The units are indeed the elements without zeros on the diagonal: 1) the map $R \to R$, $e{ii} \mapsto e{ii}$, $e_{ij} \mapsto 0$, is a homomorphism; 2) we can use Gaussian elimination to get an inverse of an upper triangular matrix column by column, starting from the diagonal element.
The strictly upper triangular matrices form an ideal, and $1-rx$ is then clearly invertible. $J(R)$ is not nil: the "left coordinate shift", aka the infinite Jordan block $\Bbb J$ with eigenvalue 0, is not nilpotent.
$J(R)$ is indeed the set of strictly upper triangular matrices: it is clearly an ideal, so $1 - rx$ has all 1's on diagonal. $R/J(R) \cong \prod\limits_{i = 1}^\infty \Bbb Q$.
The zero divisor question is more difficult. All non-invertible matrices are zero divisors, and "Note on Sus̆kevic̆’s problem on zero divisors", Holubowski, Maciaszczyk, Zurek 2015 Th1.2 proves the following: a non-invertible element that is not a right zero divisor has a right inverse that is not upper triangular in the larger CFM ring.
Properties
[x] :x: countable.
[x] :x: orthogonally finite.
[x] :x: polynomial identity. In characteristic $0$ all identities follow from the multilinear ones. Take the standard staircase $E{11}, E{12}, E{22}, E{23}, \ldots$ (its elements can have a nonzero product only in this order).
[x] :heavy_checkmark: stable range 1. By Lam's Crash Course 1.5, $sr(R) = sr(R/J(R))$; $sr(R/J(R)) = sr(R{57}) = 1$.
[x] :x: left cohopfian. See the Sushkevich paper above Theorem 1.2.
[x] :heavy_check_mark: right cohopfian. See the Sushkevich paper above Theorem 1.1.
[x] :heavy_check_mark: lift/rad. $R/J(R)$ injects into $R$.
[x] :heavy_check_mark: semiregular.
[x] :heavy_check_mark: clean. For $0$'s on the diagonal, write $-1$ in that position of the unit and $1$ in the idempotent. Make all other entries of the idempotent zero.
[x] :x: nonzero right socle. For every right ideal $I$ generated by $\{a_m | m \in M\}$, the right ideal generated by $a_m \Bbb J$ is also nonzero and strictly contained in $I$.
[x] :heavy_check_mark: nonzero left socle. It is the first row. Clearly then :x: f.g. left socle.
[x] :heavy_checkmark: essential left socle. Left-multiply any element of a left ideal with nonzero $i$-th row by $e{1i}$.
[x] :heavy_checkmark: left nonsingular. By Lam II (7.13) the left singular ideal right-annihilates the left socle, but if $e{*k}$ occur with nonzero coefficients in an element supposed to be singular, there would be a sum of form $\sum\limits_{i = 1}^k \lambdai e{1i}$ that it would not annihilate.
[x] :x: Armendariz. This passes to subrings, but in $R_{44}$ it does not hold.
[x] :heavy_check_mark: right/left Ore ring. Even though the ring is not cohopfian, by the above all two-sided regular elements are invertible, so $aS \cap sR = aS \cap R \not = 0$.
[x] $R{64}$ ($\Bbb Q[[x^2,x^3]]$) $\subset R$. Same as above.
~~- [ ] $R{167}$ ($\Bbb Z[x_0, x1, \ldots]$) $\subset R$. Express $\Bbb N = \bigsqcup\limits{i=1}^\infty N_i$, where all $Ni$ are also infinite. In $x{i-1}$, this $N_i$ is the index set that we should put an infinite Jordan block $^1$ [^1] on, and for $n \not \in Ni$, we set $e{nn} = 1$. The resulting $x$'s generate an algebra isomorphic to $\Bbb Q[x_0, x_1, \ldots]$.~~ This construction is flawed, it contains zero divisors (a difference of two Jordan blocks annihilates a third one). (03 May 2024)
[x] $R_{168}$ (Eventually constant sequences in $\Bbb Z$) $\subset R$. Subring of the diagonal.
Remarks
I think the singular ideals (at least the left one) can be described using Lam II (7.14d) p252 using the CFM.
$^1$ $N_i$'s are not contiguous. If $N_i = \{a_1, \ldots, a_j, \ldots\}$ with $a_1 < \ldots < aj < \ldots$, by this "Jordan block" I mean the sum $\sum\limits{j=1}^\infty e_{aj, a{j+1}}$.
[^1]: GitHub Markdown footnotes do not render TeX: $123$.
Description
(Or, following https://math.stackexchange.com/questions/1372406/, maybe "The subring of elements of $R_{15}$ that stabilize a given complete flag".)
The units are indeed the elements without zeros on the diagonal: 1) the map $R \to R$, $e{ii} \mapsto e{ii}$, $e_{ij} \mapsto 0$, is a homomorphism; 2) we can use Gaussian elimination to get an inverse of an upper triangular matrix column by column, starting from the diagonal element. The strictly upper triangular matrices form an ideal, and $1-rx$ is then clearly invertible. $J(R)$ is not nil: the "left coordinate shift", aka the infinite Jordan block $\Bbb J$ with eigenvalue 0, is not nilpotent.
$J(R)$ is indeed the set of strictly upper triangular matrices: it is clearly an ideal, so $1 - rx$ has all 1's on diagonal. $R/J(R) \cong \prod\limits_{i = 1}^\infty \Bbb Q$.
The zero divisor question is more difficult. All non-invertible matrices are zero divisors, and "Note on Sus̆kevic̆’s problem on zero divisors", Holubowski, Maciaszczyk, Zurek 2015 Th1.2 proves the following: a non-invertible element that is not a right zero divisor has a right inverse that is not upper triangular in the larger CFM ring.
Properties
Relations
Remarks
I think the singular ideals (at least the left one) can be described using Lam II (7.14d) p252 using the CFM.
$^1$ $N_i$'s are not contiguous. If $N_i = \{a_1, \ldots, a_j, \ldots\}$ with $a_1 < \ldots < aj < \ldots$, by this "Jordan block" I mean the sum $\sum\limits{j=1}^\infty e_{aj, a{j+1}}$. [^1]: GitHub Markdown footnotes do not render TeX: $123$.