Timeline for Euler totient function and unramified extension of $\mathbb{Q}_p$. A clarification.
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2016 at 16:26 | vote | accept | Luigi M | ||
| Feb 14, 2016 at 16:01 | answer | added | Gregory Grant | timeline score: 3 | |
| Feb 14, 2016 at 15:59 | comment | added | Gregory Grant | @LuigiM Cool, I think you have it straightened out now. Which brings us to the final problem. How do we post an official answer so this question can be removed from the queue? | |
| Feb 14, 2016 at 15:58 | comment | added | Luigi M | @GregoryGrant yes and in fact the point of my question is that I was (mis)reading primitive roots of unity everywhere, that's why I was confused. | |
| Feb 14, 2016 at 15:57 | comment | added | Gregory Grant | In fact did you read the proof in that document you link to? I think it's clarified pretty well in the proof what exactly they're talking about. | |
| Feb 14, 2016 at 15:57 | comment | added | peter a g | The finite field $k=\mathbb F_{p^n}$ is the splitting field of $x^{p^n-1} -1$ - it is the extension of degree $n$ over $\mathbb F_p$ (and the finite field $k$ is the residue field of the corresponding unramified extension of $\mathbb Q_p$). | |
| Feb 14, 2016 at 15:57 | comment | added | Luigi M | and I guess this is different from what I intend. So a primitive root of order $p^n-1$ what is supposed to be? a Generator of the multiplicative group of $\mathbb{F}_{p^n}$? | |
| Feb 14, 2016 at 15:54 | comment | added | Gregory Grant | Sorry I meant it doesn't say to add a primitive $n$th root of one, it say specifically to add "a primitive root of order $p^n-1$". So I am guessing a factor of $x^{p^n-1}-1$ that has degree $n$. Does that make more sense now? | |
| Feb 14, 2016 at 15:52 | comment | added | Luigi M | Sorry but I really don't get the meaning of both comments. What is a primitive root of $n$? If you mean order $n$ I'm aware I'm attaching a primitive root of order $p^n-1$ and not $n$. | |
| Feb 14, 2016 at 15:49 | comment | added | peter a g | Well, take $n=1$... In that case, the residue field $\mathbb F_p$ contains all the roots of $x^{p-1}-1 = 0$; hence the polynomial factors completely in $\mathbb Q_p$. | |
| Feb 14, 2016 at 15:41 | history | asked | Luigi M | CC BY-SA 3.0 |