What is the prime spectrum of a Cauchy series ring? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:55:31Z http://mathoverflow.net/feeds/question/12072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12072/what-is-the-prime-spectrum-of-a-cauchy-series-ring What is the prime spectrum of a Cauchy series ring? Pete L. Clark 2010-01-17T04:15:30Z 2011-11-30T09:28:50Z <p>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 $0$.</p> <p>This brings up the following [idle] question: what are the maximal ideals of $\mathcal{C}(k)$? The prime ideals?</p> <p>My vague recollection had been that <code>$\mathfrak{m}_0$</code> was the unique maximal ideal of $\mathcal{C}(k)$, but this is evidently not the case: for every $n$, there is a maximal ideal <code>$\mathfrak{m}_n$</code> consisting of sequences whose $n$th coordinate is $0$, the residue field being $k$ again. It is easy to see though that <code>$\mathfrak{m}_0$</code> is the unique maximal ideal containing the <s>prime</s> ideal $\mathfrak{c} = \bigoplus_{i=1}^{\infty} k$. (Edit: $\mathfrak{c}$ is prime iff the norm is trivial.)</p> <p>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 <code>$\mathfrak{m}_n$</code>. 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? </p> <p><hr /></p> <p>Addendum: </p> <p>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. <code>$\{x^n\}$</code> for any $x \in k$ with $0 &lt; |x| &lt; 1$.</p> <p><hr /></p> <p>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? </p> http://mathoverflow.net/questions/12072/what-is-the-prime-spectrum-of-a-cauchy-series-ring/12076#12076 Answer by Emerton for What is the prime spectrum of a Cauchy series ring? Emerton 2010-01-17T05:35:09Z 2010-01-17T05:35:09Z <p>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 $\mathfrak c$. </p> <p>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 $\mathfrak p$).</p> <p>So, barring a blunder, this seems to show that the maximal ideals of $\mathcal C(k)$ are precisely the $\mathfrak m_n$ and $\mathfrak m_0$.</p> <p>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.</p> http://mathoverflow.net/questions/12072/what-is-the-prime-spectrum-of-a-cauchy-series-ring/12607#12607 Answer by Emerton for What is the prime spectrum of a Cauchy series ring? Emerton 2010-01-22T04:55:34Z 2011-11-30T09:28:50Z <p>In this answer I will treat the case in which $|\text{ }|$ is not discrete.</p> <p>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 $0 &lt; |x| &lt; 1$. 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 $k^{\infty}$. </p> <p>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).</p> <p>Furthermore, this pull-back is injective.</p> <p>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 $\mathcal U$.</p> <p>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 $\mathcal U_2$.<br> Choose $x \in k$ such $0 &lt; | x | &lt; 1,$ 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 $\mathfrak m_{\mathcal U_1}$.</p> <p>Thus $\mathfrak m_{\mathcal U_1}$ and $\mathfrak m_{\mathcal U_2}$ have distinct pull-backs. </p> <p>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).</p>