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In studying the papers of Abbes and Saito on ramification theory in the imperfect residue field case, I come to the following questions. Let $\overline{\mathbb{Q}}_p\supset\mathbb{Q}{}^{t}_p\supset \mathbb{Q}{}^{nr}_p$ be an algebraic closure of the field $\mathbb{Q}_p$ of $p$-adics with its maximal tamely ramified and unramified subextensions. Let $\overline{\mathbb{Z}}_p\supset\mathbb{Z}{}^{t}_p\supset \mathbb{Z}{}^{nr}_p$ be the rings of integers of these fields. My question is:

What do the quotient rings $\overline{\mathbb{Z}}_p/(p)$, $\mathbb{Z}{}^{t}_p/(p)$ and $\mathbb{Z}{}^{nr}_p/(p)$ look like?

In fact I'm interested in all quotients $\overline{\mathbb{Z}}_p/(\pi_a)$, $\mathbb{Z}{}^{t}_p/(\pi_a)$ and $\mathbb{Z}{}^{nr}_p/(\pi_a)$ for an element $\pi_a$ of given rational valuation $a$, but I guess that the mod $p$ quotients are more tractable and give a first idea.

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Since $(p)$ is the generator of the maximal ideal of ${\bf Z}_p^{\rm nr}$, you have ${\bf Z}_p^{\rm nr}/(p)=\bar{\bf F}_p$. –  Damian Rössler Apr 5 '13 at 10:34
    
Trivial observation : since $x_n=\root n\of p$ is in what you call $\mathbf{Z}_p^t$, its image $\bar x_n\in\mathbf{Z}_p^t/(p)$ will satisfy ${\bar x_n}^m=0$ for $m=n$ and no smaller $m$. –  Chandan Singh Dalawat Apr 5 '13 at 10:55
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... for $n>0$ prime to $p$. Moreover, $\mathbf{Z}_p^t$ is generated as a $\mathbf{Z}_p^{nr}$-algebra by the family of all such $x_n$, –  Chandan Singh Dalawat Apr 5 '13 at 11:10
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More trivial observations: $\bar{\mathbb{Z}}_p/(p) \simeq \mathcal{O}_{\mathbb{C}_p}/(p) \simeq \mathcal{O}_{\mathbb{C}_p}/(\pi)$ for every $\pi$ with $p$-adic valuation in $p^\mathbb{Z}$, the first iso induced by completion and the second one by powers of Frobenius. It's zero-dimensional, local, non-Noetherian, the maximal ideal is idempotent and nil, the residue field is $\bar{\mathbb{F}}_p$. –  Torsten Schoeneberg Apr 5 '13 at 12:53
    
Thanks to all three! I more or less saw that $\mathbb{Z}_p^t/(p)$ is local with nilpotent maximal ideal, but I couldn't go further and I guess that one can't say much more than that. If the comments are turned into an answer, I'll accept it. –  Matthieu Romagny Apr 5 '13 at 18:18

2 Answers 2

up vote 0 down vote accepted

This is an extended comment.

Let $\Gamma=\mathrm{Gal}(\overline{\mathbf{Q}}_p|\mathbf{Q}_p)$. Your question pertains to the filtration $V\subset T\subset\Gamma$, where $T$ is the inertia subgroup and $V$ the ramification subgroup, so that $\overline{\mathbf{Q}}_p^T$ is the maximal unramified extension of $\mathbf{Q}_p$ and $\overline{\mathbf{Q}}_p^V$ is the maximal tamely ramified extension of $\mathbf{Q}_p$. The quotient $\Gamma/T$ is $\hat{\mathbf{Z}}$, and the quotient $T/V$ can be identified with the group of roots of $1$ of order prime to $p$.

But the filtration (in the upper numbering) on $\Gamma$ is much finer, and it is indexed by the reals $\geqslant-1$; we have $\Gamma^0=T$ and $\Gamma^{0+}=V$. This filtration is separated in the sense that the intersection of all $\Gamma^u$ (for $u\geqslant-1$) is trivial, so $\overline{\mathbf{Q}}_p$ is the inductive limit of the various $K^u=\overline{\mathbf{Q}}_p^{\Gamma^u}$. As a result the $\overline{\mathbf{F}}_p$-algebra $\overline{\mathbf{Z}}_p/p\overline{\mathbf{Z}}_p$ is the inductive limit (increasing union) of $\mathfrak{o}^u/p\mathfrak{o}^u$, where $\mathfrak{o}^u$ is the ring of integers of $K^u$, and $u>0$

It might be interesting to be able to say something sensible about each individual quotient $\mathfrak{o}^u/p\mathfrak{o}^u$ (and also about the quotients corresponding to $\Gamma^{u+}$, the union of all $\Gamma^v$ such that $v>u$).

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This was answered in the comments, so I'll just sum up:

$(p)$ is the maximal ideal in $\mathbf{Z}_p^{nr}$, so we have an isomorphism $\mathbf{Z}_p^{nr}/(p) \to \overline{\mathbf{F}_p}$. The reverse isomorphism is given by the Teichmüller lift (a multiplicative monoid map to $\mathbf{Z}_p^{nr}$).

$\mathbf{Z}_p^t \cong \mathbf{Z}_p^{nr}[\{p^{1/n}\}_{(n,p)=1}]$, so $\mathbf{Z}_p^t/(p)$ is isomorphic to $\overline{\mathbf{F}_p}[\{x_n\}_{(n,p)=1}]/(x_1, \{x_{nk}^k-x_n\}_{(n,p)=1,k>1})$. This is a local non-reduced non-Noetherian ring. Its nilradical, generated by all $x_n$, is maximal, idempotent, and nil, but not nilpotent. Any finite set of elements is contained in a principal ideal.

I do not know an explicit structure for $\overline{\mathbf{Z}_p}/(p)$, but it has basically the same general ring-theoretic properties as the previous case (except for the fact that every element is a $p$th power).

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The Fontaine-Wintenberger theorem identifies $\overline{\mathbf{Z}_p}/p$ with its characteristic $p$ counterpart, i.e., with $R/t$, where $R$ is the ring of integers in an algebraic closure of $\mathbf{F}_p ((t))$. –  anon Apr 6 '13 at 5:09
    
@S. Carnahan: Thanks! And I guess that in your presentation of $\mathbf{Z}_p^t/(p)$, the integer $k$ must be prime to $p$ also. –  Matthieu Romagny Apr 6 '13 at 11:32
    
@anon: you're right, I knew that but somehow it clicked only after you pointed it out. Thank you! –  Matthieu Romagny Apr 6 '13 at 11:35
    
@Matthieu: since $n$ and $nk$ are both prime to $p$... –  Chandan Singh Dalawat Apr 6 '13 at 11:40
    
@Chandan Singh: of course you're right, I just meant that the notation could be a bit more precise. –  Matthieu Romagny Apr 6 '13 at 12:17

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