)$ is infinite, then it always has an ideal consisting of transformations with finite dimensional images. That was answered here.

quid
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rschwieb
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  • can you give an explaination in detail.We hadnt been taught simnple groups or whatever it is......I think it cannot be wrong because this question is from Herstein Itself – Bahubali Mar 08 '18 at 18:53
  • @user439583 I can't think of what details I would add to the answer I linked to. It's already sufficient to get you to a proof. – rschwieb Mar 08 '18 at 19:50
  • I edited your question as OP had posted a corrected version of the question and I merged. – quid Mar 08 '18 at 20:01