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}$$\frac{R[x,y]}{<xy -1>} \cong R[x,1/x]$
This is laurent series ,ie, power series
As far as i understand, $\forall x^n \space\space \space(x^n)^{-1} exists$
Is this correct?For the above isomorphism, thus if one can show the left quotient ring doesnt have some of the properties of laurent series then its not isomorphic
Or if kernel is $<x^2-y^2-1>$ for $\phi : R[x,y] \rightarrow \frac{R[x,y]}{<xy -1>}$
Then it is isomorphic
Evidently, either i haven't understood enough or am not able to justify either to come to some conclusion.
$\textbf{Solution (case $R=\mathbb R$):}$
Let’s define the isomorphism $\phi:\mathbb R[u,v]\to\mathbb R[x,y] $, such that $\phi(u)=x+y, \phi(v)=x-y, \phi(1)=1$ (it’s easy to construct the inverse). Now $x^2-y^2-1=(x-y)(x+y)-1=\phi(uv-1)$. Now the morphism $$\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.