Do there exist commutative rings $A$ and $B$ and multiplicative subsets $S\subseteq A$, $T\subseteq B$ such that $A\not\simeq B$ but $S^{-1}A \simeq B$ and $T^{-1} B\simeq A$?

This question comes from a deleted claim in an answer of Qiaochu Yuan. The original claim says that (the proper class of isomorphism classes of) commutative rings equipped with the relation $A\ge B$ iff $B$ is a localization of $A$ is a (large) poset. The claim is clearly equivalent to the negative answer to the question. However, I do not find any reason that such examples do not exist; neither can I construct an example.


1 Answer 1


Example. Let $k$ be a field, and let $K = k(x_1,x_2,\ldots)$ be the fraction field of $k[x_1,x_2,\ldots]$. Let $$A = K[y_1,y_2,\ldots],$$ and $$B = A[y_1^{-1}].$$ Then $B$ is a localisation of $A$. If we further localise at the multiplicative set $S = K[y_1]\setminus\{0\} \subseteq B$, we get a ring that is isomorphic to $A$ by shifting all the indices by $1$: \begin{align*} S^{-1} B &\stackrel\sim\longrightarrow A\\ x_i &\longmapsto x_{i+1},\\ y_i &\longmapsto \left\{\begin{array}{ll}x_1 & \text{if } i = 1,\\ y_{i-1} & \text{if } i > 1.\end{array}\right.\\ \end{align*} To see that $A$ and $B$ are not isomorphic (even as rings), note that the set $A^\times \cup \{0\}$ is closed under addition, and the same is not true in $B$. $\square$

Remark. There are no examples of finite presentation (in a ridiculously general setting):

Lemma. Let $R$ be a ring, and assume $A$ and $B$ are $R$-algebras of finite presentation. Assume there exist multiplicative sets $S \subseteq A$ and $T \subseteq B$ and $R$-linear isomorphisms \begin{align*} S^{-1}A \stackrel\sim\longrightarrow B, & & T^{-1}B \stackrel\sim\longrightarrow A. \end{align*} Then $A$ and $B$ are isomorphic (as $R$-algebras).

Proof. Consider the composite map $\phi \colon A \to A$ obtained by $$A \to S^{-1} A \stackrel\sim\to B \to T^{-1}B \stackrel\sim\to A.$$ Note that this map is the localisation at the inverse image of $ST \subseteq B$ in $A$. To avoid confusion, denote the codomain of $\phi$ by $A'$. Since the natural map $$A' \otimes_A A' \to A'$$ is an isomorphism (this is always true for localisations), we conclude that the map $$\phi^* \colon \operatorname{Spec} A \to \operatorname{Spec A}$$ induced by $\phi$ is a monomorphism. By EGA IV$_4$, Proposition 17.9.6 (you can think of this as a fancy version of the Ax-Grothendieck theorem), this implies that $\phi^*$ is an automorphism, so $\phi$ is an isomorphism. But a localisation that is an isomorphism is the identity (i.e. $ST$ consists of units of $A$). Then $S$ contains only units of $A$, so $S^{-1}A = A$. $\square$

Remark. We actually only used that one of them was of finite presentation.

Remark. One may object that the example above is not $K$-linear (and therefore it violates two of the assumptions of the lemma). However, it is $k$-linear (or $\mathbb Z$-linear...), so in that sense it only violates the finite presentation assumption.

  • 1
    $\begingroup$ Maybe I should use the suggestive notation $A = K[x_0, x_{-1},x_{-2}, \ldots]$. $\endgroup$ Commented Jun 20, 2016 at 7:10
  • 2
    $\begingroup$ Are there finitely generated examples? (Not finitely presented.) $\endgroup$
    – HeinrichD
    Commented Oct 31, 2016 at 8:38
  • $\begingroup$ Could you please elaborate on why the natural map $A' \otimes_A A' \to A'$ is an isomorphism (or give a reference) ? $\endgroup$
    – user111524
    Commented May 19, 2018 at 19:33
  • $\begingroup$ @user111524: An inverse is given by $a/s \mapsto a/s \otimes 1$. To see this is an inverse, note that $a/s \otimes b/t$ is equal to $ab/s \otimes 1/t$ and that this is equal to $(ab/st)t \otimes 1/t = ab/st \otimes 1$. $\endgroup$
    – jdc
    Commented Feb 18, 2021 at 2:46

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