Mixing
Galois group of a certain splitting field
$begingroup$
Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.
Here are the steps that I have taken.
- The minimum polynomial is $x^4-6x^2+7$.
- The roots of this are $pm sqrt{3 pm sqrt{2}}$.
- I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.
Any hints appreciated, Thanks!
abstract-algebra galois-theory minimal-polynomials
edited Dec 26 '18 at 21:46
user26857
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
$endgroup$
add a comment |
$begingroup$
Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.
Here are the steps that I have taken.
- The minimum polynomial is $x^4-6x^2+7$.
- The roots of this are $pm sqrt{3 pm sqrt{2}}$.
- I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.
Any hints appreciated, Thanks!
abstract-algebra galois-theory minimal-polynomials
edited Dec 26 '18 at 21:46
user26857
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
$endgroup$
add a comment |
$begingroup$
Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.
Here are the steps that I have taken.
- The minimum polynomial is $x^4-6x^2+7$.
- The roots of this are $pm sqrt{3 pm sqrt{2}}$.
- I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.
Any hints appreciated, Thanks!
abstract-algebra galois-theory minimal-polynomials
edited Dec 26 '18 at 21:46
user26857
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
$endgroup$
Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.
Here are the steps that I have taken.
- The minimum polynomial is $x^4-6x^2+7$.
- The roots of this are $pm sqrt{3 pm sqrt{2}}$.
- I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.
Any hints appreciated, Thanks!
abstract-algebra galois-theory minimal-polynomials
abstract-algebra galois-theory minimal-polynomials
edited Dec 26 '18 at 21:46
user26857
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
edited Dec 26 '18 at 21:46
user26857
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
edited Dec 26 '18 at 21:46
user26857
39.3k124183
edited Dec 26 '18 at 21:46
user26857
39.3k124183
edited Dec 26 '18 at 21:46
user26857
39.3k124183
39.3k124183
asked Dec 24 '18 at 22:12
sarafisarafi
884
asked Dec 24 '18 at 22:12
sarafisarafi
884
asked Dec 24 '18 at 22:12
sarafisarafi
884
884
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $G=Gal(K/mathbb{Q})$.
Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.
The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.
Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
$endgroup$
add a comment |
$begingroup$
Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:
- If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.
- If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.
- If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)
The proof is not relevant to answering my question so is discarded.
An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.
An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.
My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.
We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
$alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.
It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
$sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).
But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.
answered Jan 3 at 22:15
sarafisarafi
884
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $G=Gal(K/mathbb{Q})$.
Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.
The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.
Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
$endgroup$
add a comment |
$begingroup$
Let $G=Gal(K/mathbb{Q})$.
Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.
The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.
Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
$endgroup$
add a comment |
$begingroup$
Let $G=Gal(K/mathbb{Q})$.
Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.
The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.
Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
$endgroup$
Let $G=Gal(K/mathbb{Q})$.
Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.
The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.
Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
answered Dec 24 '18 at 22:26
MaxMax
13.3k11040
13.3k11040
add a comment |
add a comment |
$begingroup$
Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:
- If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.
- If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.
- If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)
The proof is not relevant to answering my question so is discarded.
An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.
An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.
My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.
We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
$alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.
It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
$sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).
But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.
answered Jan 3 at 22:15
sarafisarafi
884
$endgroup$
add a comment |
$begingroup$
Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:
- If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.
- If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.
- If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)
The proof is not relevant to answering my question so is discarded.
An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.
An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.
My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.
We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
$alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.
It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
$sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).
But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.
answered Jan 3 at 22:15
sarafisarafi
884
$endgroup$
add a comment |
$begingroup$
Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:
- If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.
- If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.
- If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)
The proof is not relevant to answering my question so is discarded.
An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.
An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.
My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.
We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
$alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.
It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
$sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).
But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.
answered Jan 3 at 22:15
sarafisarafi
884
$endgroup$
Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:
- If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.
- If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.
- If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)
The proof is not relevant to answering my question so is discarded.
An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.
An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.
My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.
We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
$alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.
It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
$sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).
But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.
answered Jan 3 at 22:15
sarafisarafi
884
answered Jan 3 at 22:15
sarafisarafi
884
answered Jan 3 at 22:15
sarafisarafi
884
answered Jan 3 at 22:15
sarafisarafi
884
884
add a comment |
add a comment |
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