Locally isomorphic algebras over a Dedekind domain

Let $$R$$ be a Dedekind domain. Let $$A$$ and $$B$$ be two finitely generated domains over $$R$$. Assume that for every maximal ideal $$\mathfrak{p}\subset R$$ the $$R_{\mathfrak{p}}$$-algebras $$A_{\mathfrak{p}}$$ and $$B_{\mathfrak{p}}$$ are isomorphic. Are $$A$$ and $$B$$ isomorphic?

• Does subscript $\mathfrak{p}$ stand for localization or completion on the adic topology? Apr 20, 2020 at 21:15
• @kneidell localization. Apr 20, 2020 at 21:21

1 Answer

Counterexample. Let $$R$$ be a Dedekind domain with $$\operatorname{Cl}(R) \neq 0$$. Let $$I \subseteq R$$ be an ideal that is not principal (in algebraic geometry language, let $$\mathscr L$$ be a nontrivial line bundle), and let $$J = R$$ be the trivial ideal (let $$\mathcal O$$ be the trivial line bundle). Then $$I_{\mathfrak p} \cong R_\mathfrak p \cong J_\mathfrak p$$ for every $$\mathfrak p$$.

Then $$A = \operatorname{Sym}^*(I^\vee)$$ and $$B = R[t] = \operatorname{Sym}^*(J^\vee)$$ are not isomorphic. Geometrically, this is saying that the geometric vector bundles $$\mathbf V(\mathscr L) = \operatorname{Spec}(\operatorname{Sym}^* \mathscr L^\vee)$$ and $$\mathbf V(\mathcal O) = \mathbf A^1_R$$ are not isomorphic as $$R$$-schemes. For example, $$\mathbf V(\mathcal O)$$ has a pair of disjoint sections $$0, 1 \colon \operatorname{Spec} R \rightrightarrows \mathbf V(\mathcal O)$$, but a nontrivial line bundle $$\mathscr L$$ does not have two disjoint sections (the difference gives a nowhere vanishing section, which is an isomorphism $$\mathcal O \stackrel\sim\to \mathscr L$$). But they are locally isomorphic at every prime since every vector bundle over a local ring is trivial.

(In principle you could unwind this argument algebraically if you want: $$B$$ has a surjection to $$R \times R$$ by $$f(t) \mapsto (f(0),f(1))$$, but $$A$$ does not admit such a surjection. Basic geometric operations like 'take the difference of two sections' and 'a nowhere vanishing section is an isomorphism $$\mathcal O \stackrel\sim\to \mathscr L$$' become Hopf algebra stuff, so you have to do some work.)

Remark. If you want an example over a PID, just take any PID $$R$$ that has a finite extension $$R \subseteq R'$$ of Dedekind domains such that $$\operatorname{Cl}(R') \neq 0$$ (such an extension exists in many cases, e.g. if $$R = \mathbf Z$$ or $$R = k[t]$$ for any field $$k$$). Take $$I, J \subseteq R'$$ and $$R' \to A$$ and $$R' \to B$$ as above. They are still isomorphic around any $$\mathfrak p \subseteq R$$, because a line bundle on $$\operatorname{Spec} R'$$ is trivialised on the finite set of primes above $$\mathfrak p$$.

Given an isomorphism $$\phi \colon A \stackrel\sim\to B$$ of $$R$$-algebras, the integral closure $$R'$$ of $$R$$ in $$A$$ and $$B$$ is preserved by $$\phi$$, hence up to composing with an $$R$$-automorphism of $$R'$$ we may assume $$\phi$$ is an $$R'$$-algebra isomorphism, which is impossible by the argument above.

• are there counterexamples if the class group is trivial? Apr 20, 2020 at 21:23
• One thing you can do in many cases is take an extension $R \subseteq R'$ of Dedekind domains with $\operatorname{Cl}(R') \neq 0$ and proceed as in my answer. Then you have to convince yourself that an isomorphism of $R$-algebras can be upgraded to an isomorphism of $R'$-algebras, which should be fine for example if $R = k[t]$ and $R'$ is an affine part of an elliptic curve (use that maps $\mathbf A^1 \to E$ are constant). Apr 21, 2020 at 0:59
• There might be some subtlety in the "for example because..." part: The left hand sides "should be" just sets, not $R$-modules, so to show that they are not "equal", i.e. bijective, one should not invoke the module structure. On the geometric side, I believe that the same can be formulated as: we would like to show that $\mathbf{V}(\mathscr{L})$ and $\mathbf{V}(\mathcal{O})$ are not isomorphic as actual $R$-schemes, while the argument shows that they are not isomorphic as (geometric) line bundles. Apr 21, 2020 at 4:56
• @PavelČoupek: ok, you're right. The trivial line bundle has two disjoint sections $B \rightrightarrows R$, but no nontrivial line bundle does. Apr 21, 2020 at 5:10