What is the prime spectrum of a Cauchy series ring? 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? 

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 $0 < |x| < 1$.

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?  
 A: Suppose that we have an ideal $I$ of ${\mathcal C}(k)$ that is not contained in any one of the $\mathfrak m\_n.$  Then for each $n$, the ideal $I$ contains an element
$(k_i)$ with $k_n \neq 0.$ Multiplying by the element $k_n^{-1} \underline{e}_n$,
where $\underline{e}_n$ denotes the element which is $0$ in every coordinate
except $n$, we find that $\underline{e}_n \in I$ for every $n$.  Thus $I$
contains $\mathfrak c$.  
In particular, if $I$ is maximal, then it coincides with $\mathfrak m_0$
(by your characterization of $\mathfrak m_0$ as the unique maximal ieal containing
$\mathfrak p$).
So, barring a blunder, this seems to show that the maximal ideals of $\mathcal C(k)$
are precisely the $\mathfrak m_n$ and $\mathfrak m_0$.
The quotient $\mathcal C(k)/\mathfrak c$ is a local ring (by the uniquess of $\mathfrak m_0$),
but I don't see anything else to say about it yet.
A: 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).
