Mixing
Characteristic and primitive roots of unity
$begingroup$
Let $F$ be a finite field.
If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.
I believe the converse is true, too, but I can’t prove either direction?
I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.
But how does this relate to the characteristic not dividing $n$?
abstract-algebra galois-theory finite-fields
asked Jan 18 at 8:49
the manthe man
772715
$endgroup$
add a comment |
$begingroup$
Let $F$ be a finite field.
If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.
I believe the converse is true, too, but I can’t prove either direction?
I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.
But how does this relate to the characteristic not dividing $n$?
abstract-algebra galois-theory finite-fields
asked Jan 18 at 8:49
the manthe man
772715
$endgroup$
$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
1
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25
add a comment |
$begingroup$
Let $F$ be a finite field.
If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.
I believe the converse is true, too, but I can’t prove either direction?
I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.
But how does this relate to the characteristic not dividing $n$?
abstract-algebra galois-theory finite-fields
asked Jan 18 at 8:49
the manthe man
772715
$endgroup$
Let $F$ be a finite field.
If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.
I believe the converse is true, too, but I can’t prove either direction?
I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.
But how does this relate to the characteristic not dividing $n$?
abstract-algebra galois-theory finite-fields
abstract-algebra galois-theory finite-fields
asked Jan 18 at 8:49
the manthe man
772715
asked Jan 18 at 8:49
the manthe man
772715
asked Jan 18 at 8:49
the manthe man
772715
asked Jan 18 at 8:49
the manthe man
772715
asked Jan 18 at 8:49
the manthe man
772715
772715
$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
1
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25
add a comment |
$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
1
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25
$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
1
1
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
$endgroup$
add a comment |
$begingroup$
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
$endgroup$
fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
answered Jan 18 at 8:54
EnkiduEnkidu
1,36719
1,36719
add a comment |
add a comment |
$begingroup$
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
$endgroup$
add a comment |
$begingroup$
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
$endgroup$
add a comment |
$begingroup$
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
$endgroup$
Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
answered Jan 18 at 8:56
WuestenfuxWuestenfux
4,7291513
4,7291513
add a comment |
add a comment |
$begingroup$
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
$endgroup$
add a comment |
$begingroup$
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
$endgroup$
add a comment |
$begingroup$
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
$endgroup$
If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.
The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
answered Jan 18 at 10:15
ServaesServaes
25.9k33996
25.9k33996
add a comment |
add a comment |
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$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23
1
$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25