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$(m,n)=1$, then cyclotomic field extension(non-standard) necessarily disjoint?
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Let $F$ be a field. There are two cases, either $char(F)=0$ or $(char(F),m)=1$ where $m$ will be specified.
$F[m]$ is defined to be splitting field of $x^m-1$.(In either case, $F[m]$ is galois extension due to separability.)
Let $F[m_1],F[m_2]$ be two such field s.t. $(m_1,m_2)=Z$.(i.e. They are coprime.)
$textbfQ:$ Do I know $F[m_1]cap F[m_2]=F$? If not, what is counter example? Note that this is not about $F=Q$. $F=Q$'s case is clear as there is Minkowski lower bound implies on discriminant which will indicate at least one prime ramify. Ramification of prime indicates discriminant non-coprime which is contradiction for $F=Q$. Other than $Q$, do I still have linearly disjointness?
abstract-algebra number-theory
asked Jul 21 at 22:24
user45765
2,1942718
add a comment |Â
up vote
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Let $F$ be a field. There are two cases, either $char(F)=0$ or $(char(F),m)=1$ where $m$ will be specified.
$F[m]$ is defined to be splitting field of $x^m-1$.(In either case, $F[m]$ is galois extension due to separability.)
Let $F[m_1],F[m_2]$ be two such field s.t. $(m_1,m_2)=Z$.(i.e. They are coprime.)
$textbfQ:$ Do I know $F[m_1]cap F[m_2]=F$? If not, what is counter example? Note that this is not about $F=Q$. $F=Q$'s case is clear as there is Minkowski lower bound implies on discriminant which will indicate at least one prime ramify. Ramification of prime indicates discriminant non-coprime which is contradiction for $F=Q$. Other than $Q$, do I still have linearly disjointness?
abstract-algebra number-theory
asked Jul 21 at 22:24
user45765
2,1942718
1
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
1
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
1
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $F$ be a field. There are two cases, either $char(F)=0$ or $(char(F),m)=1$ where $m$ will be specified.
$F[m]$ is defined to be splitting field of $x^m-1$.(In either case, $F[m]$ is galois extension due to separability.)
Let $F[m_1],F[m_2]$ be two such field s.t. $(m_1,m_2)=Z$.(i.e. They are coprime.)
$textbfQ:$ Do I know $F[m_1]cap F[m_2]=F$? If not, what is counter example? Note that this is not about $F=Q$. $F=Q$'s case is clear as there is Minkowski lower bound implies on discriminant which will indicate at least one prime ramify. Ramification of prime indicates discriminant non-coprime which is contradiction for $F=Q$. Other than $Q$, do I still have linearly disjointness?
abstract-algebra number-theory
asked Jul 21 at 22:24
user45765
2,1942718
Let $F$ be a field. There are two cases, either $char(F)=0$ or $(char(F),m)=1$ where $m$ will be specified.
$F[m]$ is defined to be splitting field of $x^m-1$.(In either case, $F[m]$ is galois extension due to separability.)
Let $F[m_1],F[m_2]$ be two such field s.t. $(m_1,m_2)=Z$.(i.e. They are coprime.)
$textbfQ:$ Do I know $F[m_1]cap F[m_2]=F$? If not, what is counter example? Note that this is not about $F=Q$. $F=Q$'s case is clear as there is Minkowski lower bound implies on discriminant which will indicate at least one prime ramify. Ramification of prime indicates discriminant non-coprime which is contradiction for $F=Q$. Other than $Q$, do I still have linearly disjointness?
abstract-algebra number-theory
asked Jul 21 at 22:24
user45765
2,1942718
asked Jul 21 at 22:24
user45765
2,1942718
asked Jul 21 at 22:24
user45765
2,1942718
asked Jul 21 at 22:24
user45765
2,1942718
2,1942718
1
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
1
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
1
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12
add a comment |Â
1
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
1
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
1
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12
1
1
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
1
1
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
1
1
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12
add a comment |Â
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1
What about the finite field F_5 and m1 = 8 and m2 = 3? My impression is that since 8 and 3 don't divide 5 - 1 = 4, the polynomials don't split in F_5, but since they do divide 25 - 1, they will split in F_25. Do you believe this example?
– CJD
Jul 21 at 22:33
@CJD Yes. Why $8notvert 4$ implies the polynomial does not split in $F_5$ here? What is the reason for that statement? I did not thought that way. I thought $F_5[8]$ is a degree $4$ extension. If this degree 4 polynomial splits, then $4|5$ but this is not possible. Hence it does not split. Similarly for $F_5[3]$.
– user45765
Jul 21 at 22:40
1
It looks like this link might be helpful: en.wikipedia.org/wiki/Root_of_unity_modulo_n You're not right that the splitting field of $x^8 - 1$ over $F_5$ is degree 4, it's degree 2. I don't see an instant way to prove that, but the idea is that an element satisfying $x^8 - 1$ is an element of multiplicative order dividing 8, and since the multiplicative group of $F_5^2$ is cyclic of order 24, the degree 2 extension is the smallest extension in which the polynomial splits. (I don't claim these claims are obvious or that I've given a full proof.)
– CJD
Jul 21 at 22:48
1
You must have $Phi_8=X^4+1$ splitting into two quadratics over $Bbb F_5$, indeed, since $X^2+1=(X+2)(X+3)$ you get $X^4+1=(X^2+2)(X^2+3)$.
– Lubin
Jul 22 at 1:12