Only DVR's with quotient field $\mathbb{Q}$? Let $p \in \mathbb{Z}$ be a prime number. I know how to show  that $$\{r \in \mathbb{Q}: r = {a\over{b}},\text{ }a,b \in \mathbb{Z},\text{ }p\text{ doesn't divide }b\}$$ is a DVR with quotient field $\mathbb{Q}$. My question is though, are these the only DVR's with quotient field $\mathbb{Q}$?
 A: Yes, as a DVR is a  local ring there is only one prime number that is not invertible. (Note that a nontrivial ideal cannot include two distinct prime numbers and each non-unit is contained in some maximal ideal.)
In more detail: 
Every non-zero subring of $\mathbb{Q}$ must contain $\mathbb{Z}$ and since a field is not a DVR the ring must not equal $\mathbb{Q}$.
Let $\mathbb{Z} \subset R  \subsetneq  \mathbb{Q}$ be a DVR. Being a DVR $M=R \setminus R^{\times}$ is an ideal (the unique maximal ideal). Suppose there are distinct prime numbers  $p,q \in M$. Yet, the $\mathbb{Z}$-ideal generated by $p,q$ is $\mathbb{Z}$ and so constains $1$. Yet this $\mathbb{Z}$-ideal is contained in $M$ which does not contain $1$, a contradiction. 
Thus, all but one prime number $p_0$ are not in $M$ and thus in $R^{\times}$. 
This means that $b \in R^{\times} $ for every integer $b $ coprime to $p_0$ and $R$ contains a ring of the form given in OP. 
It remains to show that $R$ is not larger. Suppose $\frac{a}{b} \in R$ with $p_0 \mid b$ say $b = p_0 b'$ and $p \nmid a$. Then $a \in R^{\times}$ and $\frac{b'}{a} \in R$. So $ \frac{1}{p_0} = \frac{a}{b} \frac{b'}{a}  \in R$. And $p_0 \in R^{\times}$. 
Yet then all prime numbers are invertible and thus $R = \mathbb{Q}$, a contradiction. 
