Determine Gal$(mathbb{Q}(sqrt[8]{7},i)/mathbb{Q}(sqrt{7}))$ and...
$begingroup$
Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.
(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.
(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.
I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
$$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!
abstract-algebra galois-theory
asked Jan 4 at 20:49
AlexAlex
757
$endgroup$
add a comment |
$begingroup$
Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.
(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.
(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.
I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
$$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!
abstract-algebra galois-theory
asked Jan 4 at 20:49
AlexAlex
757
$endgroup$
add a comment |
$begingroup$
Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.
(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.
(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.
I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
$$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!
abstract-algebra galois-theory
asked Jan 4 at 20:49
AlexAlex
757
$endgroup$
Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.
(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.
(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.
I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
$$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!
abstract-algebra galois-theory
abstract-algebra galois-theory
asked Jan 4 at 20:49
AlexAlex
757
asked Jan 4 at 20:49
AlexAlex
757
edited Jan 4 at 23:12
Alex
asked Jan 4 at 20:49
AlexAlex
757
asked Jan 4 at 20:49
AlexAlex
757
asked Jan 4 at 20:49
AlexAlex
757
757
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
is Galois over $F$, being the splitting field of $x^4-a$. Your first
example has $a=sqrt7$.
The Galois group of $K/F_1$ has order $8$, and computing it is very similar
to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
dihedral of order $8$.
I don't think $K/F_2$ is Galois.
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
is Galois over $F$, being the splitting field of $x^4-a$. Your first
example has $a=sqrt7$.
The Galois group of $K/F_1$ has order $8$, and computing it is very similar
to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
dihedral of order $8$.
I don't think $K/F_2$ is Galois.
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
add a comment |
$begingroup$
If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
is Galois over $F$, being the splitting field of $x^4-a$. Your first
example has $a=sqrt7$.
The Galois group of $K/F_1$ has order $8$, and computing it is very similar
to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
dihedral of order $8$.
I don't think $K/F_2$ is Galois.
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
add a comment |
$begingroup$
If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
is Galois over $F$, being the splitting field of $x^4-a$. Your first
example has $a=sqrt7$.
The Galois group of $K/F_1$ has order $8$, and computing it is very similar
to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
dihedral of order $8$.
I don't think $K/F_2$ is Galois.
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
is Galois over $F$, being the splitting field of $x^4-a$. Your first
example has $a=sqrt7$.
The Galois group of $K/F_1$ has order $8$, and computing it is very similar
to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
dihedral of order $8$.
I don't think $K/F_2$ is Galois.
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 4 at 21:20
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
add a comment |
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
You only need $operatorname{char}(F) ne 2$
$endgroup$
– Kenny Lau
Jan 4 at 22:08
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
$endgroup$
– Alex
Jan 5 at 7:31
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@Alex You asked what the Galois group of a non-Galois extension is....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:34
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
@LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
$endgroup$
– Alex
Jan 5 at 7:38
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
$begingroup$
Exams have misprints, just as all other types of writing. @Alex
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:40
add a comment |
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