For a prime ring $R$, you can define its "*Martindale ring of quotients*" $Q(R)$. See for example:


*Martindale, Wallace S. III*, [**Prime rings satisfying a generalized polynomial identity**](http://dx.doi.org/10.1016/0021-8693(69)90029-5), J. Algebra 12, 576-584 (1969). [ZBL0175.03102](https://zbmath.org/?q=an:0175.03102).


The center of this ring is called the "*extended centroid*" of $R$. It is known that it is a field.


I was wondering if anyone knows what is the extended centroid of $R = K\left<x\_1,x\_2,\dots,x\_n\right>$, the free associative algebra over a field $K$? Is it just $K$ in this case?


In the literature (as the title of the article above suggests), this construction is usually considered in the case of rings with a polynomial identity. I cannot find any mention of the construction for free algebras.



What is the extended centroid of a free algebra?
I'm no expert, but I think it follows from Theorem 11 and the rest of the discussion in Section 6 of


*Bergman, George M.; Lewin, Jacques*, [**The semigroup of ideals of a fir is (usually) free**](http://dx.doi.org/10.1112/jlms/s2-11.1.21), J. Lond. Math. Soc., II. Ser. 11, 21-31 (1975). [ZBL0275.16003](https://zbmath.org/?q=an:0275.16003)


that the extended centroid of $K\langle x\_1,\dots,x\_n\rangle$ is just $K$, as you speculated.


