For a vector space of finite dimension, the ring of linear operators End(V) is directly finite i.e. $AB = 1$ implies $BA=1$ for linear operators $A$ and $B$. Is this also true in infinitely many dimensions? If not, what is a counterexample for this? Or is it known to be true in some specific cases, for example in the context of Hilbert spaces?
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No, in fact the endomorphism ring of an infinite dimensional vector space is probably the simplest and most commonly given example of a ring that isn't directly finite (a.k.a. Dedekind-finite.)
See for example this which describes the obvious "shift" transformations that make up a pair $a,b$ such that $ab=1$ but $ba\neq 1$.
In fact, the DaRT query for non directly-finite rings only has two positive hits at the moment. The other one is far from trivial to describe.
rschwieb
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Okay, I see there are indeed lots of examples. The DaRT page seems to suggest every endomorphism ring of an infinite vector space is not directly finite. How might one prove this (rather than proving it for specific examples)? – Krup'a Dec 13 '19 at 20:41
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@LucasL. The proof is the same for all dimensions. Fix a basis $b_i$ indexed by a well-ordered index set $\kappa$. For the initial segment of basis elements with $i<\aleph_0$, define $\phi(b_i)=b_{i+1}$, and for the rest of the basis elements, leave them fixed. Then define another map which sends $\psi(b_0)=0$ and $\psi(b_i)=b_{i-1}$ for $i<\aleph_0$ and maps the remaining basis elements to themselves. Then $\psi\circ\phi=1$ but $\phi\circ\psi\neq 1$, in particular, it sends $b_0$ to $0$. – rschwieb Dec 13 '19 at 20:56