Symmetry of unique generator property

Question

In this article:

Canfell, M. J. "Completion of Diagrams by Automorphisms and Bass′ First Stable Range Condition." Journal of algebra 176.2 (1995): 480-503.

the author defines a ring $R$ to have the unique generator property on right ideals if for all pairs $a,b\in R$, $aR=bR$ implies $a=bu$ for some unit $u\in R$. (In other articles it is just abbreviated to UGP. In fact the definition seems to go back all the way to Kaplansky.)

The author comments that they do not know if the condition is left-right symmetric. In my brief search, I was unable to unearth any resources declaring that this had been decided one way or the other since then.

Is anyone aware if it is now known if UGP is or isn't symmetric?

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Update :

Also in

Khurana, Dinesh, and T. Y. Lam. "Rings with internal cancellation." Journal of Algebra 284.1 (2005): 203-235.

the authors there remark that it is unknown at that time.

Answer 1

Lam informed me that, as far as he knew, this problem was still open. However, the example below shows that the condition is not left-right symmetric.

Let $$ R=\mathbb{F}_2\langle a,b,c\, :\, a^2=ab=ac=ba=b^2=bc=0,\ ca=b,\ cb=a\rangle. $$

The ideal generated by $a$ and $b$ is nilpotent of index $2$; call that ideal $J=\mathbb{F}_2a+\mathbb{F}_2b$. We see that $R/J\cong \mathbb{F}_2[c]$ is an integral domain and Jacobson semisimple, so $J$ is the Jacobson radical of $R$. As the only unit in $\mathbb{F}_2[c]$ is $1$, and an element of $R$ is a unit if and only if its image is a unit modulo the Jacobson radical, we see that $U(R)=1+J$.

Note that $Ra=Rb$ (since $ca=b$ and $cb=a$). However, $U(R)a=(1+J)a=\{a\}\not\ni b$, so $R$ is not a left UGP ring.

Now, suppose that $rR=sR$ for some elements $r,s\in R$. (We can quickly reduce to the case that they are nonzero.) The relations allow us to write $r$ uniquely in the form $r=f(c)+\alpha a+\beta b$ for unique polynomial $f(x)\in \mathbb{F}_2[x]$ and unique $\alpha,\beta\in \mathbb{F}_2$. Similarly, $s=g(c)+\gamma a+\delta b$.

Now, looking at $rR=sR$ modulo $J$, we get $f(c)\mathbb{F}_2[c]=g(c)\mathbb{F}_2[c]$. As $\mathbb{F}_2[c]$ is an integral domain, it has the UGP property. Its only unit is $1$, and so $f(c)=g(c)$.

Write $r=st$ for some $t\in R$. If $f(c)\neq 0$, then $t\in 1+J$, and hence $t$ is a unit. If $f(c)=0$, then $r,s\in J$, and so $$ r\mathbb{F}_2 =rR=sR=s\mathbb{F}_2. $$ Thus $r=s$ in this case.