Galois group of $x^3+x+1$ over $mathbb{Q}$ is isomorphic to $S_3 $
$begingroup$
Let E be splitting field of $x^3+x+1 in mathbb{Q}[X]$ over $mathbb{Q}$.
Proof that Gal{E/$mathbb{Q}$}$cong S_3$. Specify all extension fields L with $mathbb{Q} subset L subset E$ and the degree [L : $mathbb{Q}$] of the extensions.
Set $f(x):= x^3+x+1 in mathbb{Q}[X]$. It's easy to see that f is irreducible over $mathbb{Q}$.
At first, I wanted to determine E. With Cardano formula I got the complex roots
$$x_2 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} + i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}quadtext{and}quad x_3 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} - i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}$$
Since f has degree 3, the roots $x_2, x_3$ must be simple. So it exists another real root $x_1$ of f. Therefore we can write $f(x)= (x-x_1)(x-x_2)(x-x_3)$ in $E[x] Rightarrow f$ is separable and E is splitting field of $f Rightarrow E/mathbb{Q}$ is galois.
We got that: $[E:K] = |Gal(E/K)| Leftrightarrow E/K$ is galois. So $|Gal(E/ mathbb{Q})| = [E:mathbb{Q}]$.
Now i don't know how to go on. Can i somehow get the degree of E/$mathbb{Q}$ by means of $x_1, x_2, x_3$? And how can i easily calculate $x_1$ or isn't that necessary to solve the problem?
Thanks a lot!
abstract-algebra galois-theory extension-field galois-extensions
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
$endgroup$
|
show 4 more comments
$begingroup$
Let E be splitting field of $x^3+x+1 in mathbb{Q}[X]$ over $mathbb{Q}$.
Proof that Gal{E/$mathbb{Q}$}$cong S_3$. Specify all extension fields L with $mathbb{Q} subset L subset E$ and the degree [L : $mathbb{Q}$] of the extensions.
Set $f(x):= x^3+x+1 in mathbb{Q}[X]$. It's easy to see that f is irreducible over $mathbb{Q}$.
At first, I wanted to determine E. With Cardano formula I got the complex roots
$$x_2 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} + i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}quadtext{and}quad x_3 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} - i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}$$
Since f has degree 3, the roots $x_2, x_3$ must be simple. So it exists another real root $x_1$ of f. Therefore we can write $f(x)= (x-x_1)(x-x_2)(x-x_3)$ in $E[x] Rightarrow f$ is separable and E is splitting field of $f Rightarrow E/mathbb{Q}$ is galois.
We got that: $[E:K] = |Gal(E/K)| Leftrightarrow E/K$ is galois. So $|Gal(E/ mathbb{Q})| = [E:mathbb{Q}]$.
Now i don't know how to go on. Can i somehow get the degree of E/$mathbb{Q}$ by means of $x_1, x_2, x_3$? And how can i easily calculate $x_1$ or isn't that necessary to solve the problem?
Thanks a lot!
abstract-algebra galois-theory extension-field galois-extensions
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
$endgroup$
3
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
1
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24
|
show 4 more comments
$begingroup$
Let E be splitting field of $x^3+x+1 in mathbb{Q}[X]$ over $mathbb{Q}$.
Proof that Gal{E/$mathbb{Q}$}$cong S_3$. Specify all extension fields L with $mathbb{Q} subset L subset E$ and the degree [L : $mathbb{Q}$] of the extensions.
Set $f(x):= x^3+x+1 in mathbb{Q}[X]$. It's easy to see that f is irreducible over $mathbb{Q}$.
At first, I wanted to determine E. With Cardano formula I got the complex roots
$$x_2 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} + i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}quadtext{and}quad x_3 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} - i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}$$
Since f has degree 3, the roots $x_2, x_3$ must be simple. So it exists another real root $x_1$ of f. Therefore we can write $f(x)= (x-x_1)(x-x_2)(x-x_3)$ in $E[x] Rightarrow f$ is separable and E is splitting field of $f Rightarrow E/mathbb{Q}$ is galois.
We got that: $[E:K] = |Gal(E/K)| Leftrightarrow E/K$ is galois. So $|Gal(E/ mathbb{Q})| = [E:mathbb{Q}]$.
Now i don't know how to go on. Can i somehow get the degree of E/$mathbb{Q}$ by means of $x_1, x_2, x_3$? And how can i easily calculate $x_1$ or isn't that necessary to solve the problem?
Thanks a lot!
abstract-algebra galois-theory extension-field galois-extensions
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
$endgroup$
Let E be splitting field of $x^3+x+1 in mathbb{Q}[X]$ over $mathbb{Q}$.
Proof that Gal{E/$mathbb{Q}$}$cong S_3$. Specify all extension fields L with $mathbb{Q} subset L subset E$ and the degree [L : $mathbb{Q}$] of the extensions.
Set $f(x):= x^3+x+1 in mathbb{Q}[X]$. It's easy to see that f is irreducible over $mathbb{Q}$.
At first, I wanted to determine E. With Cardano formula I got the complex roots
$$x_2 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} + i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}quadtext{and}quad x_3 = sqrt[3]{- frac{1}{2} + sqrt{frac{31}{108}}} - i sqrt[3]{frac{1}{2} + sqrt{frac{31}{108}}}$$
Since f has degree 3, the roots $x_2, x_3$ must be simple. So it exists another real root $x_1$ of f. Therefore we can write $f(x)= (x-x_1)(x-x_2)(x-x_3)$ in $E[x] Rightarrow f$ is separable and E is splitting field of $f Rightarrow E/mathbb{Q}$ is galois.
We got that: $[E:K] = |Gal(E/K)| Leftrightarrow E/K$ is galois. So $|Gal(E/ mathbb{Q})| = [E:mathbb{Q}]$.
Now i don't know how to go on. Can i somehow get the degree of E/$mathbb{Q}$ by means of $x_1, x_2, x_3$? And how can i easily calculate $x_1$ or isn't that necessary to solve the problem?
Thanks a lot!
abstract-algebra galois-theory extension-field galois-extensions
abstract-algebra galois-theory extension-field galois-extensions
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
edited Dec 19 '18 at 12:22
Jyrki Lahtonen
110k13171378
110k13171378
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
asked Dec 19 '18 at 11:33
Zorro_CZorro_C
82
82
3
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
1
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24
|
show 4 more comments
3
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
1
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24
3
3
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
1
1
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24
|
show 4 more comments
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3
$begingroup$
See this duplicate and the article of Keith Conrad here, where he determines the Galois group for all cubic (and quartic) polynomials. Then it is clear how to "go on".
$endgroup$
– Dietrich Burde
Dec 19 '18 at 11:47
1
$begingroup$
For the first part, see math.stackexchange.com/a/2809524/589
$endgroup$
– lhf
Dec 19 '18 at 11:49
$begingroup$
Is it possible that the Galois group of a polynomial of degree 3 has cardinality $6$ ?
$endgroup$
– NewMath
Dec 19 '18 at 11:56
$begingroup$
@NewMath Yes, $S_3$ has $6$ elements. Please click on the link given by lhf (and read his answer).
$endgroup$
– Dietrich Burde
Dec 19 '18 at 12:00
$begingroup$
Vieta's formulas make it particularly easy to find the $n$th zero of a degree $n$ polynomial if you already know $n-1$ of them.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 12:24