Is $\text{Frac}(O_K)=K$?

Question

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

Answer 1

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 $x = p^{-e}\cdot(p^ex) \in \operatorname{Frac}(O_K)$.

Answer 2

Let $x\in K$. Then we have $x\in\mathcal{O}$, if $|x|\leq 1$. Otherwise $x^{-1}\in\mathcal{O}_{K}$, since $|x^{-1}| = |x|^{-1} < 1$ for $|x| > 1$. So $x\in\mathrm{Frac}(\mathcal{O}_{K})$ and hence $K\subset\mathrm{Frac}(\mathcal{O}_{K})$.

Answer 3

You've been given answers using the $p$-adic valuation, but this can also be thought of as a general fact of integral closures. I claim the following:

Let $A$ be a domain with quotient field $F$. Let $K/F$ be an algebraic extension and let $B$ be the integral closure of $A$ in $K$. Then $K$ is the quotient field of $B$.

Now, $K/\mathbb Q_p$ is finite therefore algebraic, so the above applies to $A = \mathbb Z_p$ and $B = \mathcal O_K$.

So let's prove the general statement. It's clear that the quotient field of $B$ is contained in $K$, so we need only show the converse. Let $a \in K$. Since $K/F$ is algebraic, let $f \in F[x]$ be the (monic) minimal polynomial, and say its degree is $n$. We can find some nonzero $d \in A$ such that $df \in A[x]$. For instance, $d$ could be the product of all the denominators of the coefficients of $f$. Now, consider the polynomial $d^n f(d^{-1} x)$. If $f = \sum_{i=1}^n a_i x^i$ with $a_n = 1$ then $d^n f(d^{-1} x)$ is $\sum_{i=1}^n d^{n - i} a_i x^i$, which is a monic polynomial in $A[x]$. Furthermore, evaluating $d^n f(d^{-1} x)$ at $da$ yields $d^n f(a) = 0$. As $d^n f(d^{-1} x) \in A[x]$ is monic, $da \in K$ is therefore integral over $A$. Then as $B$ is defined as the integral closure of $A$ in $K$, we must have $da \in B$. Hence, $a = b/d$ for some $b \in B$. Of course $b/d$ is in the quotient field of $B$ so we have shown that $K$ is contained in the quotient field of $B$. Hence, they are equal. In fact, we've shown the stronger fact that the denominators in $K$ can be taken to be in $A$, i.e. that $K$ is the localization $(A - 0)^{-1} B$.

As mentioned above, this applies directly to $A = \mathbb Z_p$, $F = \mathbb Q_p$ and the finite extension $K/\mathbb Q_p$. In that case, $B = \mathcal O_K$ and we conclude that the quotient field of $\mathcal O_K$ is $K$.