Re: Ring with nilpotent elements of arbitrary degree

Seán O'Leathlóbhair wrote:

Not needed for my nilpotent puzzle but the next question is can we
relax the condition still further and allow infinitely many non-zeros
in a row or column. We would need some restriction that ensured that
the sums in the product converged, and what may be harder, that the
product obeyed the restriction itself.

--
Seán O'Leathlóbhair


Something like the ring of continuous linear mappings from
L^2(N) to L^2(N)? The sequences with a single one and all other
entries equal to zero don't form a (Hamel) basis, but they do
span a dense subspace, so their image determine a continuous
mapping uniquely (but can't be chosen arbitrarily as the norms
of unit vectors must be bounded). Anyway the point of this
argument was simply to show that the resulting infinite matrix faithfully describes the mapping.

It's been a while since I had to work with topological vector
spaces of any sort, so there may be something wrong with the above,
but it might work.

Cheers,

Jyrki
.