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

What is the prime spectrum of a Cauchy series ring?

2010-01-17T04:15:30Z

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?

Answer by Emerton for What is the prime spectrum of a Cauchy series ring?

2010-01-17T05:35:09Z

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.

Answer by Emerton for What is the prime spectrum of a Cauchy series ring?

2010-01-22T04:55:34Z

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).