1
$\begingroup$

Let $R$ be a noncommutative ring (with unit). Then a "fully noncommutative" (for a lack of better wording) monomial over $R$ in the single noncommutative indeterminate $X$ of degree $d$ is given by a finite sum of words of the form $$ r_0 X r_1 X...X r_{d-1} X r_d,$$ where $r_0,...,r_d\in R$. Then one constructs the ring of "fully noncommutative" polynomials in $X$ in the obvious way. Analogously, one gets "fully noncommutative" polynomials and formal power series over $R$ in several or infinitely many noncommutative variables. (Note that this construction should not be confused with $R\langle X,Y\rangle$, where the variables do not commute with each other but are assumed to commute with the $R$-coefficients.)

(Q1) Is there a standard notation for these rings?

Currently I'm using the notation $R\rangle X\langle$, resp. $R\rangle X,Y\langle$, as I need to distinguish between the latter and $R[S,T]$ and $R\langle X,Y\rangle$. But it looks... odd and I have no desire to introduce possibly non-standard notation. Another possibility is to use $\langle R,X\rangle$ (since the elements are words), but it is not particularly suggestive of what's supposed to be a ring of polynomials and reminds more of other things (it's unfortunately overused notation).

Moreover, I imagine such rings have been well studied in the literature, but I was unable to locate them (one usually finds only the $R\langle X,Y\rangle$-type).

(Q2) What are some standard references on "$R\rangle X\langle$" ?

Thanks for your time!

$\endgroup$
  • 1
    $\begingroup$ I think if I had to come up with a notation, I would use $\mathbb{Z}\langle R,X\rangle$ (possibly replacing $\mathbb{Z}$ by whatever does commute with everything, if something does). By this I mean "the ring freely generated over $\mathbb{Z}$ by the elements of $R$ (subject to their usual relations) and the extra element $X$ (with no relation involving it)". $\endgroup$ – Gro-Tsen Nov 29 '16 at 14:40
  • $\begingroup$ Thanks! I have a feeling this is probably as standard notation as it gets. $\endgroup$ – M.G. Nov 29 '16 at 15:34
2
$\begingroup$

I would write $R \sqcup \mathbb{Z}[X]$, since the ring you describe is in fact just this coproduct in the category of rings, or $R * \mathbb{Z}[X]$ (since people do not like the coproduct symbol for some reason when applied to the category of rings and rather speak of about free products). If you want to have something similar to $R[X]$, I would suggest $R\{X\}$ or $R|X|$ or $R \lfloor X \rfloor$. I don't know if this is standard notation, probably not. When you use a notation for this ring in a paper or talk, you will probably have to explain it anyway.

$\endgroup$
  • $\begingroup$ NB: I also don't know any references in the literature. I am looking forward to read other's answers on this. $\endgroup$ – HeinrichD Nov 29 '16 at 14:25
  • $\begingroup$ Heinrich, danke dir für die superschnelle Antwort! In the mean while, I will try to search for "free products of rings" or similar. BTW, I cannot use $R\{X\}$ b/c I already have convergent power series. I forgot to mention this in my original post, sorry. $\endgroup$ – M.G. Nov 29 '16 at 14:28
  • $\begingroup$ You are right, this conflicts with the common notation for convergent power series rings. On the other hand, if $R$ is just a ring, there is no meaning for this. And if $R$ is a topological ring, then $R$ is not a ring, but only has an underlying ring, so that even then no confusion can arise if not people would drop forgetful functors from the notation as usual. Anyway, I've added $R|X|$ and $R \lfloor X \rfloor$. What do you think? $\endgroup$ – HeinrichD Nov 29 '16 at 14:34
  • $\begingroup$ I really love the floor/ceil idea! And $R\lfloor\lfloor X\rfloor\rfloor$ for the power series version remains looking nifty! I doubt whatever the standard notation is can look that well, so I will most likely start using that instead. Very nice! $\endgroup$ – M.G. Nov 29 '16 at 15:33
  • $\begingroup$ +1 for $R * \mathbb{Z}\left[X\right]$. This is a notation I have seen used (e.g., §2.2 in Crawley-Boevey's Noncommutative algebra 1). And the slickest definition of "polynomials with coefficients not commuting with the indeterminate" is probably as a free product. I would not use $R \sqcup \mathbb{Z}\left[X\right]$, since $\sqcup$ means disjoint union to me. $\endgroup$ – darij grinberg Jul 7 '17 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.