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

Any hints?

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Hint: the ideals of $R$ are just those generated by elements of $A$ that aren't units in $R$ (you have to prove this, of course). –  Brett Frankel Apr 28 '12 at 1:43
    
Is the hint helpful or would you like more detail? –  Brett Frankel Apr 28 '12 at 1:56
    
I have it typed up (sans a few routine verifications better left to the reader) but I don't want to spill the beans if you're making progress on your own. –  Brett Frankel Apr 28 '12 at 1:57
    
Thx. I was not in front of my computer hours ago. Your arguement is helpful. –  CC_Azusa Apr 28 '12 at 3:05
    
@CC_Azusa If you want an element-free approach, you can see my answer that I edited below. –  user38268 Apr 29 '12 at 4:24
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2 Answers

up vote 8 down vote accepted

Define the multiplicative set

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

Recall 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.

$\hspace{6in} \square$

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This is all true. The relevant step is to show that the contraction map is an ideal in $A$, which is equivalent to what I did in my solution. (I looked at all numerators, but looking at those numerators which appear with denominator 1 will give you the same ideal.) –  Brett Frankel Apr 28 '12 at 2:52
    
@BrettFrankel I have finished working out the details. I was wrong about a few things in the beginning, but here it the final version. By the way I did the bit after proving the isomorphism a little different from yours, perhaps you would like to take a look at that and give some feedback? Thanks. –  user38268 Apr 29 '12 at 3:31
    
You're right, $S$ is the set of units in $R$, not in $A$. I missed that before. Actually, you don't need all units, just a multiplicative subset containing 1. But the set you described will do just fine. –  Brett Frankel Apr 29 '12 at 3:38
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This all looks OK to me. Kudos to you for writing it all up, since it takes both enthusiasm and courage to post solutions in an area you're just learning. It seems to me, from reading your solution, that you understand everything line-by-line, but you don't yet have a strong intuitive feel for what's happening here, as evidenced by the fact that you're using a fair amount of machinery for a very elementary problem; and the machinery of extensions, contractions, and the corollary you quote don't really make the problem any simpler. –  Brett Frankel Apr 29 '12 at 3:46
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Yeah, it's good to do that. But it's also good to figure out what all those details mean. I love the elegance of arrow-theoretic arguments, but if you want to really grok something, sometimes you just need to get your hands dirty and work with elements. –  Brett Frankel Apr 29 '12 at 3:52
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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 $s\in S$}\}$, 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.

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In fact, I do have a question here. How to show the set of numerators of $I$ is closed under addition? –  CC_Azusa Apr 28 '12 at 4:16
    
Convince yourself that the ideal I described above is the same as the ideal of numerators that appear with 1 in the denominator. That ideal is clearly closed under addition. –  Brett Frankel Apr 28 '12 at 18:38
    
Let $\dfrac{r}{s}\in I$. Then $r\in A$. Thus there is $k\in \mathbb Z$ such that $r=kt$. This implies $\dfrac{r}{s}=\dfrac{k}{s}\dfrac{t}{1}$. But how $\dfrac{k}{s}\in R$? –  Anupam Nov 20 '13 at 2:27
    
@Anupam It's possible that $\frac{r}{s}\in R$ because that is the definition of $R=S^{-1}A$. $\frac{r}{s}$ need not be in $A$. –  Brett Frankel Nov 20 '13 at 18:15
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