Cancellation problem: $R\not\cong S$ but $R[t]\cong S[t]$ (Danielewski surfaces)

First of all, a good friend of mine wrote a nice exposition of this topic! Also, there is a very nice overview by Kraft: "Challenging problems on affine $n$-space".

R $\not\cong$ S

There are two methods I found of proving this:

Topological method (Danielewski, Fieseler, tom Dieck). Let X, Y be the surfaces in $\mathbf{C}^3$ defined by $xy-(1-z^2)$ and $x^2y-(1-z^2)$, respectively. Then, their fundamental groups at infinity are not isomorphic, in particular $$\pi_1^\infty(X) \cong \mathbf{Z}/2\mathbf{Z}, \qquad \pi_1^\infty(Y) \cong \mathbf{Z}/4\mathbf{Z}.$$ Danielewsi and Fieseler showed similar statements for the first homology groups at infinity; the former showed $H_1^\infty(X) \cong \mathbf{Z}/2\mathbf{Z}$ and that 4 divides the order of $H_1^\infty(Y)$, while Fieseler was able to compute them. Tom Dieck computed the $\pi_1^\infty$.

Algebraic method (Makar-Limanov). For $A$ a $\mathbf{C}$-algebra, recall that a $\mathbf{C}$-endomorphism $\delta\colon A \to A$ is a derivation if it satisfies the Leibniz rule. Denote by $A^\delta$ the kernel of a derivation $\delta$. We call a derivation $\delta$ locally nilpotent if for every $a \in A$ there is an $n$ such that $\delta^n(a) = 0$, and denote the set of these $\delta$ as $\operatorname{LND}(A)$. Finally, the AK-invariant denoted $\operatorname{AK}(A)$ of $A$ is the intersection of the kernels of all locally nilpotent derivations, i.e., $$\operatorname{AK}(A) := \bigcap_{\delta \in \operatorname{LND}(A)} A^\delta.$$ Then, Makar-Limanov showed that $\operatorname{AK}(R) = \mathbf{C}$, while $\operatorname{AK}(S) = \mathbf{C}[x]$. The former is simple to show, since $$x \mapsto 0, \quad y \mapsto -2z, \quad z \mapsto x$$ is a locally nilpotent derivation with kernel $\mathbf{C}[x]$, and since we can replace $x \leftrightarrow y$ to get a locally nilpotent derivation with kernel $\mathbf{C}[y]$. The other computation is a bit more difficult; see Makar-Limanov's "On the group of automorphisms of a surface $x^ny = P(z)$". There is a nice textbook discussion of this in Rowen's Graduate algebra: commutative view, starting at p. 201.

R[t] $\cong$ S[t]

The idea is as follows, following Rem. 1.5 in Fieseler. Both $X$ and $Y$ can be realized as $\mathbf{G}_a$-principal bundles over the line $V$ with two origins; see one of Prof. Speyer's answers on MathOverflow (which is for a slightly different defining equation, but a similar glueing argument should work). Then, in the cartesian square $$\require{AMScd} \begin{CD} X \times_V Y @>>> Y\\ @VVV @VVV\\ X @>>> V \end{CD} $$ all maps are projections of $\mathbf{G}_a$-principal bundles. Since $X$ and $Y$ are both affine, $H^1(X,\mathbf{G}_a) = H^1(Y,\mathbf{G}_a) = 0$, and so by Hilbert's theorem 90 (see Serre's "Espaces fibrés algébriques," §2.3), $X \times \mathbf{C} \cong X \times_V Y \cong Y \times \mathbf{C}$ is a trivial $\mathbf{G}_a$-principal bundle over both $X$ and $Y$.

I would love to know if the last isomorphism can be written out explicitly!