Mixing
Finding the Normal Basis of Cyclotomic field
$begingroup$
So let $p$ be a prime number and $zeta_p$ the p-th roots of unity. I want to proof that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is the normal basis of $mathbb{Q}(zeta_p)/mathbb{Q}$ (which I hope is true ...).
First of all, I know that $B$ is a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$ since
- the elements of $B$ are $mathbb{Q}$-linearly independent, because $zeta_p$ is a $p$-th primitive roots of unity,
- and $B$ has exactly $p-1$ elements and the degree of $mathbb{Q}(zeta_p)$ over $mathbb{Q}$ is $p-1$ too, because the Galois group of $mathbb{Q}(zeta_p)/mathbb{Q}$ is isomorphic to $(mathbb{Z}/pmathbb{Z})^times$ which has $p-1$ elements.
Now, I have trouble proving that $B$ is indeed a normal basis. According to the definition of a normal basis, if we let $sigma_i in text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$, I would have to find an $a in mathbb{Q}(zeta_p)$ such that
$$
{sigma_1(a),sigma_2(a), dots, sigma_{p-1}(a)}
$$
forms a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$.
Since $text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$ is cyclic, there is a generating automorphism which I will just call $sigma$. Then, just reformulating the definition, we would have
$$
{sigma(a),sigma^{2}(a), dots, sigma^{p-1}(a)} text{.}
$$
Now, I have troubles determining this element $a$.
Right now, I believe that $a = zeta_p$ in which case, we would have
$${sigma(zeta_p),sigma^{2}(zeta_p), dots, sigma^{p-1}(zeta_p)}text{.}$$
So, my questions are
- Am I right with my assumption to set $a = zeta_p$?
- And if so, how do I proceed with my proof in order to show that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is a normal basis?
abstract-algebra field-theory roots-of-unity cyclotomic-fields
asked Jan 17 at 21:36
mattmatt
857
$endgroup$
add a comment |
$begingroup$
So let $p$ be a prime number and $zeta_p$ the p-th roots of unity. I want to proof that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is the normal basis of $mathbb{Q}(zeta_p)/mathbb{Q}$ (which I hope is true ...).
First of all, I know that $B$ is a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$ since
- the elements of $B$ are $mathbb{Q}$-linearly independent, because $zeta_p$ is a $p$-th primitive roots of unity,
- and $B$ has exactly $p-1$ elements and the degree of $mathbb{Q}(zeta_p)$ over $mathbb{Q}$ is $p-1$ too, because the Galois group of $mathbb{Q}(zeta_p)/mathbb{Q}$ is isomorphic to $(mathbb{Z}/pmathbb{Z})^times$ which has $p-1$ elements.
Now, I have trouble proving that $B$ is indeed a normal basis. According to the definition of a normal basis, if we let $sigma_i in text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$, I would have to find an $a in mathbb{Q}(zeta_p)$ such that
$$
{sigma_1(a),sigma_2(a), dots, sigma_{p-1}(a)}
$$
forms a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$.
Since $text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$ is cyclic, there is a generating automorphism which I will just call $sigma$. Then, just reformulating the definition, we would have
$$
{sigma(a),sigma^{2}(a), dots, sigma^{p-1}(a)} text{.}
$$
Now, I have troubles determining this element $a$.
Right now, I believe that $a = zeta_p$ in which case, we would have
$${sigma(zeta_p),sigma^{2}(zeta_p), dots, sigma^{p-1}(zeta_p)}text{.}$$
So, my questions are
- Am I right with my assumption to set $a = zeta_p$?
- And if so, how do I proceed with my proof in order to show that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is a normal basis?
abstract-algebra field-theory roots-of-unity cyclotomic-fields
asked Jan 17 at 21:36
mattmatt
857
$endgroup$
add a comment |
$begingroup$
So let $p$ be a prime number and $zeta_p$ the p-th roots of unity. I want to proof that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is the normal basis of $mathbb{Q}(zeta_p)/mathbb{Q}$ (which I hope is true ...).
First of all, I know that $B$ is a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$ since
- the elements of $B$ are $mathbb{Q}$-linearly independent, because $zeta_p$ is a $p$-th primitive roots of unity,
- and $B$ has exactly $p-1$ elements and the degree of $mathbb{Q}(zeta_p)$ over $mathbb{Q}$ is $p-1$ too, because the Galois group of $mathbb{Q}(zeta_p)/mathbb{Q}$ is isomorphic to $(mathbb{Z}/pmathbb{Z})^times$ which has $p-1$ elements.
Now, I have trouble proving that $B$ is indeed a normal basis. According to the definition of a normal basis, if we let $sigma_i in text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$, I would have to find an $a in mathbb{Q}(zeta_p)$ such that
$$
{sigma_1(a),sigma_2(a), dots, sigma_{p-1}(a)}
$$
forms a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$.
Since $text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$ is cyclic, there is a generating automorphism which I will just call $sigma$. Then, just reformulating the definition, we would have
$$
{sigma(a),sigma^{2}(a), dots, sigma^{p-1}(a)} text{.}
$$
Now, I have troubles determining this element $a$.
Right now, I believe that $a = zeta_p$ in which case, we would have
$${sigma(zeta_p),sigma^{2}(zeta_p), dots, sigma^{p-1}(zeta_p)}text{.}$$
So, my questions are
- Am I right with my assumption to set $a = zeta_p$?
- And if so, how do I proceed with my proof in order to show that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is a normal basis?
abstract-algebra field-theory roots-of-unity cyclotomic-fields
asked Jan 17 at 21:36
mattmatt
857
$endgroup$
So let $p$ be a prime number and $zeta_p$ the p-th roots of unity. I want to proof that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is the normal basis of $mathbb{Q}(zeta_p)/mathbb{Q}$ (which I hope is true ...).
First of all, I know that $B$ is a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$ since
- the elements of $B$ are $mathbb{Q}$-linearly independent, because $zeta_p$ is a $p$-th primitive roots of unity,
- and $B$ has exactly $p-1$ elements and the degree of $mathbb{Q}(zeta_p)$ over $mathbb{Q}$ is $p-1$ too, because the Galois group of $mathbb{Q}(zeta_p)/mathbb{Q}$ is isomorphic to $(mathbb{Z}/pmathbb{Z})^times$ which has $p-1$ elements.
Now, I have trouble proving that $B$ is indeed a normal basis. According to the definition of a normal basis, if we let $sigma_i in text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$, I would have to find an $a in mathbb{Q}(zeta_p)$ such that
$$
{sigma_1(a),sigma_2(a), dots, sigma_{p-1}(a)}
$$
forms a $mathbb{Q}$-basis of $mathbb{Q}(zeta_p)$.
Since $text{Gal}(mathbb{Q}(zeta_p)/mathbb{Q})$ is cyclic, there is a generating automorphism which I will just call $sigma$. Then, just reformulating the definition, we would have
$$
{sigma(a),sigma^{2}(a), dots, sigma^{p-1}(a)} text{.}
$$
Now, I have troubles determining this element $a$.
Right now, I believe that $a = zeta_p$ in which case, we would have
$${sigma(zeta_p),sigma^{2}(zeta_p), dots, sigma^{p-1}(zeta_p)}text{.}$$
So, my questions are
- Am I right with my assumption to set $a = zeta_p$?
- And if so, how do I proceed with my proof in order to show that $ B = { zeta_p, zeta_{p}^{2}, dots, zeta_{p}^{p-1} }$ is a normal basis?
abstract-algebra field-theory roots-of-unity cyclotomic-fields
abstract-algebra field-theory roots-of-unity cyclotomic-fields
asked Jan 17 at 21:36
mattmatt
857
asked Jan 17 at 21:36
mattmatt
857
asked Jan 17 at 21:36
mattmatt
857
asked Jan 17 at 21:36
mattmatt
857
asked Jan 17 at 21:36
mattmatt
857
857
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $zeta_p^j$ ($1le jle p-1$) are the zeros of the $p$-th cyclotomic
polynomial $Phi_p(X)=X^{p-1}+X^{p-2}+cdots+X+1$, which is well-known to be
irreducible over $Bbb Q$. Thus the Galois group of $Bbb Q(zeta_p)$ acts
transitively on the zeros of $Phi_p(X)$. Thus there is a Galois group
element $sigma_j$ with $sigma_j(zeta_p)=zeta_p^j$. This is unique:
its action on $zeta_p$ determines its action on all of $Bbb Q(zeta_p)$.
So $B={sigma_1(zeta_p),sigma_2(zeta_p),cdots,sigma_{p-1}(zeta_p)}$
really is a normal basis.
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
add a comment |
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$begingroup$
The $zeta_p^j$ ($1le jle p-1$) are the zeros of the $p$-th cyclotomic
polynomial $Phi_p(X)=X^{p-1}+X^{p-2}+cdots+X+1$, which is well-known to be
irreducible over $Bbb Q$. Thus the Galois group of $Bbb Q(zeta_p)$ acts
transitively on the zeros of $Phi_p(X)$. Thus there is a Galois group
element $sigma_j$ with $sigma_j(zeta_p)=zeta_p^j$. This is unique:
its action on $zeta_p$ determines its action on all of $Bbb Q(zeta_p)$.
So $B={sigma_1(zeta_p),sigma_2(zeta_p),cdots,sigma_{p-1}(zeta_p)}$
really is a normal basis.
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
add a comment |
$begingroup$
The $zeta_p^j$ ($1le jle p-1$) are the zeros of the $p$-th cyclotomic
polynomial $Phi_p(X)=X^{p-1}+X^{p-2}+cdots+X+1$, which is well-known to be
irreducible over $Bbb Q$. Thus the Galois group of $Bbb Q(zeta_p)$ acts
transitively on the zeros of $Phi_p(X)$. Thus there is a Galois group
element $sigma_j$ with $sigma_j(zeta_p)=zeta_p^j$. This is unique:
its action on $zeta_p$ determines its action on all of $Bbb Q(zeta_p)$.
So $B={sigma_1(zeta_p),sigma_2(zeta_p),cdots,sigma_{p-1}(zeta_p)}$
really is a normal basis.
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
add a comment |
$begingroup$
The $zeta_p^j$ ($1le jle p-1$) are the zeros of the $p$-th cyclotomic
polynomial $Phi_p(X)=X^{p-1}+X^{p-2}+cdots+X+1$, which is well-known to be
irreducible over $Bbb Q$. Thus the Galois group of $Bbb Q(zeta_p)$ acts
transitively on the zeros of $Phi_p(X)$. Thus there is a Galois group
element $sigma_j$ with $sigma_j(zeta_p)=zeta_p^j$. This is unique:
its action on $zeta_p$ determines its action on all of $Bbb Q(zeta_p)$.
So $B={sigma_1(zeta_p),sigma_2(zeta_p),cdots,sigma_{p-1}(zeta_p)}$
really is a normal basis.
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
The $zeta_p^j$ ($1le jle p-1$) are the zeros of the $p$-th cyclotomic
polynomial $Phi_p(X)=X^{p-1}+X^{p-2}+cdots+X+1$, which is well-known to be
irreducible over $Bbb Q$. Thus the Galois group of $Bbb Q(zeta_p)$ acts
transitively on the zeros of $Phi_p(X)$. Thus there is a Galois group
element $sigma_j$ with $sigma_j(zeta_p)=zeta_p^j$. This is unique:
its action on $zeta_p$ determines its action on all of $Bbb Q(zeta_p)$.
So $B={sigma_1(zeta_p),sigma_2(zeta_p),cdots,sigma_{p-1}(zeta_p)}$
really is a normal basis.
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 17 at 22:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
add a comment |
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $sigma(zeta_p)$ is also a root of $Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $zeta_p$ determines its action on all of $mathbb{Q}(zeta_p)$? Thank you.
$endgroup$
– matt
Jan 17 at 23:48
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
I mean that $sigma(zeta)$ is also a zero of $Phi_p$, and all such zeroes arise. If $sigma$ takes $zeta$ to $zeta'$, then it takes $a_0+a_1zeta+a_2zeta^2+cdots$ to $a_0+a_1zeta'+a_2zeta'^2+cdots$ where the $a_iinBbb Q$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:49
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
$begingroup$
Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $zeta_p$ as a zero, contradicting irreducibility of $Phi_p(X)$.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 21:52
add a comment |
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