# Quotient of non-commuting indeterminates isomorphic to commuting indeterminates

Let $$K$$ be a field. I am trying to prove that the quotient of the space $$K\langle x,y\rangle$$ of two non-commuting indeterminates $$x,y$$ with $$I=(xy-yx)$$, that is, the two-sided ideal generated by $$xy-yx$$ is isomorphic to $$K[x,y]$$, i.e. $$K\langle x,y\rangle/I\cong K[x,y]$$

So I defined the most natural morphism from $$K\langle x,y\rangle$$ to $$K[x,y]$$: $$f(x,y)\mapsto f(x,y)$$. Everything works, but I am trying to determine the kernel. It is immediate that $$I$$ is included in the kernel but I am trying to prove the converse inclusion. I can see that if $$f(x,y)$$ is a polynomial that belongs to the kernel and $$cx^ny^m$$ (or $$cy^mx^n$$) is a term of $$f$$, then $$cy^mx^n$$ (or $$cx^ny^m$$) must also be a term of $$f$$. So now it suffices to show that for all $$m,n$$ we have that $$x^ny^m-y^mx^n\in I$$. The problem is I cannot see how to do that; even for $$n=2,m=1$$, I am having a problem: $$x^2y-yx^2$$ must be written as $$f(x,y)(xy-yx)g(x,y)$$ but how?

• Not that it helps at all to solve your problem, but the ideal $(p(x,y))$ generated by a polynomial $p(x,y)$ has elements of the form $l(x,y)p(x,y)+p(x,y)r(x,y)+f(x,y)p(x,y)g(x,y)$, and not only $f(x,y)p(x,y)g(x,y)$. We have \begin{align*}x^2y-yx^2&=x(xy)-yx^2\\&=x(xy-yx)+xyx-yx^2\\&=x(xy-yx)+(xy)x-(yx)x\\&=x(xy-yx)+(xy-yx)x\end{align*} – Luiz Cordeiro Apr 11 at 13:11

By induction on $$n$$ you can show that $$y^nx-xy^n\in I$$. For, we have $$y^nx = y^{n-1}xy + y^{n-1}(yx-xy) = y^{n-1}xy \quad\textrm{mod } I.$$ Then by induction on $$m$$ you can show that $$y^nx^m-x^my^n\in I$$. For, we have $$y^nx^m = y^nx^{m-1}x = x^{m-1}y^nx = x^my^n \quad \textrm{mod } I.$$ Using this we can show that the linear map $$K[x,y]\to K\langle x,y\rangle/I$$ sending basis element $$x^my^n$$ to $$x^my^n\textrm{ mod }I$$ is actually an algebra homomorphism, and it is clearly inverse to the natural map $$K\langle x,y\rangle/I\to K[x,y]$$.

This type of problem usually arises from the usual constructions of polynomial rings and such.

Here's a categorical way to define polynomial rings: Fix a commutative ring $$R$$.

The polynomial ring $$R(x_1,\ldots,x_n)$$ on $$n$$ non-commuting variables is an $$R$$-algebra together with an inclusion $$\left\{x_1,\ldots,x_n\right\}\hookrightarrow R(x_1,\ldots,x_n)$$ which satisfies the following property: For any $$R$$-algebra $$A$$ and any $$a_1,\ldots,a_n\in A$$, there exists a unique $$R$$-algebra homomorphism $$T\colon R(x_1,\ldots,x_n)\to A$$ such that $$T(x_i)=a_i$$ for all $$i$$.

The map $$T$$ above is given by: take any polynomial $$f(x_1,\ldots,x_n)$$, and substitute any instance of $$x_i$$ by $$a_i$$, and evaluate in $$A$$; thus it is usually denoted $$T(f(x_1,\ldots,x_n))=f(a_1,\ldots,a_n)$$.

This description does not guarantee that $$R(x_1,\ldots,x_n)$$ exists (so you have to perform the usual procedure, by considering "formal sums of producs among the variables $$x_i$$" and so on).

But this description determines $$R(x_1,\ldots,x_n)$$ up to isomorphism! It is a universal property.

In any case, the general idea is that $$R(x_1,\ldots,x_n)$$ is the largest $$R$$-algebra generated by $$n$$ elements, without any further relations among them.

The polynomial ring on commuting variables is similar:

The polynomial ring $$R[x_1,\ldots,x_n]$$ on $$n$$ commuting variables is an $$R$$-algebra together with an inclusion $$\left\{x_1,\ldots,x_n\right\}\hookrightarrow R[x_1,\ldots,x_n]$$ which satisfies the following property: For any $$R$$-algebra $$A$$ and any commuting $$a_1,\ldots,a_n\in A$$, there exists a unique $$R$$-algebra homomorphism $$T\colon R(x_1,\ldots,x_n)\to A$$ such that $$T(x_i)=a_i$$ for all $$i$$.

This is the universal property for $$R[x_1,\ldots,x_n]$$, so it determines $$R[x_1,\ldots,x_n]$$ up to isomorphism.

This means that in order to prove that an $$R$$-algebra $$\mathscr{A}$$, together with some elements $$\alpha_1,\ldots,\alpha_n\in \mathscr{A}$$, is isomorphic to $$R[x_1,\ldots,x_n]$$, it is enough to prove that it satisfies the universal property: For any algebra $$A$$ and elements $$a_1,\ldots,a_n\in A$$, there is a unique algebra homomorphism $$\mathscr{A}\to A$$ sending $$\alpha_i\mapsto a_i$$.

Therefore, the idea is that $$R[x_1,\ldots,x_n]$$ is the largest algebra generated by $$n$$ commuting elements.

So if we take $$R(x_1,\ldots,x_n)$$, the largest algebra generated by $$x_1,\ldots,x_n$$, and simply impose that these elements $$x_i$$ commute (which corresponds to taking the quotient by $$(x_ix_j-x_jx_i:i,j=1,\ldots,n)$$), then we should get $$R[x_1,\ldots,x_n]$$.

Formally, we can prove that $$R(x,y)/(xy-yx)$$ satisfies the universal property of $$R[x,y]$$: Let $$A$$ be an $$R$$-algebra and $$a,b\in A$$ two commuting elements. By the universal property of $$R(x,y)$$, there is a homomorphism $$T\colon R(x,y)\to A$$ sending $$T(x)=a$$ and $$T(y)=b$$. Note that $$T(xy-yx)=T(x)T(y)-T(y)T(x)=ab-ba=0$$ because $$a$$ and $$b$$ commute. So $$xy-yx\in\ker T$$, and we can factor $$T$$ through the quotient: The quotient $$R(x,y)/(xy-yx)$$ has elements $$\overline{x}$$ and $$\overline{y}$$ (namely, the classes of $$x$$ and $$y$$), and there is a homomorphism $$T'\colon R(x,y)/(xy-yx)\to A$$, such that $$T'(\overline{x})=a$$ and $$T'(\overline{y})=b$$.

This is one part of the universal property: Existence of a homomorphism; The uniqueness of $$T'$$ with this property follows from the uniqueness of $$T$$.

Therefore $$R(x,y)/(xy-yx)$$ satisfies the same universal property as, and is therefore isomorphic to, $$R[x,y]$$.