Determine $[K:mathbb{Q}]$ and show that $Gal(K/mathbb{Q})$ is non-abelian.
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Question: Let $K$ be a splitting field for $x^5-2$ over $mathbb{Q}.$
$(1)$ Determine $[K:mathbb{Q}].$
$(2)$ Show that $Gal(K/mathbb{Q})$ is non-abelian.
For $(1),$ note that roots of $x^5-2$ are
$$sqrt[5]{2}, sqrt[5]{2}omega, sqrt[5]{2}omega^2, sqrt[5]{2}omega^3, sqrt[5]{2}omega^4$$
where $omega$ is the $5$th root of unity, that is,
$$omega = e^{frac{2pi i}{5}}.$$
So we have
$$K = mathbb{Q}(sqrt[5]{2}, omega),$$
implying that
$$[K:mathbb{Q}] = [K:mathbb{Q}(omega)] cdot [mathbb{Q}(omega):mathbb{Q}] = 5 times 4 = 20.$$
For $(2),$ I know that $Gal(K/mathbb{Q})$ is isomorphic to the Dihedral group with $20$ elements.
Since such Dihedral group is non-abelian, so is $Gal(K/mathbb{Q}).$
However, I do now know how to prove it.
abstract-algebra proof-verification field-theory galois-theory
asked Nov 24 at 5:10
Idonknow
2,282748111
|
show 5 more comments
up vote
2
down vote
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Question: Let $K$ be a splitting field for $x^5-2$ over $mathbb{Q}.$
$(1)$ Determine $[K:mathbb{Q}].$
$(2)$ Show that $Gal(K/mathbb{Q})$ is non-abelian.
For $(1),$ note that roots of $x^5-2$ are
$$sqrt[5]{2}, sqrt[5]{2}omega, sqrt[5]{2}omega^2, sqrt[5]{2}omega^3, sqrt[5]{2}omega^4$$
where $omega$ is the $5$th root of unity, that is,
$$omega = e^{frac{2pi i}{5}}.$$
So we have
$$K = mathbb{Q}(sqrt[5]{2}, omega),$$
implying that
$$[K:mathbb{Q}] = [K:mathbb{Q}(omega)] cdot [mathbb{Q}(omega):mathbb{Q}] = 5 times 4 = 20.$$
For $(2),$ I know that $Gal(K/mathbb{Q})$ is isomorphic to the Dihedral group with $20$ elements.
Since such Dihedral group is non-abelian, so is $Gal(K/mathbb{Q}).$
However, I do now know how to prove it.
abstract-algebra proof-verification field-theory galois-theory
asked Nov 24 at 5:10
Idonknow
2,282748111
$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16
|
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question: Let $K$ be a splitting field for $x^5-2$ over $mathbb{Q}.$
$(1)$ Determine $[K:mathbb{Q}].$
$(2)$ Show that $Gal(K/mathbb{Q})$ is non-abelian.
For $(1),$ note that roots of $x^5-2$ are
$$sqrt[5]{2}, sqrt[5]{2}omega, sqrt[5]{2}omega^2, sqrt[5]{2}omega^3, sqrt[5]{2}omega^4$$
where $omega$ is the $5$th root of unity, that is,
$$omega = e^{frac{2pi i}{5}}.$$
So we have
$$K = mathbb{Q}(sqrt[5]{2}, omega),$$
implying that
$$[K:mathbb{Q}] = [K:mathbb{Q}(omega)] cdot [mathbb{Q}(omega):mathbb{Q}] = 5 times 4 = 20.$$
For $(2),$ I know that $Gal(K/mathbb{Q})$ is isomorphic to the Dihedral group with $20$ elements.
Since such Dihedral group is non-abelian, so is $Gal(K/mathbb{Q}).$
However, I do now know how to prove it.
abstract-algebra proof-verification field-theory galois-theory
asked Nov 24 at 5:10
Idonknow
2,282748111
Question: Let $K$ be a splitting field for $x^5-2$ over $mathbb{Q}.$
$(1)$ Determine $[K:mathbb{Q}].$
$(2)$ Show that $Gal(K/mathbb{Q})$ is non-abelian.
For $(1),$ note that roots of $x^5-2$ are
$$sqrt[5]{2}, sqrt[5]{2}omega, sqrt[5]{2}omega^2, sqrt[5]{2}omega^3, sqrt[5]{2}omega^4$$
where $omega$ is the $5$th root of unity, that is,
$$omega = e^{frac{2pi i}{5}}.$$
So we have
$$K = mathbb{Q}(sqrt[5]{2}, omega),$$
implying that
$$[K:mathbb{Q}] = [K:mathbb{Q}(omega)] cdot [mathbb{Q}(omega):mathbb{Q}] = 5 times 4 = 20.$$
For $(2),$ I know that $Gal(K/mathbb{Q})$ is isomorphic to the Dihedral group with $20$ elements.
Since such Dihedral group is non-abelian, so is $Gal(K/mathbb{Q}).$
However, I do now know how to prove it.
abstract-algebra proof-verification field-theory galois-theory
abstract-algebra proof-verification field-theory galois-theory
asked Nov 24 at 5:10
Idonknow
2,282748111
asked Nov 24 at 5:10
Idonknow
2,282748111
edited Nov 24 at 5:13
asked Nov 24 at 5:10
Idonknow
2,282748111
asked Nov 24 at 5:10
Idonknow
2,282748111
asked Nov 24 at 5:10
Idonknow
2,282748111
2,282748111
$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16
|
show 5 more comments
$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16
$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16
|
show 5 more comments
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$|Bbb Q(omega):Bbb Q|=4$.
– Lord Shark the Unknown
Nov 24 at 5:12
Consider the action of the Galois group as permutations of the roots.
– Lord Shark the Unknown
Nov 24 at 5:12
You can find two elements in the Galois group that don't commute.
– Seewoo Lee
Nov 24 at 5:12
@LordSharktheUnknown Thanks for pointing out my mistake. Corrected it.
– Idonknow
Nov 24 at 5:14
The Galois group is not dihedral.
– Lord Shark the Unknown
Nov 24 at 5:16