Linear independence of $sqrt{2}, sqrt[3]{2}, sqrt[5]{2}$ over $mathbb{Q}$
$begingroup$
In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.
There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:
"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."
So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!
abstract-algebra field-theory
asked Dec 29 '18 at 2:08
AlexAlex
757
$endgroup$
add a comment |
$begingroup$
In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.
There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:
"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."
So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!
abstract-algebra field-theory
asked Dec 29 '18 at 2:08
AlexAlex
757
$endgroup$
add a comment |
$begingroup$
In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.
There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:
"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."
So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!
abstract-algebra field-theory
asked Dec 29 '18 at 2:08
AlexAlex
757
$endgroup$
In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.
There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:
"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."
So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!
abstract-algebra field-theory
abstract-algebra field-theory
asked Dec 29 '18 at 2:08
AlexAlex
757
asked Dec 29 '18 at 2:08
AlexAlex
757
asked Dec 29 '18 at 2:08
AlexAlex
757
asked Dec 29 '18 at 2:08
AlexAlex
757
asked Dec 29 '18 at 2:08
AlexAlex
757
757
add a comment |
add a comment |
1 Answer
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$begingroup$
$x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.
Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.
By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.
Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.
Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.
The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.
An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
$endgroup$
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
add a comment |
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$begingroup$
$x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.
Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.
By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.
Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.
Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.
The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.
An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
$endgroup$
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
add a comment |
$begingroup$
$x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.
Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.
By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.
Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.
Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.
The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.
An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
$endgroup$
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
add a comment |
$begingroup$
$x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.
Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.
By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.
Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.
Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.
The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.
An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
$endgroup$
$x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.
Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.
By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.
Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.
Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.
The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.
An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
edited Dec 29 '18 at 2:34
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
answered Dec 29 '18 at 2:26
reunsreuns
21.5k21352
21.5k21352
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
add a comment |
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
$endgroup$
– Alex
Dec 29 '18 at 2:55
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
$begingroup$
@Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
$endgroup$
– reuns
Dec 29 '18 at 5:08
1
1
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
$begingroup$
@renus Aren't they irreducible by Eisenstein's Criterion?
$endgroup$
– Alex
Dec 29 '18 at 5:10
add a comment |
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