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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?

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  • $\begingroup$ 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*} $\endgroup$ – Luiz Cordeiro Apr 11 at 13:11
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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 $K\langle x,y\rangle/I\to K[x,y]$.

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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, $R[x,y]$.

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