Proof that $\mathbb Q_p$ is unique up to unique isomorphism preserving the absolute values

Question

On pages 58-59 of Gouvea's $p-$adic Numbers: An Introduction, he gives the following proof that the field $\mathbb Q_p$, constructed using equivalence classes of Cauchy sequences, is unique up to unique isomorphism preserving the absolute values. I'll type out the full theorem and proof he gives; I've been able to double-check everything except his claim that the isomorphism he constructed is unique "by construction."

Theorem 3.2.13 For each prime $p\in\mathbb Z$ there exists a field $\mathbb Q_p$ with a non-archimedean absolute value $|\ |_p$, such that:

1. there exists an inclusion $\mathbb Q\hookrightarrow \mathbb Q_p$, and the absolute value induced by $|\ |_p$ on $\mathbb Q$ via this inclusion is the $p-$adic absolute value;

2. the image of $\mathbb Q$ under this inclusion is dense in $\mathbb Q_p$ w.r.t. $|\ |_p$; and

3. $\mathbb Q_p$ is complete w.r.t. $|\ |_p$.

The field $\mathbb Q_p$ satisfying $(i)$, $(ii)$, and $(iii)$ is unique up to unique isomorphism preserving the absolute values.

PROOF: We've done it all except the uniqueness statement. To get that, suppose we have another such field $K$. Then we can think of the inclusion $\mathbb Q\hookrightarrow K$ as a map defined on a dense subset of $\mathbb Q_p$. Since this map has to preserve the absolute values of any element of $\mathbb Q$, it is continuous. Now, any map defined on a dense subset which is continuous can be extended uniquely to the whole field, so that we get a map $\mathbb Q_p\rightarrow K$ which is the unique continuous extension of the inclusion of $\mathbb Q$ in $K$. It is now easy to check that it is an isomorphism that preserves the absolute values, and its uniqueness is clear by construction.

I'm not following his claim that the map's uniqueness is clear by construction. We have to show that any other absolute value-preserving isomorphism $F:\mathbb Q\rightarrow K$ is equivalent to the isomorphism constructed above. I see that $F$ has to be continuous since it preserves absolute values, and if we knew that $F$ extended the inclusion $\mathbb Q\hookrightarrow K$ then we would be done, but I don't see why $F$ has to extend this inclusion. Please advise.

Answer 1

For a field $K$ of characteristic zero, there is only one morphism $\mathbb{Q}\to K$.

The basic claim here is that this morphism extends like so: $\mathbb{Q}\to\mathbb{Q}_p \to K$. This requires the use of properties 1 and 3. Uniqueness is immediate from property 2.

If $K$ has all the same properties, then we have also shown that $\mathbb{Q}\to\mathbb{Q}_p$ extends to $\mathbb{Q}\to K \to \mathbb{Q}_p$. But this shows precisely that the two maps from $\mathbb{Q}$ to $\mathbb{Q}_p$ and $K$ are isomorphic.