Timeline for What is the quotient group $\mathfrak{q}^2/\mathfrak{p}^2 \mathbb{Z}_p$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 26, 2023 at 15:12 | comment | added | Vik78 | That looks right to me | |
| May 26, 2023 at 15:10 | comment | added | MAS | @Vik78, thank you for the link. Since the quotient $\mathfrak{q}^2/\mathfrak{p}^2 \mathbb Z_p$ is trivial, its torsion subgroup is also trivial. That is, the order of the torsion subgroup of $\mathfrak{q}^2/\mathfrak{p}^2 \mathbb Z_p$ is $1$. Am I right ? | |
| May 26, 2023 at 3:16 | comment | added | Vik78 | Sorry, I misspoke. I believe any totally ramified extension of $\mathbb{Q}_p$ has monogenic ring of integers over $\mathbb{Z}_p$, generated by any uniformizer. See here: math.stackexchange.com/questions/117973/… | |
| May 26, 2023 at 2:50 | comment | added | MAS | @Vik78, thank you. But isn't $K_{\mathfrak{p}} \simeq \mathbb Q_p(\zeta_p)$ because $\mathfrak{p} \mid p$. So $K_{\mathfrak{p}}$ should be totally ramified because the other one is so. | |
| May 26, 2023 at 2:35 | history | edited | MAS | CC BY-SA 4.0 |
added 2 characters in body
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| May 26, 2023 at 2:30 | comment | added | MAS | @LSpice, my apology, yes, I mean $\zeta_p^p=1$ | |
| May 25, 2023 at 17:21 | comment | added | LSpice | You mean $\zeta_p^p = 1$, not $\zeta^{p - 1} = 1$, right? | |
| May 25, 2023 at 16:18 | comment | added | Vik78 | Completing at $\mathfrak{p}$ means that any element of $K$ whose $\mathfrak{p}$-valuation is 1 maps to a uniformizer in the completion. $K_\mathfrak{p}$ is an unramified extension of $\mathbb{Q}_p$, so its ring of integers is monogenic over $\mathbb{Z}_p$. So I think $\mathfrak{p}^2 \mathbb{Z}_p = \mathfrak{q}^2$. | |
| May 25, 2023 at 16:11 | comment | added | Will Sawin | Why do you think it is nontrivial? | |
| May 25, 2023 at 15:56 | history | asked | MAS | CC BY-SA 4.0 |