# Ideals of $\mathscr{I}(V) = \text{Hom}(V,V)$

I have been trying this notoriously difficult problem for quite some time but i haven't made any progress.

Let $\mathscr{I}(V)$ denote the set of all homomorphisms $f : V \to V$. That is $\mathscr{I}(V) = \text{Hom}(V,V)$.

• Suppose $V$ is a vector space over a field $K$ and $\text{dim}_{K}(V)>1$, then prove that $\mathscr{I}(V)$ has no proper two sided ideals. This means we have to show that $\mathscr{I}(V)$ has no two sided ideals other than $(0)$ and $\mathscr{I}(V)$. Next, does the conclusion of the above problem remain true if $V$ is infinite dimensional.

Moreover, since Field's are the ubiquitous algebraic structures to have trivial ideals i tried thinking of proving $\mathscr{I}(V)$ to be a field. But i haven't yet taken a single step forward in terms of this problem.

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possible duplicate of Why is the ring of matrices over a field simple? –  Arturo Magidin Mar 11 '11 at 19:45
it isnt a field. for instance if $V$ is finite dimensional, it's isomorphic to a matrix ring. i think the minimal (non-trivial) right and left ideals are the matrices with one non-zero row (or column). any two sided ideal is both a right and left ideal hence contains one of each of these, which will generate the whole ring –  yoyo Mar 11 '11 at 19:52

The first part, for finite dimensional spaces, amounts to showing that the ring of $n\times n$ matrices with coefficients in $K$ is simple. This is covered in this previous question and answer.

For infinite dimensional spaces, I have a two word hint: finite support.

You are incorrect on your final paragraph: $\mathscr{I}(V)$ cannot be a field for $n\gt 1$, since it always nontrivial idempotents (projections onto nontrivial proper subspaces), nonzero nilpotent elements (the left-shift operator on a basis), and zero divisors (any non-invertible map is a zero divisor: compose with an appropriate projection onto a complement of the image).

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Imagine that you have a non-zero map $\phi:V \to V$, say with $V$ finite dimensional. Now ask yourself what other maps can you get by pre- and post-composing with other endomorphisms, and then taking linear combinations.

Once you understand the finite dimensional case well, you will be ready for the infinite dimensional case. (Arturo's hint will play an important role!)

Note: In order to prepare yourself for the infinite dimensional case, you will be better off trying to prove simplicity in the finite dimensional case using arguments with endomorphisms, rather than shifting to the language of matrices and relying on matrix manipulations. (At least, if you do the latter, try to translate what you have done back into the language of endomorphisms and reinterpret the argument in that language.)

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