Timeline for Is the extension field by zeros of $x^{2m}-p^{2(m-1)}=0$ over $\mathbb Q_p$ totally ramified?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 15 at 8:25 | comment | added | Chris Wuthrich | There are only three quadratic extensions in this case and one is unramified. | |
| Mar 15 at 4:20 | comment | added | MAS | @ChrisWuthrich, why is the compositum of two distinct quadratic extensions of $\mathbb{Q}_p$ not totally ramified? What is the argument here? | |
| Mar 14 at 9:07 | comment | added | Chris Wuthrich | For $m=2$. the composition of two distinct quadratic extensions of $\mathbb{Q}_p$ with odd $p$ is never totally ramified. So if $p\equiv 3 \pmod{4}$ then we have $e=f=2$. | |
| Mar 14 at 3:34 | history | edited | MAS | CC BY-SA 4.0 |
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| Mar 14 at 3:30 | comment | added | MAS | @PaulBroussous, sorry for the mistake. $v_p(\sqrt p)=1/2 \notin \mathbb Z$. So $\sqrt p \notin \mathbb Q_p$. | |
| Mar 13 at 18:50 | comment | added | Paul Broussous | If $\sqrt{p}$ existed in ${\mathbb Q}_p$, what would be its $p$-valuation ?! | |
| Mar 13 at 11:55 | comment | added | MAS | @Wojowu, Okay, think about it later. At first, I want to be assured of the total degree of extension by $f(x)$. I want to know if there are some conditions that give totally ramified extensions over $\mathbb{Q}_p$ or over its extension | |
| Mar 13 at 11:42 | comment | added | Wojowu | $p\mid m$ is not enough, I'm not sure of the exact criterion because of roots of unity already in $\mathbb Q_p$. I'm also not sure about ramification over $\mathbb Q_p(\zeta_{2m})$ - it will be generated by just a root of $p$, but $p$ is no longer a uniformizer. I can think about it later. | |
| Mar 13 at 11:16 | comment | added | MAS | @Wojowu, thanks. Yes, I remember the definition, for a totally ramified extension, every subsextension must be totally ramified. So for $p \mid m$, adjoining $m$-th roots gives totally ramified extensions, I guess. But for $p \nmid m$, I guess $\mathbb{Q}_p(f(x)=0)$ can be totally ramified over the base field $\mathbb{Q}_p(\zeta_m)$ instead over $\mathbb{Q}_p$? | |
| Mar 13 at 11:02 | comment | added | Wojowu | In that case the extension will not be totally ramified in general, even if you only adjoin roots for one of the factors $f_i$. It will have a subextension given by adjoining $m$-th roots of unity, which is unramified if $p\nmid m$. | |
| Mar 13 at 10:57 | history | edited | MAS | CC BY-SA 4.0 |
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| Mar 13 at 10:55 | comment | added | MAS | @Wojowu, I am adjoining all roots of the polynomial | |
| Mar 13 at 10:43 | comment | added | Wojowu | Do you mean to adjoin just one root of this polynomial, or all the roots? | |
| Mar 13 at 10:40 | history | asked | MAS | CC BY-SA 4.0 |