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I would like to understand why the two rings $$ R={\mathbb{C}[x,y,z]}/{(xy - (1 - z^2))} \\ S=\mathbb{C}[x,y,z]/{(x^2y - (1 - z^2))} $$ are not isomorphic, but $R[t]\cong S[t]$. This example is discussed in this document by Hochster. He references a paper by Danielewski, but it is a preprint from 1989, so I could not locate it anywhere. In fact, this example was also given by Timothy Wagner in this MSE thread, but no proof was supplied.

Finally, this MSE thread seems to refer to this problem (though possibly with different formulation), but unfortunately it hasn't even received a single comment in the last 3.5 years.

Can someone either find a precise reference (preferably online, or at least published in some book/journal) or supply a proof for this claim?

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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!

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    $\begingroup$ Wow, there is so much new stuff that I can learn from here. Thanks a lot to you and to your friend Bob Lutz! :) $\endgroup$ – Prism Jul 10 '15 at 20:22
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M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring, Proc. Amer. Math. Soc. 34 (1972), 81-82, online

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  • $\begingroup$ At first sight, this example is different. But I suspect that you can make a change of variables. $\endgroup$ – Martin Brandenburg Jul 10 '15 at 11:30

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