About the Artin symbol
$begingroup$
Let $L$ be a finite abelian extension of $mathbb{Q}$ and let $m$ be a positive integer such that $Lsubsetmathbb{Q}(zeta)$, where $zeta$ is a primitive $m$-th root of unity. Let $a$ be an integer coprime to $m$. Then the Artin symbol $(frac{L}{a})$ is the automorphism of $L$ obtained by restricting to $L$ the automorphism $phi$ of $mathbb{Q}(zeta)$ determined by $(zetamapstozeta^a)$.
My question is, why is, $phi(L)subset L$?
class-field-theory
asked Dec 27 '18 at 12:30
saisai
1376
$endgroup$
|
show 4 more comments
$begingroup$
Let $L$ be a finite abelian extension of $mathbb{Q}$ and let $m$ be a positive integer such that $Lsubsetmathbb{Q}(zeta)$, where $zeta$ is a primitive $m$-th root of unity. Let $a$ be an integer coprime to $m$. Then the Artin symbol $(frac{L}{a})$ is the automorphism of $L$ obtained by restricting to $L$ the automorphism $phi$ of $mathbb{Q}(zeta)$ determined by $(zetamapstozeta^a)$.
My question is, why is, $phi(L)subset L$?
class-field-theory
asked Dec 27 '18 at 12:30
saisai
1376
$endgroup$
$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
2
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
1
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00
|
show 4 more comments
$begingroup$
Let $L$ be a finite abelian extension of $mathbb{Q}$ and let $m$ be a positive integer such that $Lsubsetmathbb{Q}(zeta)$, where $zeta$ is a primitive $m$-th root of unity. Let $a$ be an integer coprime to $m$. Then the Artin symbol $(frac{L}{a})$ is the automorphism of $L$ obtained by restricting to $L$ the automorphism $phi$ of $mathbb{Q}(zeta)$ determined by $(zetamapstozeta^a)$.
My question is, why is, $phi(L)subset L$?
class-field-theory
asked Dec 27 '18 at 12:30
saisai
1376
$endgroup$
Let $L$ be a finite abelian extension of $mathbb{Q}$ and let $m$ be a positive integer such that $Lsubsetmathbb{Q}(zeta)$, where $zeta$ is a primitive $m$-th root of unity. Let $a$ be an integer coprime to $m$. Then the Artin symbol $(frac{L}{a})$ is the automorphism of $L$ obtained by restricting to $L$ the automorphism $phi$ of $mathbb{Q}(zeta)$ determined by $(zetamapstozeta^a)$.
My question is, why is, $phi(L)subset L$?
class-field-theory
class-field-theory
asked Dec 27 '18 at 12:30
saisai
1376
asked Dec 27 '18 at 12:30
saisai
1376
asked Dec 27 '18 at 12:30
saisai
1376
asked Dec 27 '18 at 12:30
saisai
1376
asked Dec 27 '18 at 12:30
saisai
1376
1376
$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
2
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
1
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00
|
show 4 more comments
$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
2
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
1
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00
$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
2
2
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
1
1
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00
|
show 4 more comments
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$begingroup$
what are your thoughts?
$endgroup$
– mathworker21
Dec 27 '18 at 12:31
2
$begingroup$
Hint: what does it mean for an extension to be abelian?
$endgroup$
– Wojowu
Dec 27 '18 at 12:34
1
$begingroup$
@Wojowu: Thanks for the hint. $phi$ restricts to a $mathbb{Q}$-homomorphism of $Ltophi(L)$. Since $L/mathbb{Q}$ is Galois, $phi(L)subset L$. Is this correct? Also, it seems to me that $mathbb{Q}(zeta)/mathbb{Q}$ is not always a cyclic Galois extension, right?
$endgroup$
– sai
Dec 27 '18 at 12:51
$begingroup$
This is correct. Cyclotomic extensions are always Galois (note the conjugates of $zeta$ are powers of $zeta$)
$endgroup$
– Wojowu
Dec 27 '18 at 12:53
$begingroup$
Also, I think the Artin symbol $(frac{L}{a})$ depends not only on $a$, but also on $m$, am I right?
$endgroup$
– sai
Dec 27 '18 at 13:00