Let $k$ be a field, and let $| \ |$ be a norm on $k$. The norm induces a metric. To construct the completion $\hat{k}$ as a normed field, the standard recipe is to take the quotient of the ring $\mathcal{C}(k)$ of all Cauchy sequences in $k$ -- viewed as a subring of $k^{\infty} = \prod\_{i=1}^{\infty} k$ -- by the maximal ideal $\mathfrak{m}\_0$ of all sequences converging to $0$.


This brings up the following [idle] question: what are the maximal ideals of $\mathcal{C}(k)$? The prime ideals?


My vague recollection had been that $\mathfrak{m}\_0$ was the unique maximal ideal of $\mathcal{C}(k)$, but this is evidently not the case: for every $n$, there is a maximal ideal $\mathfrak{m}\_n$ consisting of sequences whose $n$th coordinate is $0$, the residue field being $k$ again. It is easy to see though that $\mathfrak{m}\_0$ is the unique maximal ideal containing the ~~prime~~ ideal $\mathfrak{c} = \bigoplus\_{i=1}^{\infty} k$. (Edit: $\mathfrak{c}$ is prime iff the norm is trivial.)


Now this reminds me of filters. The prime ideals of the (zero-dimensional) ring $k^{\infty}$ correspond precisely to the ultrafilters on $\mathbb{Z}^+$. The principal ultrafilter of all sets containing $n$ pulls back to the maximal ideal $\mathfrak{m}\_n$. Since every nonprincipal ultrafilter contains the Frechet filter of cofinite sets, it follows that it pulls back to $\mathfrak{m}\_0$. But is it true that every maximal ideal of $\mathcal{C}(k)$ is pulled back from a prime (= maximal) ideal of $k^{\infty}$? If so, is this an instance of a general theorem? 




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Addendum: 


Note that in the case that the norm is trivial -- so that the induced metric is the discrete metric -- a sequence converges iff it is eventually constant, so a sequence converges to $0$ iff it has only finitely many nonzero terms: $\mathfrak{c} = \mathfrak{m}\_0$. The converse also holds: for any nontrivial norm there exist nowhere zero sequences converging to $0$, e.g. $\{x^n\}$ for any $x \in k$ with $0 < |x| < 1$.




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Once the original question is worked out, I am also curious about generalizations. What is the analogue for the ring of minimal Cauchy filters in an arbitrary topological ring? 



What is the prime spectrum of a Cauchy series ring?
In this answer I will treat the case in which $|\text{ }|$ is not discrete.


I first claim that $\mathfrak m\_0$ is not the restriction of any proper ideal in
$k^{\infty}.$ Indeed, choose $x \in k$ such that $0 < |x| < 1$. Then $(x^i)$
is an element of $\mathfrak m\_0$ which is invertible in $k^{\infty}$ (with
inverse equal to $(x^{-i})$, and so $\mathfrak m\_0$ generates the unit ideal
of $k^{\infty}$. 


This doesn't contradict anything; the maximal ideals of $k^{\infty}$ pull-back
to prime ideals in $\mathcal C(k)$ which are simply not maximal (as often happens
with maps of rings).


Furthermore, this pull-back is injective.


To see this, we first introduce some notation; namely, we
let $\mathfrak m\\_{\mathcal U}$ denote the prime ideal of $k^{\infty}$ corresponding to
the non-principal ultra-filter ${\mathcal U}$,and recall that $\mathfrak m\\_{\mathcal U}$ is defined as follows: an element $(x\_i)$ lies in $\mathfrak m\\_{\mathcal U}$ if and only
if $\{i \, | \, x\_i = 0\}$ lies in in $\mathcal U$.


Now suppose that $\mathcal U\_1$ and $\mathcal U\_2$ are two distinct non-principal ultra-filters. Let $A$ be a set lying in $\mathcal U\_1$, but not in $\mathcal U\_2$.
Then $A^c$, the complement of $A$, lies in $\mathcal U\_2$.  

Choose $x \in k$ such $0 < | x | < 1,$ and let $x\_i = x^i$ if $i \in A$ and
$x\_i = 0$ if $i \not\in A$. Then $(x\_i)$ is an element of $\mathcal C(k)$,
in fact of $\mathfrak m\_0$, and it lies in $\mathfrak m\\_{\mathcal U\_2}$
but not in $\mathfrak m\\_{\mathcal U\_1}$.


Thus $\mathfrak m\\_{\mathcal U\_1}$ and $\mathfrak m\\_{\mathcal U\_2}$ have distinct pull-backs. 


So the map 
Spec $k^{\infty} \rightarrow $ Spec $\mathcal C(k)$
is injective
and dominant (since it comes from an injective map of rings), but is not surjective.
Choosing the valuation $|\text{ }|$ allows us to add to Spec $k^{\infty}$ (which is the
Stone-Cech compactification of $\mathbb Z\\_+$) an extra point dominating all the
other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives
us a definitive way to compute limits (provided we begin with a Cauchy sequence).


