Isomorphism Quotient Polynomial Rings 2 Var

polynomial-ringsring-isomorphism

Examine Whether:

$$
\frac{\mathbb{R}[x,y]}{<x^2-y^2-1>} \cong \frac{\mathbb{R}
[x,y]}{<xy -1>}
$$

Background:

  • 2nd year math undergrad
  • Currently doing introductory ring theory

What I Know\Questions:

  • $ R[x,y] \cong R[x][y] \cong R[y][x]$

  • $f(x,y) \in R[x,y],\space\space\space f_i \in R[x]$
    $f(x,y) = f_0(x) + f_1(x)y + f_2(x)y^2 + …$

  • Factorization concept of polynomial rings in 1 var doesnt translate for multiple variables

  • $<xy-1>$ is additive identity
    $xy-1 =0 \implies x=\frac{1}{y}$

  • $\frac{R[x,y]}{<xy -1>} \cong R[x,1/x]$
    This is laurent series ,ie, power series
    As far as i understand, $\forall x^n \space\space \space(x^n)^{-1} exists$

    Is this correct?

  • For the above isomorphism, thus if one can show the left quotient ring doesnt have some of the properties of laurent series then its not isomorphic
    Or if kernel is $<x^2-y^2-1>$ for $\phi : R[x,y] \rightarrow \frac{R[x,y]}{<xy -1>}$
    Then it is isomorphic

Evidently, either i haven't understood enough or am not able to justify either to come to some conclusion.

Best Answer

$\textbf{Solution (case $R=\mathbb R$):}$

Let’s define the isomorphism $\phi:\mathbb R[u,v]\to\mathbb R[x,y] $, such that $\phi(u)=x+y, \phi(v)=x-y, \phi(1)=1$ (it’s easy to construct the inverse). Now $x^2-y^2-1=(x-y)(x+y)-1=\phi(uv-1)$. Now the morphism $$\psi=\pi\circ\phi:\mathbb R[u,v]\to \mathbb R[x,y]\to\frac{\mathbb R[x,y]}{\langle x^2-y^2-1\rangle},$$ is surjective (because it’s the composition of two surjective homomorphism) and if $\psi(p(u,v))=0$ we have, for the injectivity of $\phi$, that $\phi(p(u,v))=p(x+y,x-y)\in\langle x^2-y^2-1\rangle \implies p(x+y,x-y)=(x^2-y^2-1)q(x,y)=((x+y)(x-y)-1)q(x,y)$

now we just have to apply $\phi^{-1}$: $$p(u,v)=\phi^{-1}(p(x+y,x-y))=\phi^{-1}\bigg(((x+y)(x-y)-1)q(x,y)\bigg)=(uv-1)\phi^{-1}(q(u,v)) \implies p(u,v)\in \langle uv-1\rangle .$$ we just proved that $\langle uv-1\rangle\supset\ker\psi$ (the other inclusion its trivial, hence $\langle uv-1\rangle=\ker\psi$).

Now we have just to use the Fundamental theorem on homomorphisms which guarantees the existence of isomorphism :$$\bar\psi:\frac{\mathbb R[u,v]}{\langle uv-1\rangle}\to\frac{\mathbb R[x,y]}{\langle x^2-y^2-1\rangle}$$ (the surjectivity follows from the surjectivity of $\psi$, while the injectivity from $\langle uv-1\rangle=\ker\psi$ ).

$\textbf{Inverse of } \phi:$

It is immediate to verify that $\phi^{-1}:\mathbb R[x,y]\to\mathbb R[u,v]$ is such that $\phi^{-1}(x)=(u+v)/2,\;\phi^{-1}(y)=(u-v)/2$

$\textbf{if $R$ is a generic field? } $

The solution works even if $R\neq \mathbb R$, but we needed the inverse of 2 to construct the inverse of $\phi$. But if char($R$)=2 then there is no isomorphism:

In $\frac{\mathbb R[x,y]}{\langle x^2-y^2-1\rangle}$ there is a non-trivial idempotent element $(x+y)^2=x^2+y^2+2xy=x^2+y^2=x^2-y^2=1.$ However in $\frac{\mathbb R[x,y]}{\langle xy-1\rangle}$ every element has a unique writing in the form (using $xy=1$): $$a_0+\sum^n a_i x^i+\sum^m b_iy^i, \text{ with } a_i,b_i\in \mathbb F_2, $$ and there is no idempotent element because (wlog $n\ge m$ with $a_n\neq 1$): $$(a_0+\sum^n a_i x^i+\sum^m b_iy^i)^2=a_n^2x^{2n}+(\text{other terms with lower degree in $x$})\implies (\ldots)^2\neq 1.$$ hence there is no non-trivial idempotent element.

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