The standard formal modeling for polynomials is the polynomial ring $R[X_1,...,X_n]$ which is a monoid ring $R[\mathbb{N}^n]$ over an rng $R$.

Under this construction, it is possible to commute $X_i$'s so that $X_iX_j=X_jX_i$.

However, over a non-commutative ring, we need more than this.

For example, say we want to find a solution $(x_1,x_2,x_3)$ of a equation $aX_1X_2^2X_3 + bX_3X_1X_2=0$ over a non-commutative ring.

If we want to find $3$-tuples in the center of the ring, then the standard polynomial ring can be used to find the solutions. However, in general, we don't want $X_i$'s to commute. For example, we want $(X_1X_2^2X_3)(X_3X_1)=X_1X_2^2X_3^2X_1$. Moreover, if there is a formal model for polynomials, it always form a ring.

Consquently, rather than $\mathbb{N}^n$, we need somewhat more complex structure.

Say, $R[M]$ is a monoid ring over $R$ and it is a correct model of polynomial rings.

Then, $M$ should have information about how variables are ordered and what the powers of each variable in a term are. What would $M$ be?

Is there a way to construct polynomials in non-commutative rings?

  • $\begingroup$ One could construct a noncommutative polynomial ring over a noncommutative ring. I'd venture to guess it wouldn't be a monoid ring, though. $\endgroup$ Commented Jun 1, 2015 at 18:52
  • $\begingroup$ @MattSamuel How do you know that it can be constructed? $\endgroup$
    – Rubertos
    Commented Jun 1, 2015 at 18:54
  • $\begingroup$ I have an idea of how to construct it. I haven't completely fleshed it out but I see no reason why it wouldn't work. One has quite a bit of freedom in constructing rings. $\endgroup$ Commented Jun 1, 2015 at 18:56
  • $\begingroup$ en.wikipedia.org/wiki/Free_algebra (or search for "noncommutative polynomials") $\endgroup$
    – sdcvvc
    Commented Jun 1, 2015 at 18:59
  • 1
    $\begingroup$ @sdcvvc The free algebra is not quite what we want. The article defines it over a commutative ring. I'm thinking more along the lines of the tensor algebra of a free bimodule. $\endgroup$ Commented Jun 1, 2015 at 19:11

1 Answer 1


The tensor "algebra" of a free $(R,R)$-bimodule should work. I put the term in quotes because it is not an $R$-algebra, but rather an $R$-ring (i.e. multiplication by elements of $R$ makes sense but $R$ is not central).

This ring is the direct sum of tensor powers of the free bimodule, where we use the bimodule tensor product. Multiplication is given by taking tensor products. Considering the free bimodule to be generated by the variables $x_1,x_2,\ldots,x_n$, this ring can be explicitly described additively as the bimodule generated by products of the form

$$r_1x_{i_1}r_2x_{i_2}\cdots x_{i_{n-1}}r_nx_{i_n}r_{n+1}$$ Technically this is only a generating set if $R$ is unital, but the modification for nonunital rings, while possibly notationally ugly, is not a big deal.

This behaves like a polynomial ring in that the variables can be specialized by substituting elements of $R$, and this substitution yields a homomorphism into $R$.


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