Prove that $eta - omega notin mathbb{Q}$ where $omega$ and $eta$ are two differents n-th primitive roots $in...
$begingroup$
Let $n in mathbb{N}$ be a natural number, and be $omega$ and $eta$ two differents n-th primitive roots in $mathbb{C}$.
Prove that $eta - omega notin mathbb{Q}$
My attempt was to follow the false line of the following :
If i'd to prove that $sqrt-2 - sqrt-5 notin mathbb{Q}$, i'd try something by contradiction like : $$sqrt-2 - sqrt-5 = alpha, alpha in mathbb{Q}$$
$$sqrt-2 = sqrt-5 + alpha$$
$$ -2 = alpha^{2} + 2alphasqrt-5 -5$$
But then $-2,alpha^{2},-5 in mathbb{Q}$ which leads to $sqrt-5 in mathbb{Q}$,false.
So here i'd like to re-write $$eta = omega + alpha , alpha in mathbb{Q} $$
And raise to the n-th power sothat $eta in mathbb{Q}$, but then i'm unable to find some contradiction due to the difficulties in seeing the terms of the newton binomial $(omega + alpha )^{n}$.
Is this the right approach ?
Any help or tip would be appreciated,
Thanks a lot
number-theory elementary-number-theory complex-numbers arithmetic primitive-roots
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
$endgroup$
|
show 12 more comments
$begingroup$
Let $n in mathbb{N}$ be a natural number, and be $omega$ and $eta$ two differents n-th primitive roots in $mathbb{C}$.
Prove that $eta - omega notin mathbb{Q}$
My attempt was to follow the false line of the following :
If i'd to prove that $sqrt-2 - sqrt-5 notin mathbb{Q}$, i'd try something by contradiction like : $$sqrt-2 - sqrt-5 = alpha, alpha in mathbb{Q}$$
$$sqrt-2 = sqrt-5 + alpha$$
$$ -2 = alpha^{2} + 2alphasqrt-5 -5$$
But then $-2,alpha^{2},-5 in mathbb{Q}$ which leads to $sqrt-5 in mathbb{Q}$,false.
So here i'd like to re-write $$eta = omega + alpha , alpha in mathbb{Q} $$
And raise to the n-th power sothat $eta in mathbb{Q}$, but then i'm unable to find some contradiction due to the difficulties in seeing the terms of the newton binomial $(omega + alpha )^{n}$.
Is this the right approach ?
Any help or tip would be appreciated,
Thanks a lot
number-theory elementary-number-theory complex-numbers arithmetic primitive-roots
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
$endgroup$
1
$begingroup$
$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:11
2
$begingroup$
To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:17
1
$begingroup$
But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:21
1
$begingroup$
For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
$endgroup$
– jacopoburelli
Dec 28 '18 at 21:23
3
$begingroup$
Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:28
|
show 12 more comments
$begingroup$
Let $n in mathbb{N}$ be a natural number, and be $omega$ and $eta$ two differents n-th primitive roots in $mathbb{C}$.
Prove that $eta - omega notin mathbb{Q}$
My attempt was to follow the false line of the following :
If i'd to prove that $sqrt-2 - sqrt-5 notin mathbb{Q}$, i'd try something by contradiction like : $$sqrt-2 - sqrt-5 = alpha, alpha in mathbb{Q}$$
$$sqrt-2 = sqrt-5 + alpha$$
$$ -2 = alpha^{2} + 2alphasqrt-5 -5$$
But then $-2,alpha^{2},-5 in mathbb{Q}$ which leads to $sqrt-5 in mathbb{Q}$,false.
So here i'd like to re-write $$eta = omega + alpha , alpha in mathbb{Q} $$
And raise to the n-th power sothat $eta in mathbb{Q}$, but then i'm unable to find some contradiction due to the difficulties in seeing the terms of the newton binomial $(omega + alpha )^{n}$.
Is this the right approach ?
Any help or tip would be appreciated,
Thanks a lot
number-theory elementary-number-theory complex-numbers arithmetic primitive-roots
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
$endgroup$
Let $n in mathbb{N}$ be a natural number, and be $omega$ and $eta$ two differents n-th primitive roots in $mathbb{C}$.
Prove that $eta - omega notin mathbb{Q}$
My attempt was to follow the false line of the following :
If i'd to prove that $sqrt-2 - sqrt-5 notin mathbb{Q}$, i'd try something by contradiction like : $$sqrt-2 - sqrt-5 = alpha, alpha in mathbb{Q}$$
$$sqrt-2 = sqrt-5 + alpha$$
$$ -2 = alpha^{2} + 2alphasqrt-5 -5$$
But then $-2,alpha^{2},-5 in mathbb{Q}$ which leads to $sqrt-5 in mathbb{Q}$,false.
So here i'd like to re-write $$eta = omega + alpha , alpha in mathbb{Q} $$
And raise to the n-th power sothat $eta in mathbb{Q}$, but then i'm unable to find some contradiction due to the difficulties in seeing the terms of the newton binomial $(omega + alpha )^{n}$.
Is this the right approach ?
Any help or tip would be appreciated,
Thanks a lot
number-theory elementary-number-theory complex-numbers arithmetic primitive-roots
number-theory elementary-number-theory complex-numbers arithmetic primitive-roots
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
asked Dec 28 '18 at 21:05
jacopoburellijacopoburelli
1707
1707
1
$begingroup$
$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:11
2
$begingroup$
To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:17
1
$begingroup$
But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:21
1
$begingroup$
For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
$endgroup$
– jacopoburelli
Dec 28 '18 at 21:23
3
$begingroup$
Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:28
|
show 12 more comments
1
$begingroup$
$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:11
2
$begingroup$
To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:17
1
$begingroup$
But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:21
1
$begingroup$
For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
$endgroup$
– jacopoburelli
Dec 28 '18 at 21:23
3
$begingroup$
Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:28
1
1
$begingroup$
$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:11
$begingroup$
$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:11
2
2
$begingroup$
To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:17
$begingroup$
To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:17
1
1
$begingroup$
But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:21
$begingroup$
But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:21
1
1
$begingroup$
For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
$endgroup$
– jacopoburelli
Dec 28 '18 at 21:23
$begingroup$
For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
$endgroup$
– jacopoburelli
Dec 28 '18 at 21:23
3
3
$begingroup$
Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:28
$begingroup$
Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 21:28
|
show 12 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Because $|eta|=|omega|=1$ and $etaneqomega$ we have $0<|eta-omega|leq2$, and switching $eta$ and $omega$ if necessary gives, without loss of generality, that $0<eta-omegaleq2$. Suppose now that $eta-omegainBbb{Q}$. Because $eta$ and $omega$ are integral over $Bbb{Z}$, so is $eta-omega$ and hence $eta-omegainBbb{Z}$. This shows that $eta-omegain{1,2}$.
If $eta-omega=1$ then $eta=frac{1}{2}pmfrac{sqrt{3}}{2}i$ and $omega=-frac{1}{2}pmfrac{sqrt{3}}{2}i$, the two $pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $eta-omega=2$ then $eta=1$ and $omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $eta-omeganotinBbb{Q}$.
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
$endgroup$
$begingroup$
I knew I was missing something simple :-)
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 6:13
$begingroup$
Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
$endgroup$
– jacopoburelli
Dec 29 '18 at 7:18
$begingroup$
To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
$endgroup$
– Servaes
Dec 29 '18 at 9:25
$begingroup$
To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
$endgroup$
– Servaes
Dec 29 '18 at 9:26
add a comment |
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$begingroup$
Because $|eta|=|omega|=1$ and $etaneqomega$ we have $0<|eta-omega|leq2$, and switching $eta$ and $omega$ if necessary gives, without loss of generality, that $0<eta-omegaleq2$. Suppose now that $eta-omegainBbb{Q}$. Because $eta$ and $omega$ are integral over $Bbb{Z}$, so is $eta-omega$ and hence $eta-omegainBbb{Z}$. This shows that $eta-omegain{1,2}$.
If $eta-omega=1$ then $eta=frac{1}{2}pmfrac{sqrt{3}}{2}i$ and $omega=-frac{1}{2}pmfrac{sqrt{3}}{2}i$, the two $pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $eta-omega=2$ then $eta=1$ and $omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $eta-omeganotinBbb{Q}$.
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
$endgroup$
$begingroup$
I knew I was missing something simple :-)
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 6:13
$begingroup$
Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
$endgroup$
– jacopoburelli
Dec 29 '18 at 7:18
$begingroup$
To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
$endgroup$
– Servaes
Dec 29 '18 at 9:25
$begingroup$
To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
$endgroup$
– Servaes
Dec 29 '18 at 9:26
add a comment |
$begingroup$
Because $|eta|=|omega|=1$ and $etaneqomega$ we have $0<|eta-omega|leq2$, and switching $eta$ and $omega$ if necessary gives, without loss of generality, that $0<eta-omegaleq2$. Suppose now that $eta-omegainBbb{Q}$. Because $eta$ and $omega$ are integral over $Bbb{Z}$, so is $eta-omega$ and hence $eta-omegainBbb{Z}$. This shows that $eta-omegain{1,2}$.
If $eta-omega=1$ then $eta=frac{1}{2}pmfrac{sqrt{3}}{2}i$ and $omega=-frac{1}{2}pmfrac{sqrt{3}}{2}i$, the two $pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $eta-omega=2$ then $eta=1$ and $omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $eta-omeganotinBbb{Q}$.
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
$endgroup$
$begingroup$
I knew I was missing something simple :-)
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 6:13
$begingroup$
Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
$endgroup$
– jacopoburelli
Dec 29 '18 at 7:18
$begingroup$
To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
$endgroup$
– Servaes
Dec 29 '18 at 9:25
$begingroup$
To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
$endgroup$
– Servaes
Dec 29 '18 at 9:26
add a comment |
$begingroup$
Because $|eta|=|omega|=1$ and $etaneqomega$ we have $0<|eta-omega|leq2$, and switching $eta$ and $omega$ if necessary gives, without loss of generality, that $0<eta-omegaleq2$. Suppose now that $eta-omegainBbb{Q}$. Because $eta$ and $omega$ are integral over $Bbb{Z}$, so is $eta-omega$ and hence $eta-omegainBbb{Z}$. This shows that $eta-omegain{1,2}$.
If $eta-omega=1$ then $eta=frac{1}{2}pmfrac{sqrt{3}}{2}i$ and $omega=-frac{1}{2}pmfrac{sqrt{3}}{2}i$, the two $pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $eta-omega=2$ then $eta=1$ and $omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $eta-omeganotinBbb{Q}$.
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
$endgroup$
Because $|eta|=|omega|=1$ and $etaneqomega$ we have $0<|eta-omega|leq2$, and switching $eta$ and $omega$ if necessary gives, without loss of generality, that $0<eta-omegaleq2$. Suppose now that $eta-omegainBbb{Q}$. Because $eta$ and $omega$ are integral over $Bbb{Z}$, so is $eta-omega$ and hence $eta-omegainBbb{Z}$. This shows that $eta-omegain{1,2}$.
If $eta-omega=1$ then $eta=frac{1}{2}pmfrac{sqrt{3}}{2}i$ and $omega=-frac{1}{2}pmfrac{sqrt{3}}{2}i$, the two $pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $eta-omega=2$ then $eta=1$ and $omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $eta-omeganotinBbb{Q}$.
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
edited Dec 28 '18 at 22:59
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
answered Dec 28 '18 at 22:39
ServaesServaes
25.9k33996
25.9k33996
$begingroup$
I knew I was missing something simple :-)
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 6:13
$begingroup$
Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
$endgroup$
– jacopoburelli
Dec 29 '18 at 7:18
$begingroup$
To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
$endgroup$
– Servaes
Dec 29 '18 at 9:25
$begingroup$
To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
$endgroup$
– Servaes
Dec 29 '18 at 9:26
add a comment |
$begingroup$
I knew I was missing something simple :-)
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 6:13
$begingroup$
Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
$endgroup$
– jacopoburelli
Dec 29 '18 at 7:18
$begingroup$
To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
$endgroup$
– Servaes
Dec 29 '18 at 9:25
$begingroup$
To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
$endgroup$
– Servaes
Dec 29 '18 at 9:26
$begingroup$
I knew I was missing something simple :-)
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– Jyrki Lahtonen
Dec 29 '18 at 6:13
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I knew I was missing something simple :-)
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– Jyrki Lahtonen
Dec 29 '18 at 6:13
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Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
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– jacopoburelli
Dec 29 '18 at 7:18
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Just two things that are not entirely clear to me, why $eta - omega in mathbb{Z}$ ? And why can you remove the absolute value so easily ?
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– jacopoburelli
Dec 29 '18 at 7:18
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To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
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– Servaes
Dec 29 '18 at 9:25
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To the second point; if $eta-omega<0$ then $omega-eta>0$, and the question is symmetric in $omega$ and $eta$.
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– Servaes
Dec 29 '18 at 9:25
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To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
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– Servaes
Dec 29 '18 at 9:26
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To the first point; both $eta$ and $omega$ are elements of $Bbb{Z}[zeta_n]$, hence so is $eta-omega$. And $Bbb{Z}[zeta_n]capBbb{Q}=Bbb{Z}$.
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– Servaes
Dec 29 '18 at 9:26
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1
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$-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials.
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– Jyrki Lahtonen
Dec 28 '18 at 21:11
2
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To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-omega)(x-eta)$ where $omega$ and $eta$ are the two primitive roots. Now, if $eta=omega+q$ for some rational number $q$, this means that $omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(omega)=p(omega+q)=p(eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so....
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– Jyrki Lahtonen
Dec 28 '18 at 21:17
1
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But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that.
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– Jyrki Lahtonen
Dec 28 '18 at 21:21
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For irreducibility of the cyclotomic polynomials over $mathbb{Q}$ could use Eisenstein's criterion, using $frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen
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– jacopoburelli
Dec 28 '18 at 21:23
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Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real.
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– Jyrki Lahtonen
Dec 28 '18 at 21:28