Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Described below are some observations I have made while fiddling around with polynomials. In addition to the two questions below, I am looking for any sort of relevant information so I can read more.

Preliminary question:

Suppose I want to consider the polynomial ring $\mathbb{C}\left[x, y\right]$, except I don't want to assume that $x$ and $y$ are commutative. I believe that the resulting structure is still a ring, but I have never seen anything dealing with such a scenario. Is there a name for such a structure/assumption? Is it even a sound thing to do?

Main question:

Denote by $\mathbb{C}[x, y]^*$ the ring $\mathbb{C}[x, y]$ with the added restriction that $xy \ne yx$ in general. Let $R$ be a non-commutative ring and suppose I have a homomorphism

$$\varphi \colon \mathbb{C} \longrightarrow Z(R)$$

Then it seems I can use some "extended" form of the substitution principle to make the substitution $x = a$ and $y = b$ for any $a, b \in R$, for it appears to be the case that the map

$$\Phi \colon \mathbb{C}[x, y]^* \longrightarrow R$$

given by

$$\Phi(\sum_{i + j = 0}^{n} a_{ij}x^iy^j) = \sum_{i+j=0}^n\varphi(a_{ij})a^ib^j$$

is a homomorphism.

Is this in fact the case or am I missing something?

share|cite|improve this question
twisted polynomial or noncommutative polynomial ring. whether it is sound is a matter of taste, but it can be done at any rate – Alexander Grothendieck Nov 17 '13 at 22:27
For a field $F$ and indeterminates $x,y$, the free $F$-algebra on $x,y$ is denoted $F\langle x,y \rangle$. In a sense it is "the most general (associative) $F$-algebra containing $x$ and $y$." The ordinary polynomial ring $F[x,y]$ can be retrieved as a quotient of the free algebra: $F[x,y]=F\langle x,y \rangle/(xy-yx)$. – rschwieb Nov 18 '13 at 14:29
up vote 3 down vote accepted

The usual notation for non-commutative free $k$-algebras is $k\langle x,y\rangle$. And what you have verified is its universal property. All this is well-known, you can find it in Bourbaki's Algèbre for sure. Notice that $k\langle x,y\rangle$ is the tensor algebra over the free module of rank $2$, wheras $k[x,y]$ is the symmetric algebra over the free module of rank $2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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