Determine the automorphism group $Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q})$
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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 Nov 19 at 4:24
Idonknow
2,267748111
|
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 Nov 19 at 4:24
Idonknow
2,267748111
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
asked Nov 19 at 4:24
Idonknow
2,267748111
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 Nov 19 at 4:24
Idonknow
2,267748111
asked Nov 19 at 4:24
Idonknow
2,267748111
edited Nov 19 at 5:36
asked Nov 19 at 4:24
Idonknow
2,267748111
asked Nov 19 at 4:24
Idonknow
2,267748111
asked Nov 19 at 4:24
Idonknow
2,267748111
2,267748111
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 $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
edited Nov 23 at 13:45
Joel Pereira
47319
answered Nov 23 at 13:26
Bilo
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 |
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 $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
edited Nov 23 at 13:45
Joel Pereira
47319
answered Nov 23 at 13:26
Bilo
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 |
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 $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
edited Nov 23 at 13:45
Joel Pereira
47319
answered Nov 23 at 13:26
Bilo
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 |
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 $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$
edited Nov 23 at 13:45
Joel Pereira
47319
answered Nov 23 at 13:26
Bilo
1089
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
answered Nov 23 at 13:26
Bilo
1089
edited Nov 23 at 13:45
Joel Pereira
47319
edited Nov 23 at 13:45
Joel Pereira
47319
edited Nov 23 at 13:45
Joel Pereira
47319
47319
answered Nov 23 at 13:26
Bilo
1089
answered Nov 23 at 13:26
Bilo
1089
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 |
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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