Methods of finding the minimal polynomial of $w = e^{2ipi/12}$. over $mathbb{Q}$
$begingroup$
So the minimal polynomial of $w = e^{2ipi/12}$ will be given by the 12th cyclotomic polynomial, where the $n$-th cyclotomic polynomial can be calculated recursively by the following formula:
$b=prod_{i}gamma_i$ and then $frac{x^n-1}{b}=gamma_n$ where the index on the product cycles through all positive integers less than $n$ s.t. $gcd(n,i)=1$
and thus
$gamma_{12} = frac{x^{12}-1}{gamma_1gamma_2gamma_3gamma_4gamma_6} = x^4-x^2+1$.
So that works technically, although I'm not a fan of doing tedious calculations if I don't have to, especially when lots of long division is involved.
Alternatively, one could note that $w^6=-1$ and thus $w^6+1=0$. Furthermore, we know the $deg(gamma_{12})=phi(12)=4$, where $phi$ is the euler toitent function. It seems like from the combination of these two things we should able to deduce what $gamma_{12}$ is, perhaps by getting the $12$-gon involved and looking at the vertices, because the roots of $gamma_{12}$ are precisely the primitive roots of $x^{12}-1$, correct?
A root $c$ of $x^{12}-1$ is primitive $iff$ $c^{12}=1$ and $c^r neq 1$ $forall r$ such that $0<r<12$. Isn't it then true that the roots of $x^4-x^2+1$ are given by $e^{2i*pi*k/12}$ where $k$ is such that $k$-mod $12$ generates $(mathbb{Z}_{12},+)$? So I could find these $k$'s, which there would be four, and then $x^4-x^2+1 = prod_{1 leq k leq 4}(1-e^{2ipi k_i/12})$?
So, I guess the set of elements that generate $mathbb{Z_{12}}$ are $1,5,7, 11$ (mod$12$).
Okay, I just got put $real((x-e^{2pi i/12})(x-e^{10pi i/12})(x-e^{14pi i/12})(x-e^{22pi*i/12})$ into wolfram alpha and got that it equals $x^4-x^2+1$ under the assumption that $x$ is positive. Now I'm confused. Why must $x$ be positive? I feel like it should be true for any number in the complex plane...
abstract-algebra
edited Dec 9 '18 at 20:38
Bernard
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
$endgroup$
add a comment |
$begingroup$
So the minimal polynomial of $w = e^{2ipi/12}$ will be given by the 12th cyclotomic polynomial, where the $n$-th cyclotomic polynomial can be calculated recursively by the following formula:
$b=prod_{i}gamma_i$ and then $frac{x^n-1}{b}=gamma_n$ where the index on the product cycles through all positive integers less than $n$ s.t. $gcd(n,i)=1$
and thus
$gamma_{12} = frac{x^{12}-1}{gamma_1gamma_2gamma_3gamma_4gamma_6} = x^4-x^2+1$.
So that works technically, although I'm not a fan of doing tedious calculations if I don't have to, especially when lots of long division is involved.
Alternatively, one could note that $w^6=-1$ and thus $w^6+1=0$. Furthermore, we know the $deg(gamma_{12})=phi(12)=4$, where $phi$ is the euler toitent function. It seems like from the combination of these two things we should able to deduce what $gamma_{12}$ is, perhaps by getting the $12$-gon involved and looking at the vertices, because the roots of $gamma_{12}$ are precisely the primitive roots of $x^{12}-1$, correct?
A root $c$ of $x^{12}-1$ is primitive $iff$ $c^{12}=1$ and $c^r neq 1$ $forall r$ such that $0<r<12$. Isn't it then true that the roots of $x^4-x^2+1$ are given by $e^{2i*pi*k/12}$ where $k$ is such that $k$-mod $12$ generates $(mathbb{Z}_{12},+)$? So I could find these $k$'s, which there would be four, and then $x^4-x^2+1 = prod_{1 leq k leq 4}(1-e^{2ipi k_i/12})$?
So, I guess the set of elements that generate $mathbb{Z_{12}}$ are $1,5,7, 11$ (mod$12$).
Okay, I just got put $real((x-e^{2pi i/12})(x-e^{10pi i/12})(x-e^{14pi i/12})(x-e^{22pi*i/12})$ into wolfram alpha and got that it equals $x^4-x^2+1$ under the assumption that $x$ is positive. Now I'm confused. Why must $x$ be positive? I feel like it should be true for any number in the complex plane...
abstract-algebra
edited Dec 9 '18 at 20:38
Bernard
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
$endgroup$
1
$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
1
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50
add a comment |
$begingroup$
So the minimal polynomial of $w = e^{2ipi/12}$ will be given by the 12th cyclotomic polynomial, where the $n$-th cyclotomic polynomial can be calculated recursively by the following formula:
$b=prod_{i}gamma_i$ and then $frac{x^n-1}{b}=gamma_n$ where the index on the product cycles through all positive integers less than $n$ s.t. $gcd(n,i)=1$
and thus
$gamma_{12} = frac{x^{12}-1}{gamma_1gamma_2gamma_3gamma_4gamma_6} = x^4-x^2+1$.
So that works technically, although I'm not a fan of doing tedious calculations if I don't have to, especially when lots of long division is involved.
Alternatively, one could note that $w^6=-1$ and thus $w^6+1=0$. Furthermore, we know the $deg(gamma_{12})=phi(12)=4$, where $phi$ is the euler toitent function. It seems like from the combination of these two things we should able to deduce what $gamma_{12}$ is, perhaps by getting the $12$-gon involved and looking at the vertices, because the roots of $gamma_{12}$ are precisely the primitive roots of $x^{12}-1$, correct?
A root $c$ of $x^{12}-1$ is primitive $iff$ $c^{12}=1$ and $c^r neq 1$ $forall r$ such that $0<r<12$. Isn't it then true that the roots of $x^4-x^2+1$ are given by $e^{2i*pi*k/12}$ where $k$ is such that $k$-mod $12$ generates $(mathbb{Z}_{12},+)$? So I could find these $k$'s, which there would be four, and then $x^4-x^2+1 = prod_{1 leq k leq 4}(1-e^{2ipi k_i/12})$?
So, I guess the set of elements that generate $mathbb{Z_{12}}$ are $1,5,7, 11$ (mod$12$).
Okay, I just got put $real((x-e^{2pi i/12})(x-e^{10pi i/12})(x-e^{14pi i/12})(x-e^{22pi*i/12})$ into wolfram alpha and got that it equals $x^4-x^2+1$ under the assumption that $x$ is positive. Now I'm confused. Why must $x$ be positive? I feel like it should be true for any number in the complex plane...
abstract-algebra
edited Dec 9 '18 at 20:38
Bernard
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
$endgroup$
So the minimal polynomial of $w = e^{2ipi/12}$ will be given by the 12th cyclotomic polynomial, where the $n$-th cyclotomic polynomial can be calculated recursively by the following formula:
$b=prod_{i}gamma_i$ and then $frac{x^n-1}{b}=gamma_n$ where the index on the product cycles through all positive integers less than $n$ s.t. $gcd(n,i)=1$
and thus
$gamma_{12} = frac{x^{12}-1}{gamma_1gamma_2gamma_3gamma_4gamma_6} = x^4-x^2+1$.
So that works technically, although I'm not a fan of doing tedious calculations if I don't have to, especially when lots of long division is involved.
Alternatively, one could note that $w^6=-1$ and thus $w^6+1=0$. Furthermore, we know the $deg(gamma_{12})=phi(12)=4$, where $phi$ is the euler toitent function. It seems like from the combination of these two things we should able to deduce what $gamma_{12}$ is, perhaps by getting the $12$-gon involved and looking at the vertices, because the roots of $gamma_{12}$ are precisely the primitive roots of $x^{12}-1$, correct?
A root $c$ of $x^{12}-1$ is primitive $iff$ $c^{12}=1$ and $c^r neq 1$ $forall r$ such that $0<r<12$. Isn't it then true that the roots of $x^4-x^2+1$ are given by $e^{2i*pi*k/12}$ where $k$ is such that $k$-mod $12$ generates $(mathbb{Z}_{12},+)$? So I could find these $k$'s, which there would be four, and then $x^4-x^2+1 = prod_{1 leq k leq 4}(1-e^{2ipi k_i/12})$?
So, I guess the set of elements that generate $mathbb{Z_{12}}$ are $1,5,7, 11$ (mod$12$).
Okay, I just got put $real((x-e^{2pi i/12})(x-e^{10pi i/12})(x-e^{14pi i/12})(x-e^{22pi*i/12})$ into wolfram alpha and got that it equals $x^4-x^2+1$ under the assumption that $x$ is positive. Now I'm confused. Why must $x$ be positive? I feel like it should be true for any number in the complex plane...
abstract-algebra
abstract-algebra
edited Dec 9 '18 at 20:38
Bernard
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
edited Dec 9 '18 at 20:38
Bernard
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
edited Dec 9 '18 at 20:38
Bernard
123k741117
edited Dec 9 '18 at 20:38
Bernard
123k741117
edited Dec 9 '18 at 20:38
Bernard
123k741117
123k741117
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
asked Dec 9 '18 at 19:43
Math is hardMath is hard
822211
822211
1
$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
1
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50
add a comment |
1
$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
1
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50
1
1
$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
1
1
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50
add a comment |
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$begingroup$
whitman.edu/Documents/Academics/Mathematics/2015/…
$endgroup$
– Samvel Safaryan
Dec 9 '18 at 19:47
1
$begingroup$
You can just calculate the coefficients using the symmetric functions of the roots, and you'll get the same answer regardless of positive or negative. It's probably just an anomaly with the algorithm Wolfram Alpha uses. I mean, trivially it holds for non-negatives by plugging in $0$, so obviously it doesn't just hold for positives.
$endgroup$
– Melody
Dec 9 '18 at 19:50