Timeline for A condition such that $p\mid\sum_{f(\theta)=0}\theta^n$ for all $n$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 31, 2021 at 3:17 | vote | accept | Milo Moses | ||
| Oct 31, 2021 at 1:28 | answer | added | David E Speyer | timeline score: 7 | |
| Oct 30, 2021 at 18:14 | comment | added | Milo Moses | @LaurentMoret-Bailly, Generally my preference is to define polynomials over fields instead of rings, but I totally see how that is confusing. It has been edited once again. | |
| Oct 30, 2021 at 18:12 | history | edited | Milo Moses | CC BY-SA 4.0 |
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| Oct 30, 2021 at 7:20 | comment | added | Laurent Moret-Bailly | @MiloMoses So why not just change $\mathbb{Q}$ to $\mathbb{Z}$ in the first line? | |
| Oct 29, 2021 at 22:59 | comment | added | Milo Moses | @OfirGorodetsky, I thought that it was clear that $f$ had integer coefficients from when I said it was monic but the question has been changed for clarity. In my main paragraph I clearly state that I already know that either $p\mid t_n$ implies either $p\mid c_n$ for all $n<\deg(f)$ or $p \mid t_n$ for all $n$. Implicit in this is that the condition is equivalent to $p\mid c_n$ for $p>\deg(f)$ since $t_0=\deg(f)$. I am well aware of Newton's formulas and I used them in my proof. | |
| Oct 29, 2021 at 22:56 | history | edited | Milo Moses | CC BY-SA 4.0 |
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| Oct 29, 2021 at 22:55 | comment | added | Ofir Gorodetsky | It seems you are assuming implicitly that, say, $f$ has integer coefficients (otherwise, why would $t_n$ be in $\mathbb{Z}$?) At least for $p>\deg f$ the condition $p \mid t_n$ for all $n$ is the same as $p \mid c_n$ for $0 \le n < \deg f$. Indeed, the coefficients of $f$ are polynomials (with rational coefficients having denominator dividing $\deg f!$) in $t_1,\ldots,t_{\deg f}$; see Newton-Girard identities for explicit formulas for these. If $p>\deg f$ the denominators shouldn't worry us. | |
| Oct 29, 2021 at 22:46 | comment | added | Milo Moses | @JeremyRouse I think I see where the failure in my intuition was. My idea is that if you have an element of $I\cap \mathbb{Z}$ you can "symmetrize" by adding all permutations of the roots, and then that should be reducible with the $t_k$ which are all multiples of $p$ so every element of $I \cap \mathbb{Z}$ is a multiple of $p$. The problem is, when you symmetrize you multiply by a factor of $d!$ so this argument does not work when $p|d$ and thus the intuition is moot. | |
| Oct 29, 2021 at 21:58 | comment | added | Milo Moses | @LSpice Sorry I meant that the roots should be counted with multiplicity; I totally see how that wasn't clear. I will use $p\mid t$ in the future, thanks for the note! | |
| Oct 29, 2021 at 21:57 | history | edited | Milo Moses | CC BY-SA 4.0 |
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| Oct 29, 2021 at 21:41 | comment | added | LSpice |
It seems that you are counting the roots without multiplicity. Is it obvious why this should be the right approach? \\ Also, TeX note: please use $p \mid t$ p \mid t, not $p | t$ p | t. I have edited accordingly.
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| Oct 29, 2021 at 21:41 | history | edited | LSpice | CC BY-SA 4.0 |
`|` -> `\mid`
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| Oct 29, 2021 at 21:02 | comment | added | Milo Moses | @JeremyRouse Thank you so much for the example! | |
| Oct 29, 2021 at 20:52 | comment | added | Jeremy Rouse | The observation you propose, that $p \mid t_{n}$ for all $n$ implies that every element of $I \cap \mathbb{Z}$ is a multiple of $p$, isn't true. For example, if $\theta_{1} = 1$, $\theta_{2} = 4$ and $\theta_{3} = 7$, then $3 \mid t_{n}$ for all $n$, but $\langle \theta_{1}, \theta_{2}, \theta_{3} \rangle = \langle 1 \rangle \subseteq \mathbb{Z}$ contains many elements that are not multiples of $3$. | |
| Oct 29, 2021 at 19:42 | history | asked | Milo Moses | CC BY-SA 4.0 |