Extend an embedding that preserves absolute values

Question

Assume that ($K$,| |) is a field with absolute value that is dense in a complete field $K'$. And assume that $f:K\to L$ is an embedding that preserves absolute value to a complete field $L$. Show that we can extend $f$ to an embedding $f':K'\to L$ which preserves absolue values as well.

I know the completion theorem that states: If ($K$,| |) is a field with absolute value then there exists a field $K'$ such that ($K'$,| |) is complete and there exists an embedding $f:K\to K'$ that preserves absolue values such that $f(K)$ is dense in $K'$.

Can I use this theorem to get that there is an embedding $h:K\to K'$ that preserves absolue values and $h(K)$ is dense in $K'$. Bue then how to finish this? Can we take $f'=f\circ h^{-1}$.

Thank you for any help.

A theorem saying a field $K'$ exists doesn't help since $K'$ is given in the problem assumption. You don't need a mapping $K \to K'$ since the assumption $K$ "is dense in" $K'$ implies $K \subseteq K'$. — aschepler, Nov 18 '22 at 18:14
Regarding the bounty: Eric Wofsey's answer is indeed the canonical answer, and pretty much the only answer, up to your prefered definition of convergence. — SomeCallMeTim, Nov 28 '22 at 12:44

score 1 · Accepted Answer · answered Nov 23 '22 at 02:03

The idea is just to extend $f$ by continuity. Since $K$ is dense in $K'$, for each $x\in K'$, there is a sequence $(x_n)$ in $K$ which converges to $x$. Since $f$ preserves the absolute value, the sequence $(f(x_n))$ will still be Cauchy, so by completeness it converges to some element $y\in L$. Define $f'(x)$ to be this element $y$.

Now you just have to verify that this $f'$ is also a homomorphism which preserves the absolute value. This is somewhat awkward from the definition above since it uses a specific choice of sequence for each $x\in K'$, so for instance, when comparing $f'(x+y)$ and $f'(x)+f'(y)$, you don't know that the chosen sequence converging to $x+y$ is the same as the sum of the chosen sequences converging to $x$ and $y$. To get around this, you can first prove that in fact the limit of $(f(x_n))$ is the same for all possible sequences $(x_n)$ that converge to $x$, so the choice of sequence does not matter.

Extend an embedding that preserves absolute values

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