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$zeta_n^k$ is primitive if and only if $(k,n) = 1$ [duplicate]
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This question already has an answer here:
primitive n-th roots of unity
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Show that for $k in mathbb{Z}$:
Is $zeta_n$ a primitive $n$-th root of unity, then $zeta_n^k$ is primitive if and only if $(k,n) = 1$.
I only need the backwards direction:
$zeta_n^k$ is primitive $Rightarrow$ $(k,n) = 1$
abstract-algebra roots-of-unity primitive-roots
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marked as duplicate by Dietrich Burde abstract-algebra
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Jan 30 at 11:39
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$begingroup$
This question already has an answer here:
primitive n-th roots of unity
3 answers
Show that for $k in mathbb{Z}$:
Is $zeta_n$ a primitive $n$-th root of unity, then $zeta_n^k$ is primitive if and only if $(k,n) = 1$.
I only need the backwards direction:
$zeta_n^k$ is primitive $Rightarrow$ $(k,n) = 1$
abstract-algebra roots-of-unity primitive-roots
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marked as duplicate by Dietrich Burde abstract-algebra
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Jan 30 at 11:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
primitive n-th roots of unity
3 answers
Show that for $k in mathbb{Z}$:
Is $zeta_n$ a primitive $n$-th root of unity, then $zeta_n^k$ is primitive if and only if $(k,n) = 1$.
I only need the backwards direction:
$zeta_n^k$ is primitive $Rightarrow$ $(k,n) = 1$
abstract-algebra roots-of-unity primitive-roots
$endgroup$
This question already has an answer here:
primitive n-th roots of unity
3 answers
Show that for $k in mathbb{Z}$:
Is $zeta_n$ a primitive $n$-th root of unity, then $zeta_n^k$ is primitive if and only if $(k,n) = 1$.
I only need the backwards direction:
$zeta_n^k$ is primitive $Rightarrow$ $(k,n) = 1$
This question already has an answer here:
primitive n-th roots of unity
3 answers
abstract-algebra roots-of-unity primitive-roots
abstract-algebra roots-of-unity primitive-roots
asked Jan 30 at 10:52
user625682
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Jan 30 at 11:39
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Jan 30 at 11:39
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1 Answer
1
active
oldest
votes
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First of all notice that $zeta_n^k$ is of the form
$$
e^{2pi ifrac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,binBbb Z$ and $a,bge1$. Thus
$$
zeta_n^k=e^{2pi ifrac {ad}{bd}}=e^{2pi ifrac d{b}}.
$$
From here we deduce that $(zeta_n^k)^b=1$ and since $b<n$ (in fact $0le kle n-1$) our root cannot be primitive.
answered Jan 30 at 11:14
JoeJoe
7,23921229
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$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
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– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all notice that $zeta_n^k$ is of the form
$$
e^{2pi ifrac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,binBbb Z$ and $a,bge1$. Thus
$$
zeta_n^k=e^{2pi ifrac {ad}{bd}}=e^{2pi ifrac d{b}}.
$$
From here we deduce that $(zeta_n^k)^b=1$ and since $b<n$ (in fact $0le kle n-1$) our root cannot be primitive.
answered Jan 30 at 11:14
JoeJoe
7,23921229
$endgroup$
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
add a comment |
$begingroup$
First of all notice that $zeta_n^k$ is of the form
$$
e^{2pi ifrac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,binBbb Z$ and $a,bge1$. Thus
$$
zeta_n^k=e^{2pi ifrac {ad}{bd}}=e^{2pi ifrac d{b}}.
$$
From here we deduce that $(zeta_n^k)^b=1$ and since $b<n$ (in fact $0le kle n-1$) our root cannot be primitive.
answered Jan 30 at 11:14
JoeJoe
7,23921229
$endgroup$
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
add a comment |
$begingroup$
First of all notice that $zeta_n^k$ is of the form
$$
e^{2pi ifrac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,binBbb Z$ and $a,bge1$. Thus
$$
zeta_n^k=e^{2pi ifrac {ad}{bd}}=e^{2pi ifrac d{b}}.
$$
From here we deduce that $(zeta_n^k)^b=1$ and since $b<n$ (in fact $0le kle n-1$) our root cannot be primitive.
answered Jan 30 at 11:14
JoeJoe
7,23921229
$endgroup$
First of all notice that $zeta_n^k$ is of the form
$$
e^{2pi ifrac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,binBbb Z$ and $a,bge1$. Thus
$$
zeta_n^k=e^{2pi ifrac {ad}{bd}}=e^{2pi ifrac d{b}}.
$$
From here we deduce that $(zeta_n^k)^b=1$ and since $b<n$ (in fact $0le kle n-1$) our root cannot be primitive.
answered Jan 30 at 11:14
JoeJoe
7,23921229
edited Jan 30 at 13:27
answered Jan 30 at 11:14
JoeJoe
7,23921229
answered Jan 30 at 11:14
JoeJoe
7,23921229
answered Jan 30 at 11:14
JoeJoe
7,23921229
7,23921229
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
add a comment |
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
$(3,9)=3$ so whats $a,b$ here? It should be $1 leq a,b$ but is your argument then still working?
$endgroup$
– user625682
Jan 30 at 12:09
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The key fact is that, since $kle n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,bge1$ and in this case $a=1$).
$endgroup$
– Joe
Jan 30 at 13:29
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
$begingroup$
The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $mathbb{C}$, so an approach which argues on the basis of primitive roots in $mathbb{C}$ doesn't really answer the question.
$endgroup$
– Peter Taylor
Jan 30 at 14:40
add a comment |
