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, J. Algebra 12, 576-584 (1969). ZBL0175.03102.

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.


1 Answer 1


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, J. Lond. Math. Soc., II. Ser. 11, 21-31 (1975). ZBL0275.16003

that the extended centroid of $K\langle x_1,\dots,x_n\rangle$ is just $K$, as you speculated.

  • $\begingroup$ It looks like you are probably right. Thanks! $\endgroup$
    – Nick
    Oct 17, 2020 at 16:02

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