Determine a Normal Basis for Galois Extension of $mathbb{Q}$ with primitive pth root of unit (p prime)
$begingroup$
Let p be a prime, $xi_p in mathbb{C}$ a primitive p-th unit root and $K = mathbb{Q}(xi_p)$.
Give a normal basis for $K/mathbb{Q}$.
I know, that a basis of $L/K$ (finite and galois) is called a normal basis if it has the form ${sigma(a)}_{sigma in text{Gal}(L/K)}$ for an $a in L$.
A procedure to determine such an element a if $|K| = infty$ is the following:
- Determine the minimal polynomial $m_{beta}(X)$ for a primitive Element $beta$ ($L = K(beta)$)
- Set $g_{sigma}(X) := prodlimits_{substack{tau in G\tau neq sigma}} (X - tau(beta))$ and $A(X) := (g_{tau^{-1}sigma})_{tau,sigma in G}$ with $G := text{Gal}(L/K)$
- Search a $gamma in K$ with $det(A(gamma)) neq 0$
- Set $a := frac{m_{beta}(gamma)}{gamma - beta}$
So far so good. I have only been able to achieve that $Phi_p(X) = frac{X^p - 1}{X-1} = X^{p-1} + . . . + X + 1$ is the minimal polynomial of $xi_p$ over $mathbb{Q}$ since p is prime. The elements of the Galois group $G:=text{Gal}(K/mathbb{Q})$ should affect the primitive root of unit $beta$ like potencies. That means it exists an isomorphism
begin{align}
k : G longrightarrow (mathbb{Z}/nmathbb{Z})^x\
sigma mapsto kappa(sigma) + nmathbb{Z}
end{align}
so that $sigma(w) = w^{kappa(sigma)}$. [$(K)^x$ means the group of invertible units in K]
Now I'm unsure how to proceed. Thanks for help.
abstract-algebra prime-numbers roots-of-unity primitive-roots galois-extensions
asked Jan 17 at 10:50
Zorro_CZorro_C
82
$endgroup$
|
show 1 more comment
$begingroup$
Let p be a prime, $xi_p in mathbb{C}$ a primitive p-th unit root and $K = mathbb{Q}(xi_p)$.
Give a normal basis for $K/mathbb{Q}$.
I know, that a basis of $L/K$ (finite and galois) is called a normal basis if it has the form ${sigma(a)}_{sigma in text{Gal}(L/K)}$ for an $a in L$.
A procedure to determine such an element a if $|K| = infty$ is the following:
- Determine the minimal polynomial $m_{beta}(X)$ for a primitive Element $beta$ ($L = K(beta)$)
- Set $g_{sigma}(X) := prodlimits_{substack{tau in G\tau neq sigma}} (X - tau(beta))$ and $A(X) := (g_{tau^{-1}sigma})_{tau,sigma in G}$ with $G := text{Gal}(L/K)$
- Search a $gamma in K$ with $det(A(gamma)) neq 0$
- Set $a := frac{m_{beta}(gamma)}{gamma - beta}$
So far so good. I have only been able to achieve that $Phi_p(X) = frac{X^p - 1}{X-1} = X^{p-1} + . . . + X + 1$ is the minimal polynomial of $xi_p$ over $mathbb{Q}$ since p is prime. The elements of the Galois group $G:=text{Gal}(K/mathbb{Q})$ should affect the primitive root of unit $beta$ like potencies. That means it exists an isomorphism
begin{align}
k : G longrightarrow (mathbb{Z}/nmathbb{Z})^x\
sigma mapsto kappa(sigma) + nmathbb{Z}
end{align}
so that $sigma(w) = w^{kappa(sigma)}$. [$(K)^x$ means the group of invertible units in K]
Now I'm unsure how to proceed. Thanks for help.
abstract-algebra prime-numbers roots-of-unity primitive-roots galois-extensions
asked Jan 17 at 10:50
Zorro_CZorro_C
82
$endgroup$
$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
1
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46
|
show 1 more comment
$begingroup$
Let p be a prime, $xi_p in mathbb{C}$ a primitive p-th unit root and $K = mathbb{Q}(xi_p)$.
Give a normal basis for $K/mathbb{Q}$.
I know, that a basis of $L/K$ (finite and galois) is called a normal basis if it has the form ${sigma(a)}_{sigma in text{Gal}(L/K)}$ for an $a in L$.
A procedure to determine such an element a if $|K| = infty$ is the following:
- Determine the minimal polynomial $m_{beta}(X)$ for a primitive Element $beta$ ($L = K(beta)$)
- Set $g_{sigma}(X) := prodlimits_{substack{tau in G\tau neq sigma}} (X - tau(beta))$ and $A(X) := (g_{tau^{-1}sigma})_{tau,sigma in G}$ with $G := text{Gal}(L/K)$
- Search a $gamma in K$ with $det(A(gamma)) neq 0$
- Set $a := frac{m_{beta}(gamma)}{gamma - beta}$
So far so good. I have only been able to achieve that $Phi_p(X) = frac{X^p - 1}{X-1} = X^{p-1} + . . . + X + 1$ is the minimal polynomial of $xi_p$ over $mathbb{Q}$ since p is prime. The elements of the Galois group $G:=text{Gal}(K/mathbb{Q})$ should affect the primitive root of unit $beta$ like potencies. That means it exists an isomorphism
begin{align}
k : G longrightarrow (mathbb{Z}/nmathbb{Z})^x\
sigma mapsto kappa(sigma) + nmathbb{Z}
end{align}
so that $sigma(w) = w^{kappa(sigma)}$. [$(K)^x$ means the group of invertible units in K]
Now I'm unsure how to proceed. Thanks for help.
abstract-algebra prime-numbers roots-of-unity primitive-roots galois-extensions
asked Jan 17 at 10:50
Zorro_CZorro_C
82
$endgroup$
Let p be a prime, $xi_p in mathbb{C}$ a primitive p-th unit root and $K = mathbb{Q}(xi_p)$.
Give a normal basis for $K/mathbb{Q}$.
I know, that a basis of $L/K$ (finite and galois) is called a normal basis if it has the form ${sigma(a)}_{sigma in text{Gal}(L/K)}$ for an $a in L$.
A procedure to determine such an element a if $|K| = infty$ is the following:
- Determine the minimal polynomial $m_{beta}(X)$ for a primitive Element $beta$ ($L = K(beta)$)
- Set $g_{sigma}(X) := prodlimits_{substack{tau in G\tau neq sigma}} (X - tau(beta))$ and $A(X) := (g_{tau^{-1}sigma})_{tau,sigma in G}$ with $G := text{Gal}(L/K)$
- Search a $gamma in K$ with $det(A(gamma)) neq 0$
- Set $a := frac{m_{beta}(gamma)}{gamma - beta}$
So far so good. I have only been able to achieve that $Phi_p(X) = frac{X^p - 1}{X-1} = X^{p-1} + . . . + X + 1$ is the minimal polynomial of $xi_p$ over $mathbb{Q}$ since p is prime. The elements of the Galois group $G:=text{Gal}(K/mathbb{Q})$ should affect the primitive root of unit $beta$ like potencies. That means it exists an isomorphism
begin{align}
k : G longrightarrow (mathbb{Z}/nmathbb{Z})^x\
sigma mapsto kappa(sigma) + nmathbb{Z}
end{align}
so that $sigma(w) = w^{kappa(sigma)}$. [$(K)^x$ means the group of invertible units in K]
Now I'm unsure how to proceed. Thanks for help.
abstract-algebra prime-numbers roots-of-unity primitive-roots galois-extensions
abstract-algebra prime-numbers roots-of-unity primitive-roots galois-extensions
asked Jan 17 at 10:50
Zorro_CZorro_C
82
asked Jan 17 at 10:50
Zorro_CZorro_C
82
asked Jan 17 at 10:50
Zorro_CZorro_C
82
asked Jan 17 at 10:50
Zorro_CZorro_C
82
asked Jan 17 at 10:50
Zorro_CZorro_C
82
82
$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
1
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46
|
show 1 more comment
$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
1
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46
$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
1
1
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46
|
show 1 more comment
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$begingroup$
Do you know a Q-basis of K?
$endgroup$
– eduard
Jan 17 at 10:58
$begingroup$
Can you tell what is $;zeta_p^k;$ , for $;kinBbb N;$ ? Do you know what the effect of any automorphism of $;K/Bbb Q;$ is over $;zeta_p;$ ? Well, there you go...
$endgroup$
– DonAntonio
Jan 17 at 11:17
$begingroup$
@DonAntonio For all $1 leq k leq p-1$ is $zeta_p^k neq 1$ since $zeta_p$ is primitive. And the effect of an automorphism over $zeta_p$ should be as I already wrote: If $sigma$ is an Automorphism of $K/mathbb{Q}$, so $sigma(zeta_p) = zeta_p^{k(sigma)}$ with $k(sigma) in (mathbb{Z}/pmathbb{Z})^x$ (means: $1 leq k(sigma) leq p-1$)
$endgroup$
– Zorro_C
Jan 17 at 14:30
$begingroup$
@eduard I think ${1, xi_p, xi_p^2, ..., xi_p^{p-1}}$ is a Basis of K? I'm right?
$endgroup$
– Zorro_C
Jan 17 at 14:30
1
$begingroup$
Yeah, I just noticed as well. Think it must be ${xi_p, ..., xi_p^{p-1}}$? And that could be even the normal base, right?
$endgroup$
– Zorro_C
Jan 17 at 14:46