Mixing
Is it true that $Gal(K/F)cong S_{n_1}times cdots S_{n_k}$?
$begingroup$
I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) cong S_n$?
Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)cong S_{n_1}times cdots S_{n_k}$? If so then what is the arguement?
For $Bbb Q(sqrt 2, sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)cong S_2 times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)cong S_3$.
group-theory field-theory galois-theory galois-extensions separable-extension
asked Jan 29 at 23:46
GimgimGimgim
31513
$endgroup$
add a comment |
$begingroup$
I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) cong S_n$?
Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)cong S_{n_1}times cdots S_{n_k}$? If so then what is the arguement?
For $Bbb Q(sqrt 2, sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)cong S_2 times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)cong S_3$.
group-theory field-theory galois-theory galois-extensions separable-extension
asked Jan 29 at 23:46
GimgimGimgim
31513
$endgroup$
$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29
add a comment |
$begingroup$
I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) cong S_n$?
Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)cong S_{n_1}times cdots S_{n_k}$? If so then what is the arguement?
For $Bbb Q(sqrt 2, sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)cong S_2 times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)cong S_3$.
group-theory field-theory galois-theory galois-extensions separable-extension
asked Jan 29 at 23:46
GimgimGimgim
31513
$endgroup$
I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) cong S_n$?
Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)cong S_{n_1}times cdots S_{n_k}$? If so then what is the arguement?
For $Bbb Q(sqrt 2, sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)cong S_2 times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)cong S_3$.
group-theory field-theory galois-theory galois-extensions separable-extension
group-theory field-theory galois-theory galois-extensions separable-extension
asked Jan 29 at 23:46
GimgimGimgim
31513
asked Jan 29 at 23:46
GimgimGimgim
31513
asked Jan 29 at 23:46
GimgimGimgim
31513
asked Jan 29 at 23:46
GimgimGimgim
31513
asked Jan 29 at 23:46
GimgimGimgim
31513
31513
$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29
add a comment |
$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29
$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29
add a comment |
2 Answers
2
active
oldest
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$begingroup$
This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $nphi(n)$ or $nphi(n)/2$.
https://mathoverflow.net/questions/143739/galois-group-of-xn-2
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $Bbb Q$ we have $Bbb Q(psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $psi_p$ so arbitrary permutation is not possible at all.
answered Jan 30 at 3:25
GimgimGimgim
31513
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $nphi(n)$ or $nphi(n)/2$.
https://mathoverflow.net/questions/143739/galois-group-of-xn-2
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $nphi(n)$ or $nphi(n)/2$.
https://mathoverflow.net/questions/143739/galois-group-of-xn-2
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
add a comment |
$begingroup$
This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $nphi(n)$ or $nphi(n)/2$.
https://mathoverflow.net/questions/143739/galois-group-of-xn-2
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $nphi(n)$ or $nphi(n)/2$.
https://mathoverflow.net/questions/143739/galois-group-of-xn-2
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 0:04
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
add a comment |
add a comment |
$begingroup$
After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $Bbb Q$ we have $Bbb Q(psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $psi_p$ so arbitrary permutation is not possible at all.
answered Jan 30 at 3:25
GimgimGimgim
31513
$endgroup$
add a comment |
$begingroup$
After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $Bbb Q$ we have $Bbb Q(psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $psi_p$ so arbitrary permutation is not possible at all.
answered Jan 30 at 3:25
GimgimGimgim
31513
$endgroup$
add a comment |
$begingroup$
After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $Bbb Q$ we have $Bbb Q(psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $psi_p$ so arbitrary permutation is not possible at all.
answered Jan 30 at 3:25
GimgimGimgim
31513
$endgroup$
After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $Bbb Q$ we have $Bbb Q(psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $psi_p$ so arbitrary permutation is not possible at all.
answered Jan 30 at 3:25
GimgimGimgim
31513
answered Jan 30 at 3:25
GimgimGimgim
31513
answered Jan 30 at 3:25
GimgimGimgim
31513
answered Jan 30 at 3:25
GimgimGimgim
31513
31513
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$begingroup$
$X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$
$endgroup$
– reuns
Jan 30 at 0:11
$begingroup$
What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$.
$endgroup$
– Mathmo123
Feb 25 at 15:29