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?

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

1 Answer 1


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.

  • $\begingroup$ are there counterexamples if the class group is trivial? $\endgroup$
    – danand
    Apr 20, 2020 at 21:23
  • $\begingroup$ 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). $\endgroup$ Apr 21, 2020 at 0:59
  • $\begingroup$ 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. $\endgroup$ Apr 21, 2020 at 4:56
  • 1
    $\begingroup$ @PavelČoupek: ok, you're right. The trivial line bundle has two disjoint sections $B \rightrightarrows R$, but no nontrivial line bundle does. $\endgroup$ Apr 21, 2020 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.