Proof that there are exactly $n$ distinct $n$th roots of unity in fields of characteristic zero?
$begingroup$
I think it's true that in a field $F$ of characteristic zero, there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $bar{F}$), that is, roots of the polynomial $x^n-1$.
I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?
I also know that in fields of characteristic zero, all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.
abstract-algebra field-theory roots-of-unity
asked Sep 14 '18 at 23:10
WillGWillG
46739
$endgroup$
|
show 1 more comment
$begingroup$
I think it's true that in a field $F$ of characteristic zero, there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $bar{F}$), that is, roots of the polynomial $x^n-1$.
I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?
I also know that in fields of characteristic zero, all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.
abstract-algebra field-theory roots-of-unity
asked Sep 14 '18 at 23:10
WillGWillG
46739
$endgroup$
6
$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01
|
show 1 more comment
$begingroup$
I think it's true that in a field $F$ of characteristic zero, there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $bar{F}$), that is, roots of the polynomial $x^n-1$.
I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?
I also know that in fields of characteristic zero, all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.
abstract-algebra field-theory roots-of-unity
asked Sep 14 '18 at 23:10
WillGWillG
46739
$endgroup$
I think it's true that in a field $F$ of characteristic zero, there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $bar{F}$), that is, roots of the polynomial $x^n-1$.
I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?
I also know that in fields of characteristic zero, all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.
abstract-algebra field-theory roots-of-unity
abstract-algebra field-theory roots-of-unity
asked Sep 14 '18 at 23:10
WillGWillG
46739
asked Sep 14 '18 at 23:10
WillGWillG
46739
asked Sep 14 '18 at 23:10
WillGWillG
46739
asked Sep 14 '18 at 23:10
WillGWillG
46739
asked Sep 14 '18 at 23:10
WillGWillG
46739
46739
6
$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01
|
show 1 more comment
6
$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01
6
6
$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01
|
show 1 more comment
1 Answer
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$begingroup$
Suppose we have a primitive $n$th root of unity, $omega$. Then for all $1 < a le n$, $(omega^a)^n = (omega^n)^a = 1^a = 1$. If $1 le a < b le n$ then $omega^a = omega^b implies omega^{b-a} = 1$, contradicting the assumption that $omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = frac nm$. Then $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) neq 1$$ Now, $gcd(p(x), q(x)) = gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have
$$frac{x^{km} - 1}{x^m-1} - (x^m-1) sum_{i=0}^{k-2} (k-1-i)x^{im} = frac{(x^{km} - 1) - (x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$
But that $r$ was carefully chosen to telescope:
$$(x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$
so that $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) =
gcd(k, x^m - 1)$$
This is non-constant iff $k$ is a multiple of the characteristic of $F$.
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
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add a comment |
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$begingroup$
Suppose we have a primitive $n$th root of unity, $omega$. Then for all $1 < a le n$, $(omega^a)^n = (omega^n)^a = 1^a = 1$. If $1 le a < b le n$ then $omega^a = omega^b implies omega^{b-a} = 1$, contradicting the assumption that $omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = frac nm$. Then $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) neq 1$$ Now, $gcd(p(x), q(x)) = gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have
$$frac{x^{km} - 1}{x^m-1} - (x^m-1) sum_{i=0}^{k-2} (k-1-i)x^{im} = frac{(x^{km} - 1) - (x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$
But that $r$ was carefully chosen to telescope:
$$(x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$
so that $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) =
gcd(k, x^m - 1)$$
This is non-constant iff $k$ is a multiple of the characteristic of $F$.
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
$endgroup$
add a comment |
$begingroup$
Suppose we have a primitive $n$th root of unity, $omega$. Then for all $1 < a le n$, $(omega^a)^n = (omega^n)^a = 1^a = 1$. If $1 le a < b le n$ then $omega^a = omega^b implies omega^{b-a} = 1$, contradicting the assumption that $omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = frac nm$. Then $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) neq 1$$ Now, $gcd(p(x), q(x)) = gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have
$$frac{x^{km} - 1}{x^m-1} - (x^m-1) sum_{i=0}^{k-2} (k-1-i)x^{im} = frac{(x^{km} - 1) - (x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$
But that $r$ was carefully chosen to telescope:
$$(x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$
so that $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) =
gcd(k, x^m - 1)$$
This is non-constant iff $k$ is a multiple of the characteristic of $F$.
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
$endgroup$
add a comment |
$begingroup$
Suppose we have a primitive $n$th root of unity, $omega$. Then for all $1 < a le n$, $(omega^a)^n = (omega^n)^a = 1^a = 1$. If $1 le a < b le n$ then $omega^a = omega^b implies omega^{b-a} = 1$, contradicting the assumption that $omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = frac nm$. Then $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) neq 1$$ Now, $gcd(p(x), q(x)) = gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have
$$frac{x^{km} - 1}{x^m-1} - (x^m-1) sum_{i=0}^{k-2} (k-1-i)x^{im} = frac{(x^{km} - 1) - (x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$
But that $r$ was carefully chosen to telescope:
$$(x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$
so that $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) =
gcd(k, x^m - 1)$$
This is non-constant iff $k$ is a multiple of the characteristic of $F$.
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
$endgroup$
Suppose we have a primitive $n$th root of unity, $omega$. Then for all $1 < a le n$, $(omega^a)^n = (omega^n)^a = 1^a = 1$. If $1 le a < b le n$ then $omega^a = omega^b implies omega^{b-a} = 1$, contradicting the assumption that $omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = frac nm$. Then $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) neq 1$$ Now, $gcd(p(x), q(x)) = gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have
$$frac{x^{km} - 1}{x^m-1} - (x^m-1) sum_{i=0}^{k-2} (k-1-i)x^{im} = frac{(x^{km} - 1) - (x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$
But that $r$ was carefully chosen to telescope:
$$(x^m-1)^2 sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$
so that $$gcdleft(frac{x^{km}-1}{x^m-1}, x^m - 1right) =
gcd(k, x^m - 1)$$
This is non-constant iff $k$ is a multiple of the characteristic of $F$.
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
answered Dec 4 '18 at 11:51
Peter TaylorPeter Taylor
8,89712341
8,89712341
add a comment |
add a comment |
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$begingroup$
If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative...
$endgroup$
– Daniel Schepler
Sep 14 '18 at 23:13
$begingroup$
Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $mathbb{R}$ as opposed to generic fields?
$endgroup$
– WillG
Sep 14 '18 at 23:57
$begingroup$
Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts?
$endgroup$
– WillG
Sep 14 '18 at 23:58
$begingroup$
Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $alpha$ iff $f$ has a repeated root at $alpha$ (that is, $(x-alpha)^2$ divides $f$).
$endgroup$
– Daniel Mroz
Sep 15 '18 at 0:00
$begingroup$
Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$.
$endgroup$
– Daniel Schepler
Sep 15 '18 at 0:01