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My question started as follows: An intermediate ring $\Bbb Z\subseteq R\subseteq\Bbb Q$ arises as a localisation. In particular it holds that $R=S_{P}^{-1}\Bbb Z$, where $S_{P}$ is the multiplicative subset of $\Bbb Z$ generated by the primes which are invertible in $R$, i.e. the set of primes $\{p:1/p\in R\}$. The crucial part to prove this is the property

$$\forall a/b\in\Bbb Q \ \ \ \ (a/b\in R \iff 1/b\in R) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$

which follows easily from the Bézout identity.

Now, I want to find an intermediate ring $\Bbb Q[x,y]\subseteq R\subseteq\Bbb Q(x,y)$ which is not a localisation of $\Bbb Q[x,y]$. So I came up with the ring $\Bbb Q[x,y,x/y]$, in which while it holds that $x/y\in R$, I managed to prove that $1/y\not\in R$. So if I assume that $R$ is a localisation of $\Bbb Q[x,y]$ at a multiplicative closed subset $S\subseteq\Bbb Q[x,y]$ then I am pretty sure that taking into account the fact that $\Bbb Q[x,y]$ is a domain we have the property $(*)$, which will end the proof. However, I cannot prove it until now. So I made up the conjecture below:

Conjecture: Let $A$ be an integral domain and $S\subseteq A$ a multiplicative subset of $A$. Then for every $a/b\in \mathrm{Frac}(A)$ it holds that $a/b\in S^{-1}A\iff 1/b\in S^{-1}A$.

Bernard
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richarddedekind
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1 Answers1

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As written, the conjecture is false.

Let $A=\mathbb{Z}$ and $S$ the set of odd numbers. Observe that $S$ is a multiplicatively closed. We also observe that $2/2\in\operatorname{Frac}(A)$ is in $S^{-1}A$ since $2/2=1/1$ and $1/1\in S^{-1}A$. Note, however, that $1/2\not\in S^{-1}A$ since if $1/2=a/s$ with $a\in A$ and $s\in S$, then $2a=s$, but the LHS is even while the RHS is odd. A contradiction.

Now, one might argue that this example is unfair since $2/2$ is not in lowest terms. However, if $A$ is not a unique factorization domain (or worse), lowest terms might not make sense.

Michael Burr
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