TITLE: Show that two $R$-algebras are not isomorphic.
QUESTION [7 upvotes]: Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.
Does anyone have any pointers on how to approach this?

REPLY [3 votes]: Proposition. Let $R$ be a UFD and $p,q\in R$ prime elements. Let $\phi:R_p\to R_q$ be a ring isomorphism. Then $\phi(p)=uq$ or $\phi(p)=uq^{-1}$ with $u\in R$ invertible.
Proof. Since $p$ is invertible in $R_p$ its image by $\phi$ is also invertible, that is, $\phi(p)=uq^s$ with $u\in R$ invertible and $s\in\mathbb Z$. Conversely, $\phi^{-1}(q)=vp^t$ with $v\in R$ invertible and $t\in\mathbb Z$. Then $q=\phi(v)u^tq^{st}$ and since $q$ is prime we get $st=1$, so $s=\pm1$.
Now let $R=\mathbb R[a,b,c,d]$, $p=a^2+b^2+c^2+d^2$ and $q=ad-bc$. Suppose that $\phi(p)=uq$ with $u\in\mathbb R\setminus\{0\}$. Then $\phi(a)^2+\phi(b)^2+\phi(c)^2+\phi(d)^2=uq$. We can write $\phi(a)=a_1/q^i$ and so on. Then we get $a_1^2+b_1^2+c_1^2+d_1^2=uq^{2m+1}$ with $m$ a nonnegative integer. Now define a monomial order on $\mathbb R[a,b,c,d]$ such that $a>b>c>d$. Then the leading term of the right hand side is $(ad)^{2m+1}$, while the leading term of the left hand side is a sum at most four squares from $\mathbb R$ times the square of a monomial. Since that coefficient can not be zero (as it could happen if one works over $\mathbb C$, for instance!), we get a contradiction.