Proof of Existence of Complete Field with Dense Subfield

Question

First, the question at hand:

Prove there exists a field $E$ which contains $F$ as a subfield, such that E has an absolute value extending the absolute value on $F$, such that $F$ is dense in $E$, and $E$ is complete.

Now here we use the field definition for an absolute value and we say a field is complete if every Cauchy sequence in $F$ defined on this absolute value converges to some value in $F$. Furthermore, we define $F$ to be dense in $E$ if given $\epsilon>0$, and an element $\alpha\in$E, there exists $a\in F$ such that $|\alpha-a|<\epsilon$.

I already know how to prove that for a field $F$ with an absolute value defined on it, the factor(quotient) ring of Cauchy sequences mod the null sequences is a field $E$ itself, and that the absolute value of $F$ can be extended to this field $E$, and indeed that this field $E$ is complete. Does this help in the question?

Thank you very much.

Answer 1

There are two main examples. First, $(\mathbb{Q},|\cdot|_{\infty})$ together with the Archimedean absolute value. It is not complete with respect to the induced metric. The completion of this valued field is $(\mathbb{R},|\cdot|_{\infty})$. Of course, $\mathbb{Q}$ is dense in $\mathbb{R}$. Secondly, $(\mathbb{Q},|\cdot|_{p})$ together with the non-Archimedean $p$-adic absolute value. The completion of this valued field is $(\mathbb{Q}_p,|\cdot|_{p})$. Again, $\mathbb{Q}$ is dense in $\mathbb{Q}_p$.

Answer 2

Hint: Take $E = \mathbb{R}$ and $F = \mathbb{Q}$, with the typical absolute value. It is up to you to show that this is such an example.

Remark: More generally one can take the "completion" of a field with an absolute value equipped, which would have all of these desired properties more or less by definition. As Dietrich Burde notes, $\mathbb{Q}_p$, the field of $p$-adic numbers, is the "completion" of $\mathbb{Q}$ with the $p$-adic absolute value.