Timeline for Algebraic $p$-adic integers mod $p$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 6, 2013 at 19:25 | comment | added | Matthieu Romagny | I believe that I can accept only one answer. It does not matter much, but this is now done. Thanks, Scott. | |
| Apr 6, 2013 at 16:07 | vote | accept | Matthieu Romagny | ||
| Apr 6, 2013 at 14:53 | comment | added | S. Carnahan♦ | I have no reason to complain about any outcome. | |
| Apr 6, 2013 at 11:59 | comment | added | Matthieu Romagny | Now I find the 2 answers below relevant and complementary. Ideally I'd like to accept both. What can I do? | |
| Apr 6, 2013 at 5:25 | answer | added | Chandan Singh Dalawat | timeline score: 1 | |
| Apr 6, 2013 at 1:57 | answer | added | S. Carnahan♦ | timeline score: 1 | |
| Apr 5, 2013 at 18:18 | comment | added | Matthieu Romagny |
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.
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| Apr 5, 2013 at 12:53 | comment | added | Torsten Schoeneberg |
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 $\bar{\mathbb{F}}_p$.
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| Apr 5, 2013 at 11:10 | comment | added | Chandan Singh Dalawat | ... 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 $x_n$, | |
| Apr 5, 2013 at 10:55 | comment | added | Chandan Singh Dalawat | 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 $m$. | |
| Apr 5, 2013 at 10:34 | comment | added | Damian Rössler | Since $(p)$ is the generator of the maximal ideal of ${\bf Z}_p^{\rm nr}$, you have ${\bf Z}_p^{\rm nr}/(p)=\bar{\bf F}_p$. | |
| Apr 5, 2013 at 9:06 | history | asked | Matthieu Romagny | CC BY-SA 3.0 |