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
|
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
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
Nov 19 at 5:11
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
1
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34
|
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
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
edited Nov 19 at 5:36
asked Nov 19 at 4:24
Idonknow
2,267747111
2,267747111
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
Nov 19 at 5:11
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
1
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34
|
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
Nov 19 at 5:11
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
1
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34
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
Nov 19 at 5:11
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
Nov 19 at 5:11
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
1
1
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
3
down vote
Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.
Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.
In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.
Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.
In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
add a comment |
up vote
3
down vote
Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.
Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.
In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
add a comment |
up vote
3
down vote
up vote
3
down vote
Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.
Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.
In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.
Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.
Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.
In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
edited Nov 23 at 13:45
Joel Pereira
47319
47319
answered Nov 23 at 13:26
Bilo
1089
1089
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
add a comment |
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 at 13:48
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 at 15:06
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Nov 19 at 5:11
@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 at 5:20
${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 at 5:31
1
@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 at 5:33
"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 at 5:34