Mixing
Determine the automorphism group $Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q})$
up vote
1
down vote
favorite
Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$
My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.
However, I do not know how to determine the automorphism group.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked 11 hours ago
Idonknow
2,205746110
|
show 1 more comment
up vote
1
down vote
favorite
Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$
My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.
However, I do not know how to determine the automorphism group.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked 11 hours ago
Idonknow
2,205746110
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
1
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$
My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.
However, I do not know how to determine the automorphism group.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
asked 11 hours ago
Idonknow
2,205746110
Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$
My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.
However, I do not know how to determine the automorphism group.
Any hint is appreciated.
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked 11 hours ago
Idonknow
2,205746110
asked 11 hours ago
Idonknow
2,205746110
edited 10 hours ago
asked 11 hours ago
Idonknow
2,205746110
asked 11 hours ago
Idonknow
2,205746110
asked 11 hours ago
Idonknow
2,205746110
2,205746110
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
1
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago
|
show 1 more comment
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
1
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
1
1
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago
|
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004520%2fdetermine-the-automorphism-group-aut-mathbbq-sqrt13-sqrt37-mathbb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago
1
@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago