
> 
> 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](https://mathoverflow.net/posts/239420/revisions) 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. 



Non-isomorphic rings that are localizations of each other
**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 $B$. $\square$




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**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$\_4$](http://www.numdam.org/item/PMIHES_1967__32__5_0), 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.


