Show that two $R$-algebras are not isomorphic. 
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?
 A: I don't know a purely algebraic proof. The following proof involves a little algebraic geometry and algebraic topology, so it may not be the intended answer to your question. In that case, please specify your context (i.e. where did you get the problem from).
Suppose $A$ and $B$ are isomorphic as $\Bbb R$-algebras. Then their corresponding $\Bbb R$-points, viewed as real manifolds, are isomorphic, as polynomial maps are continuous (and even $C^\infty$).
However, these two manifolds are: $A(\Bbb R) = \Bbb R^4 \backslash \{0\}$ and $B(\Bbb R) = \operatorname{GL}(2, \Bbb R)$. It is obvious that they are not isomorphic as topological spaces, by e.g. comparing their simplicial (co)homology groups.
A: 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.
A: Question: "Does anyone have any pointers on how to approach this?"
Answer: Your argument is correct: Define
$$A := \mathbb{R}[a,b,c,d]/(a^2+b^2+c^2+d^2), B := \mathbb{R}[a,b,c,d]/(ad-bc).$$
Formally, if there is an isomorphism of affine schemes (over $k$ - the real numbers) $\phi: T:=Spec(A) \cong S:=Spec(B)$ it would follow that $T$ and $S$ have the same $k$-rational points: The map $\phi$ would induce a 1-1 correspondence
$$\phi(k): T(k) \cong S(k)$$
and an equality of sets
$$T(k) \cong Hom_{k-alg}(A,k) \cong Hom_{k-alg}(B,k)\cong S(k),$$
but $T$ has only one $k$-rational point: $x:=(0,0,0,0)$. $S$ has a rational point $x:=(a,b,c,d)$ for any $x$ with $det(x)=0$.
Formally: If $T\cong S$ are isomorphic schemes (over $k$) it follows $h_T\cong h_S$
are isomorphic as functors (over $k$). Hence an isomorphism $\phi$ of affine schemes over $k$ does not exist.
