The field of fractions of an order is $\mathbb{Q}$

Question

Let $K$ be an algebraic number field and $L$ the algebraic integers in $K$. Assuming an order $O$ is defined as a subring of $L$ that is of finite index as a subgroup of the additive group $L$, I have to prove that the field of fractions of $O$ is $K$.

It is not hard to prove that $O$ is a $\mathbb{Z}$-module of rank $n = [K : \mathbb{Q}]$. Moreover, we know that the field of fractions of $L$ is $K$, which implies that the field of fractions of $O$ is a subfield of $K$. How can we prove that the field of fractions of $O$ is also $K$?

Answer 1

An element of $K$ has the form $a/b$ for $a,b\in L$ ($b\neq0$) since $K=\operatorname{Frac}L$. Let $k=[L:O]$, which is finite by assumption. We then have that $k{\cdot}a\in O$ and $k{\cdot}b\in O$ (this comes from applying Lagrange's theorem to the abelian group $L/O$), and therefore $a/b=(ka)/(kb)\in\operatorname{Frac}O$.

Answer 2

For any element $k = \frac{x}{y}\in K$ where $x, y \in L$, we have $k = \frac{dx}{dy}$ where $d = [L:O]$ is the index. And $dx$ and $dy$ are elements of $O$.

Answer 3

Let $m$ be the index of $O$ in $L$. Then for $x \in L$, $mx \in O$.

Now let $x/y \in K$. $x/y=mx/my$, so $K \subseteq \operatorname{Frac}(O)$.