Is $R[r]$ equal to $R$ itself? And some consequences Suppose $A$ and $B$ are integral domains with $A\subseteq B$. Let $b\in B$. Then $b$ is integral over $A$ iff $A[b]$ is finitely generated as an $A$-module.
What about in the case that $A = R$ a principal ideal domain and $B = Q = \operatorname{Quot}(R)$? Clearly $R \subseteq Q$ so if $q \in Q$ then $q$ is integral over $R$ iff $R[q]$ is finitely generated as an $R$-module. 
I am trying to use this to prove that principal ideal domains are integrally closed (I know that unique factorisation domains are integrally closed and that principal ideal domains are unique factorisation domains but I'm looking for a proof via the above theorem, if possible). 
In combination with the theorem above, I feel like the following statements should be equivalent;

(a) If $R$ is a principal ideal domain then $R$ is integrally closed;
(b) If $q \in \operatorname{Quot}(R)$ then $R[q]$ is finitely generated as an $R$-module only when $q \in R$ (i.e. if $q \in \operatorname{Quot}(R)\setminus R$ then $R[q]$ is not finitely generated as an $R$-module.)

I've attempted this in the following way, but I'm entirely unsure if this is correct or not (and I must in some way have implicitly assumed a property of PIDs else this would appear to be true in any integral domain);
Suppose that $q \in \operatorname{Quot}(R)$. Then $q$ can be written $q = rs^{-1}$ with $r, s \in R$, so that $R[q] = R[rs^{-1}]$. This is the ring of polynomials in $rs^{-1}$, so the elements look something like
$$R[rs^{-1}] = \left\lbrace \sum_{i \in \Bbb N} a_i(rs^{-1})^i : a_i \in R, rs^{-1} \in \operatorname{Quot}(R)\right\rbrace.$$
Now the terms $a_i(rs^{-1})^i = a_ir^{i}(s^{-1})^i$ and since $r, a_i \in R$, their product is in $R$, so in fact we have
\begin{align}
R[rs^{-1}] &= \left\lbrace \sum_{i \in \Bbb N} a_ir^i(s^{-1})^i : a_i, r \in R, s^{-1} \in \operatorname{Quot}(R)\right\rbrace\\
&= \left\lbrace \sum_{i \in \Bbb N} b_i(s^{-1})^i : b_i \in R, s^{-1} \in \operatorname{Quot}(R)\right\rbrace\\
&= R[s^{-1}].
\end{align}
If $s\in R$ is a unit then this is just $R$, and $R$ is clearly finitely generated as a module over itself, so suppose $s$ is not a unit in $R$. 
If $R[s^{-1}]$ were finitely generated as an $R$-module then for any element $x \in R[s^{-1}]$ we would have elements $u_i \in R[s^{-1}]$ and elements $r_i \in R$ so that $x = r_1u_1 + r_2u_2 + \dots r_nu_n$.
What property of PIDs can I now use to show that this can't be the case?
I tried some kind of inductive argument to show that $\frac{1}{s^k}$ can't be written as a sum of lower powers of $\frac{1}{s}$ with coefficients in $R$, since this would imply that $s$ is a unit, but I don't think this shows that $R[s^{-1}]$ is not finitely generated so if possible, could I have some hints as to what direction I could go in to prove something like this?
Also, what property of PIDs have I somehow implicitly built into this (potential) argument without which it would fail over more general rings?
 A: Here is a try. 
1. For simplicity, let's first assume $R=\mathbb{Z}$. Let $\frac{1}{q}\in \mathbb{Q}$ be such that $\mathbb{Z}[\frac{1}{q}]$ is a finitely generated $\mathbb{Z}$-module (you have already noted that it's sufficient to consider this cases only).
Case I: Assume $q$ is a prime number. Let $\frac{a_1}{q^{r_1}}, \cdots, \frac{a_n}{q^{r_n}}$ be a set of generators of $\mathbb{Z}[\frac{1}{q}]$ as $\mathbb{Z}$-module. We can also assume that $\text{gcd}(a_i, q)=1 $ and $r_i \geq 1, \forall i=1, \cdots, n$. Let $r > \sum_{i=1}^n r_i$. Then it's easy to see that $\frac{1}{q^r}$ can not be written as $\mathbb{Z}$-linear combination of $\frac{a_1}{q^{r_1}}, \cdots, \frac{a_1}{q^{r_1}}$.
Case II: Now assume $q$ is a composite number and $q=p_1^{s_1}\cdots p_m^{s_m}$ be a prime factorization of $q$. Then $\mathbb{Z}[\frac{1}{q}]=\mathbb{Z}[\frac{1}{p_1\cdots p_m}].$ Using a similar argument as above one can show that $\mathbb{Z}[\frac{1}{p_1\cdots p_m}]$ is a finitely generated $\mathbb{Z}$-module.
2. Now let $R$ is any PID. Then using the fact that $R$ is also an UFD, it's easy to see that the above proof goes through in this case also (e.g. "prime number" replaced by "prime/irreducible element", "prime factorization" replaced by "unique factorization in prime/irreducible elements").
Note: I do not know how you will be able to avoid the fact "PIDs are UFDs". In the argument above, I never used any particular property of PID. All I used "unique factorizations", which is a property of UFDs. This actually shows that every UFD is integrally closed.
