A subring of the field of fractions of a PID is a PID as well.

Question

Let $A$ be a P.I.D and $R$ be a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the fractional field of $A$. How to show $R$ is also a P.I.D?

Any hints?

Answer 1

Here are my preliminary thoughts on the problem. This is not a full solution but perhaps it may lead to one.

I think you can suppose without loss of generality that $A \subsetneqq R \subsetneqq \operatorname{Frac}(A)$. Now we have the inclusion map $\iota :A \rightarrow R$. Therefore I think it is possible to prove that we have an $A$ - algebra isomorphism

$$R \cong S^{-1}A$$

where $S$ is the multiplicative set consisting of all units in $A$. Supposing this is true, we have because of the containment $R \subsetneqq \operatorname{Frac}(A)$ that is there is an element of $R$ that is not a unit in $A$. Hence there are non-units in $S^{-1}A$ and so we can speak of proper ideals in here.

Then from here I think you can conclude your problem by noticing that the contraction of an ideal in $S^{-1}A$ is an ideal in $A$, and that the extension of an ideal $\mathfrak{a}$ in $A$ to one in $S^{-1}A$ is of the form $S^{-1}\mathfrak{a}$.

$\textbf{Edit:}$ I think we are now in the position to prove the full statement of your problem. Now I was wrong before about $S$ because we actually need

$$S = \{\text{multiplicative set of all elements $a \in A$ that are units in $R$}\}.$$

and not as defined before. If $S$ was defined as before to be the set of all units in $A$, then we have an isomorphism $S^{-1}A \cong A$ which is completely redundant! Now with our new $S$ to prove that $R \cong S^{-1}A$ we will use Corollary 3.3 of Atiyah - Macdonald:

${\color{blue}{\text{Corollary 3.3: If $g: A \rightarrow B$ is a ring homomorphism such that}}}$

${\color{blue}{\text{(i) $s \in S \implies g(s)$ is a unit }}}$

${\color{blue}{\text{(ii) $g(a) = 0\implies as = 0$ for some $s \in S$}}}$

${\color{blue}{\text{(iii) Every element in $B$ is of the form $g(a)g(s)^{-1}$ for some $a\in A$ and $s \in S$ }}}$

${\color{blue}{\text{then there is a unique isomorphism $h:S^{-1}A \rightarrow B$ such that $g$ factorises through the}}}$ ${\color{blue}{\text{localisation $S^{-1}A$.}}}$

In our case we have $g$ to be the inclusion map $\iota : A \hookrightarrow R$ and $R$ as the ring $B$. Clearly the first two properties are satisfied. To see that the third property is satisfied, since everything is happening inside the fraction field of $A$ let us consider an element $\frac{a}{b}$ in $R$ where $a,b \in A$. Since $a,b$ are elements in $A$ that is a PID we can compute GCDs, so we can assume that $a,b$ are coprime. Again because $A$ is a PID this means that there are elements $x,y \in A$ such that

$$ax + by = 1.$$

Now go up to $\operatorname{Frac}(A)$ and view this expression as lying in here. Then dividing by $b$ we see that

$$\frac{a}{b}x + y = \frac{1}{b}.$$

However the left hand side is in $R$ so the right is. It follows that $b$ is an element of $A$ that is a unit in $R$ $\implies b\in S$. Hence we can write $a/b \in R$ as $\iota(a)\iota(b)^{-1}$ and so since $a/b$ was arbitrary it follows by the corollary that

$$S^{-1}A \cong R.$$

Now to prove that $R$ is a PID it suffices by the isomorphism to prove that $S^{-1}A$ is a PID. Let $\mathfrak{a}$ be an ideal in $S^{-1}A$. Then a basic result about localisation by multiplicative sets says that

$$(\mathfrak{a}^{c})^{e} = \mathfrak{a}$$

where $(\mathfrak{a}^{c})^{e}$ denotes the extension of the contraction of $\mathfrak{a}$. Now $\mathfrak{a}^c$ is always an ideal in $A$; since $A$ is a PID we can write

$$\mathfrak{a}^c = (\alpha)$$

for some $\alpha \in A$. It follows that $$(\alpha)^{e} = (\mathfrak{a}^{c})^{e} = \mathfrak{a}. $$

Now I claim that $(\alpha)^{e} = S^{-1}(\alpha)$. To see this, clearly we have $S^{-1}(\alpha) \subseteq (\alpha)^{e}$. To see the other inclusion, take an element

$$\sum_{i=1}^n \frac{\alpha_i}{s_i}$$

in $(\alpha)^e$ where $s_i \in S$, $\alpha_i \in (\alpha)$. Then clearing denominators gives that this element is in $S^{-1}A$, so we have proven the other inclusion.

Hence $\mathfrak{a} = S^{-1}(\alpha)$, from which it follows immediately that $\mathfrak{a}$ is a principal ideal.

$\hspace{6in} \square$

Answer 2

If $I$ is an ideal of $R$, then you can verify that that the set of numerators of $I$, $\{r\in A: \frac{r}{s}\in I\text{ for some $s\in A$}\}$, is an ideal in $A$, thus a principal ideal, generated by some element $t$. Now show that the ideal generated by $t=\frac{t}{1}$ in $R$ is the ideal you started out with.