Non-isomorphic localized rings I want to know why $\mathbb{R}[x,y]_{x^2+y^2}\not\cong\mathbb{R}[x,y]_{xy}$?
(I mean $\mathbb{R}[x,y,(x^2+y^2)^{-1}]\not\cong \Bbb{R}[x,y,(xy)^{-1}]$)
I'm reading about weil restriction. Let $G_{m,k}$ be the multiplicative group, where $k$ is field and $\mathbb{S}:= R_{\mathbb{C}/\mathbb{R}}G_{m,\mathbb{C}}$ to be the Weil restriction of $G_{m,\mathbb{C}}$. Now, I'm trying to prove that $\mathbb{S}\not\cong G_{m,\mathbb{R}}\times G_{m,\mathbb{R}}$(as affine algebraic varieties over $\mathbb{R}$).
My attempt: I was thinking of finding some maximal ideal $m$ such that $m/m^2$ has different $\mathbb{R}-$dimension in the above two rings.
 A: If there were an isomorphism of rings $\mathbb{R}[x,y]_{x^2 + y^2} \cong \mathbb{R}[x,y]_{xy}$ then there would be an induced isomorphism of their unit groups.
Since $\mathbb{R}[x,y]$ is a UFD and $x^2 + y^2$ is irreducible, the units of $\mathbb{R}[x,y]_{x^2 + y^2}$ are precisely the elements of the form $r (x^2 + y^2)^m$ where $0 \not= r \in \mathbb{R}$ and $m \in \mathbb{Z}$.  Hence the group of units of $\mathbb{R}[x,y]_{x^2 + y^2}$ is $\mathbb{R}^{\times} \times \mathbb{Z}$.  (The first factor is the non-zero reals with multiplicative group structure, the second factor is the integers with additive group structure).
Similarly one calculates that the group of units of $\mathbb{R}[x,y]_{xy}$ is $\mathbb{R}^{\times} \times \mathbb{Z} \times \mathbb{Z}$.
The problem thus reduces to checking that there is no group isomorphism $\mathbb{R}^{\times} \times \mathbb{Z} \cong \mathbb{R}^{\times} \times \mathbb{Z} \times \mathbb{Z}$.  In fact there are no such surjections.
To check this you could proceed as follows: suppose that such a surjection exists and let $(q,a), (r, b)$ be the elements mapping to $(1, 1, 0), (1, 0, 1)$.
This implies that $(q^b/r^a, 0)$ maps to $(1, b, a)$.  Since $n$th roots exist in $\mathbb{R}$, conclude that $a,b$ are divisible by arbitrarily large integers, hence $a = b = 0$.  But then $(q, 0)$ maps to $(1,1,0)$, which by the same reasoning is absurd.
