Is $\text{Frac}(O_K)=K$? Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $O_K$.
We know that $\mathbb{Q}_p$ is the field of fractions of $p$-adic integers $\mathbb{Z}_p$ i.e., $$\mathbb{Q}_p=\text{Frac}(\mathbb{Z}_p).$$
Similarly, $O_K$ is also an integral domain with no zero divisor. So we can take field of fractions of $O_K$.
My question:
Is $\text{Frac}(O_K)=K$ ?
I believe so.
any help please

I assume that you know

*

*that the $p$-adic absolute value extends uniquely to an absolute value $|\cdot|_p$ on $K$;

*that the ring of integers $O_K$ is nothing but the elements $x\in K$ such that $|x|_p \leq 1$.

Then it is clear that $\operatorname{Frac}(O_K) = K$, as for any element $x \in K$, there exists a sufficiently large integer $e$ such that $|p^e x|_p \leq 1$ and hence $x = p^{-e}\cdot(p^ex) \in \operatorname{Frac}(O_K)$.
