[proofplan]
We prove both existence and uniqueness of the completion of a valued field $(K, |\cdot|)$. For existence, we construct $\hat{K}$ as the quotient of the ring of Cauchy sequences in $K$ by the maximal ideal of null sequences. We verify the field structure, define the extended absolute value as the limit of absolute values along a Cauchy sequence (well-defined by the reverse triangle inequality), embed $K$ via constant sequences, and prove completeness using a diagonal approximation argument. For uniqueness, we show that any two completions are connected by a unique isometric field isomorphism, using the density of the embedded copy of $K$ and the continuity of the field operations to extend the identity map on $K$.
[/proofplan]
[step:Define the ring of Cauchy sequences and the ideal of null sequences]
Let $\mathcal{C}$ denote the set of all Cauchy sequences in $(K, |\cdot|)$:
\begin{align*}
\mathcal{C} := \bigl\{(x_n)_{n \geq 1} \subset K : \text{for all } \varepsilon > 0, \text{ there exists } N \text{ such that } |x_m - x_n| < \varepsilon \text{ for all } m, n \geq N\bigr\}.
\end{align*}
Define addition and multiplication termwise: $(x_n) + (y_n) := (x_n + y_n)$ and $(x_n) \cdot (y_n) := (x_n y_n)$. These are Cauchy sequences because:
- $|(x_m + y_m) - (x_n + y_n)| \leq |x_m - x_n| + |y_m - y_n|$, so the sum is Cauchy.
- $|x_m y_m - x_n y_n| = |x_m(y_m - y_n) + (x_m - x_n)y_n| \leq |x_m||y_m - y_n| + |x_m - x_n||y_n|$. Since Cauchy sequences are bounded (there exists $M$ with $|x_n|, |y_n| \leq M$ for all $n$), the product is Cauchy.
The zero element is $(0, 0, \ldots)$ and the identity element is $(1, 1, \ldots)$. So $\mathcal{C}$ is a commutative ring with identity.
Let $\mathcal{N} := \{(x_n) \in \mathcal{C} : |x_n| \to 0\}$ be the set of null sequences. This is an ideal: if $(x_n) \in \mathcal{N}$ and $(y_n) \in \mathcal{C}$, then $|x_n y_n| = |x_n||y_n| \leq M|x_n| \to 0$, so $(x_n y_n) \in \mathcal{N}$.
[/step]
[step:Verify that $\mathcal{N}$ is a maximal ideal and $\hat{K} := \mathcal{C}/\mathcal{N}$ is a field]
We show $\mathcal{N}$ is maximal by proving $\mathcal{C}/\mathcal{N}$ is a field. Let $[(x_n)] \in \mathcal{C}/\mathcal{N}$ be non-zero, meaning $(x_n) \notin \mathcal{N}$, i.e., $|x_n| \not\to 0$.
Since $(x_n)$ is Cauchy, the sequence $(|x_n|)$ is Cauchy in $\mathbb{R}$ (by the reverse triangle inequality $\bigl||x_m| - |x_n|\bigr| \leq |x_m - x_n|$) and therefore converges to some $L \geq 0$. Since $(x_n) \notin \mathcal{N}$, we have $L > 0$.
Choose $N$ such that $|x_n| > L/2 > 0$ for all $n \geq N$. Define $y_n = x_n^{-1}$ for $n \geq N$ and $y_n = 0$ for $n < N$. The sequence $(y_n)$ is Cauchy for $n \geq N$:
\begin{align*}
|y_m - y_n| = |x_m^{-1} - x_n^{-1}| = \frac{|x_m - x_n|}{|x_m||x_n|} \leq \frac{4}{L^2} |x_m - x_n| \to 0.
\end{align*}
The product $(x_n)(y_n)$ satisfies $x_n y_n = 1$ for $n \geq N$, so $(x_n y_n) - (1, 1, \ldots)$ is eventually zero, hence a null sequence. Therefore $[(x_n)] \cdot [(y_n)] = [(1, 1, \ldots)]$, and $[(x_n)]$ is invertible in $\mathcal{C}/\mathcal{N}$.
[guided]
The construction of the multiplicative inverse is the only step where the field structure of $K$ is used (as opposed to just the ring structure). The key issue is that individual terms $x_n$ might be zero, but only finitely many can be: since $|x_n| \to L > 0$, eventually $x_n \neq 0$.
We handle the finitely many zero terms by redefining $y_n = 0$ for $n < N$. This changes the sequence by a null sequence (the difference from the "correct" inverse is supported on finitely many indices, hence tends to zero), so the equivalence class $[(y_n)]$ is independent of the choice of $N$.
The Cauchy estimate $|y_m - y_n| \leq \frac{4}{L^2}|x_m - x_n|$ uses the bound $|x_n| > L/2$ for $n \geq N$, which gives $|x_m|^{-1}|x_n|^{-1} < (L/2)^{-2} = 4/L^2$. This uniform lower bound on $|x_n|$ is crucial: without it, the denominators could blow up and the inverse sequence might not be Cauchy.
[/guided]
[/step]
[step:Define the absolute value on $\hat{K}$ and verify it is well-defined]
For $[(x_n)] \in \hat{K}$, define
\begin{align*}
\bigl|[(x_n)]\bigr|_{\hat{K}} := \lim_{n \to \infty} |x_n|.
\end{align*}
This limit exists in $\mathbb{R}$ because $(|x_n|)$ is a Cauchy sequence in $\mathbb{R}$ (by the reverse triangle inequality) and $\mathbb{R}$ is complete.
**Well-definedness:** If $(x_n) \sim (y_n)$, meaning $(x_n - y_n) \in \mathcal{N}$, then
\begin{align*}
\bigl||x_n| - |y_n|\bigr| \leq |x_n - y_n| \to 0,
\end{align*}
so $\lim |x_n| = \lim |y_n|$.
**Absolute value axioms:**
1. $|[(x_n)]|_{\hat{K}} = 0$ iff $\lim |x_n| = 0$ iff $(x_n) \in \mathcal{N}$ iff $[(x_n)] = 0$.
2. $|[(x_n)] \cdot [(y_n)]|_{\hat{K}} = \lim |x_n y_n| = \lim |x_n| \cdot \lim |y_n| = |[(x_n)]|_{\hat{K}} \cdot |[(y_n)]|_{\hat{K}}$.
3. $|[(x_n)] + [(y_n)]|_{\hat{K}} = \lim |x_n + y_n| \leq \lim (|x_n| + |y_n|) = |[(x_n)]|_{\hat{K}} + |[(y_n)]|_{\hat{K}}$.
If the original absolute value is non-archimedean (satisfies the strong triangle inequality), then $|x_n + y_n| \leq \max(|x_n|, |y_n|)$, and taking limits preserves this: $|[(x_n)] + [(y_n)]|_{\hat{K}} \leq \max(|[(x_n)]|_{\hat{K}}, |[(y_n)]|_{\hat{K}})$. So the extended absolute value is non-archimedean if the original was.
[/step]
[step:Embed $K$ isometrically via constant sequences and verify density]
Define the embedding
\begin{align*}
\iota: K &\to \hat{K} \\
x &\mapsto [(x, x, x, \ldots)].
\end{align*}
This is a field homomorphism (termwise operations on constant sequences reproduce the field operations in $K$) and is isometric: $|\iota(x)|_{\hat{K}} = \lim_{n \to \infty} |x| = |x|$.
To show $\iota(K)$ is dense in $\hat{K}$: let $[(x_n)] \in \hat{K}$ and $\varepsilon > 0$. Choose $N$ such that $|x_m - x_n| < \varepsilon$ for all $m, n \geq N$. Then
\begin{align*}
\bigl|[(x_n)] - \iota(x_N)\bigr|_{\hat{K}} = \lim_{n \to \infty} |x_n - x_N| \leq \varepsilon.
\end{align*}
So every element of $\hat{K}$ is within $\varepsilon$ of $\iota(K)$.
[/step]
[step:Prove that $\hat{K}$ is complete]
Let $(z_k)_{k \geq 1}$ be a Cauchy sequence in $\hat{K}$. We must find $z \in \hat{K}$ with $|z_k - z|_{\hat{K}} \to 0$.
Since $\iota(K)$ is dense in $\hat{K}$, for each $k$ choose $y_k \in K$ with $|z_k - \iota(y_k)|_{\hat{K}} < 1/k$. We claim $(y_k)$ is a Cauchy sequence in $K$. For $j, k \geq 1$:
\begin{align*}
|y_j - y_k| = |\iota(y_j) - \iota(y_k)|_{\hat{K}} \leq |\iota(y_j) - z_j|_{\hat{K}} + |z_j - z_k|_{\hat{K}} + |z_k - \iota(y_k)|_{\hat{K}} < \frac{1}{j} + |z_j - z_k|_{\hat{K}} + \frac{1}{k}.
\end{align*}
Given $\varepsilon > 0$, choose $N_1$ with $|z_j - z_k|_{\hat{K}} < \varepsilon/3$ for $j, k \geq N_1$, and $N_2$ with $1/k < \varepsilon/3$ for $k \geq N_2$. For $j, k \geq \max(N_1, N_2)$, $|y_j - y_k| < \varepsilon$. So $(y_k)$ is Cauchy in $K$.
Set $z := [(y_k)] \in \hat{K}$. Then
\begin{align*}
|z_k - z|_{\hat{K}} \leq |z_k - \iota(y_k)|_{\hat{K}} + |\iota(y_k) - z|_{\hat{K}} < \frac{1}{k} + |\iota(y_k) - z|_{\hat{K}}.
\end{align*}
Now $|\iota(y_k) - z|_{\hat{K}} = |[(y_k, y_k, \ldots)] - [(y_1, y_2, \ldots)]|_{\hat{K}} = \lim_{n \to \infty} |y_k - y_n|$. Since $(y_n)$ is Cauchy, for any $\varepsilon > 0$ there exists $N$ with $|y_k - y_n| < \varepsilon$ for $k, n \geq N$, giving $\lim_{n \to \infty} |y_k - y_n| \leq \varepsilon$ for $k \geq N$. So $|\iota(y_k) - z|_{\hat{K}} \to 0$, and therefore $|z_k - z|_{\hat{K}} \to 0$.
[/step]
[step:Prove uniqueness up to unique isometric isomorphism]
Let $(\hat{K}, \iota)$ and $(\hat{K}', \iota')$ be two completions of $(K, |\cdot|)$. We construct the unique isometric isomorphism $\phi: \hat{K} \to \hat{K}'$ with $\phi \circ \iota = \iota'$.
**Construction:** Every element of $\hat{K}$ is the limit of a sequence in $\iota(K)$ (by density). For $z \in \hat{K}$, choose a sequence $(x_n)$ in $K$ with $\iota(x_n) \to z$. Since $\iota$ is isometric, $(x_n)$ is Cauchy in $K$. Since $\iota'$ is also isometric, $(\iota'(x_n))$ is Cauchy in $\hat{K}'$ (with the same distances), and converges to some element $z' \in \hat{K}'$ (by completeness of $\hat{K}'$). Define $\phi(z) := z'$.
**Well-definedness:** If $(x_n)$ and $(y_n)$ are two sequences in $K$ with $\iota(x_n), \iota(y_n) \to z$, then $|x_n - y_n| = |\iota(x_n) - \iota(y_n)|_{\hat{K}} \to 0$, so $|\iota'(x_n) - \iota'(y_n)|_{\hat{K}'} = |x_n - y_n| \to 0$, giving $\lim \iota'(x_n) = \lim \iota'(y_n)$.
**Isometry:** $|\phi(z)|_{\hat{K}'} = \lim |\iota'(x_n)|_{\hat{K}'} = \lim |x_n| = \lim |\iota(x_n)|_{\hat{K}} = |z|_{\hat{K}}$.
**Ring homomorphism:** For $z, w \in \hat{K}$ with approximating sequences $(x_n)$ and $(y_n)$:
\begin{align*}
\phi(z + w) &= \lim \iota'(x_n + y_n) = \lim \iota'(x_n) + \lim \iota'(y_n) = \phi(z) + \phi(w), \\
\phi(zw) &= \lim \iota'(x_n y_n) = \lim \iota'(x_n) \cdot \lim \iota'(y_n) = \phi(z)\phi(w).
\end{align*}
**Surjectivity:** The image $\phi(\hat{K})$ contains $\iota'(K)$ (since $\phi(\iota(x)) = \iota'(x)$) and is complete (as the isometric image of a complete space). Since $\iota'(K)$ is dense in $\hat{K}'$ and $\phi(\hat{K})$ is a complete (hence closed) subset containing $\iota'(K)$, we get $\phi(\hat{K}) = \hat{K}'$.
**Uniqueness:** Any isometric field isomorphism $\psi: \hat{K} \to \hat{K}'$ with $\psi \circ \iota = \iota'$ must satisfy $\psi(\iota(x)) = \iota'(x)$ for all $x \in K$. Since $\iota(K)$ is dense in $\hat{K}$ and $\psi$ is continuous (being isometric), $\psi$ is determined on all of $\hat{K}$ by its values on $\iota(K)$. So $\psi = \phi$.
[/step]
