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Degree and basis of field extension $mathbb{Q}[sqrt{2+sqrt{5}}]$
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I want to find the degree and basis of the field extension $mathbb{Q}(sqrt{2+sqrt{5}})$.
let $alpha=sqrt{2+sqrt{5}}$.
$$alpha^2=2+sqrt{5},quad alpha^4-4alpha^2-1=0.$$
So possible minimal polynomial $f(X)=X^4-X^2-1$. $f$ is irreducible over $mathbb{Q}$ since its roots are all non rationals and so its possible factors are of degree $2$ not $1$. Looking at the roots shows irreducibility and hence $f$ is the minimal polynomial and degree of extension is $4$.
My problem is with the basis. Since $alpha^2=2+sqrt{5}$, then $sqrt{5}$ is in $mathbb{Q}(alpha)$. So $mathbb{Q}(alpha)$ has degree $4$ over $mathbb{Q}$ and $mathbb{Q}(sqrt{5})$ has degree $2$ over $mathbb{Q}$ using Tower Law gives degree $mathbb{Q}(alpha)$ over $mathbb{Q}sqrt{5}$ is $2$.
So a basis for $mathbb{Q}(alpha)$ is the product of the bases.
$$B={1,sqrt{5},alpha,alpha sqrt{5}}$$
Is the basis and method correct?
galois-theory extension-field
asked Feb 3 at 10:26
orangeorange
729316
$endgroup$
add a comment |
$begingroup$
I want to find the degree and basis of the field extension $mathbb{Q}(sqrt{2+sqrt{5}})$.
let $alpha=sqrt{2+sqrt{5}}$.
$$alpha^2=2+sqrt{5},quad alpha^4-4alpha^2-1=0.$$
So possible minimal polynomial $f(X)=X^4-X^2-1$. $f$ is irreducible over $mathbb{Q}$ since its roots are all non rationals and so its possible factors are of degree $2$ not $1$. Looking at the roots shows irreducibility and hence $f$ is the minimal polynomial and degree of extension is $4$.
My problem is with the basis. Since $alpha^2=2+sqrt{5}$, then $sqrt{5}$ is in $mathbb{Q}(alpha)$. So $mathbb{Q}(alpha)$ has degree $4$ over $mathbb{Q}$ and $mathbb{Q}(sqrt{5})$ has degree $2$ over $mathbb{Q}$ using Tower Law gives degree $mathbb{Q}(alpha)$ over $mathbb{Q}sqrt{5}$ is $2$.
So a basis for $mathbb{Q}(alpha)$ is the product of the bases.
$$B={1,sqrt{5},alpha,alpha sqrt{5}}$$
Is the basis and method correct?
galois-theory extension-field
asked Feb 3 at 10:26
orangeorange
729316
$endgroup$
$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44
add a comment |
$begingroup$
I want to find the degree and basis of the field extension $mathbb{Q}(sqrt{2+sqrt{5}})$.
let $alpha=sqrt{2+sqrt{5}}$.
$$alpha^2=2+sqrt{5},quad alpha^4-4alpha^2-1=0.$$
So possible minimal polynomial $f(X)=X^4-X^2-1$. $f$ is irreducible over $mathbb{Q}$ since its roots are all non rationals and so its possible factors are of degree $2$ not $1$. Looking at the roots shows irreducibility and hence $f$ is the minimal polynomial and degree of extension is $4$.
My problem is with the basis. Since $alpha^2=2+sqrt{5}$, then $sqrt{5}$ is in $mathbb{Q}(alpha)$. So $mathbb{Q}(alpha)$ has degree $4$ over $mathbb{Q}$ and $mathbb{Q}(sqrt{5})$ has degree $2$ over $mathbb{Q}$ using Tower Law gives degree $mathbb{Q}(alpha)$ over $mathbb{Q}sqrt{5}$ is $2$.
So a basis for $mathbb{Q}(alpha)$ is the product of the bases.
$$B={1,sqrt{5},alpha,alpha sqrt{5}}$$
Is the basis and method correct?
galois-theory extension-field
asked Feb 3 at 10:26
orangeorange
729316
$endgroup$
I want to find the degree and basis of the field extension $mathbb{Q}(sqrt{2+sqrt{5}})$.
let $alpha=sqrt{2+sqrt{5}}$.
$$alpha^2=2+sqrt{5},quad alpha^4-4alpha^2-1=0.$$
So possible minimal polynomial $f(X)=X^4-X^2-1$. $f$ is irreducible over $mathbb{Q}$ since its roots are all non rationals and so its possible factors are of degree $2$ not $1$. Looking at the roots shows irreducibility and hence $f$ is the minimal polynomial and degree of extension is $4$.
My problem is with the basis. Since $alpha^2=2+sqrt{5}$, then $sqrt{5}$ is in $mathbb{Q}(alpha)$. So $mathbb{Q}(alpha)$ has degree $4$ over $mathbb{Q}$ and $mathbb{Q}(sqrt{5})$ has degree $2$ over $mathbb{Q}$ using Tower Law gives degree $mathbb{Q}(alpha)$ over $mathbb{Q}sqrt{5}$ is $2$.
So a basis for $mathbb{Q}(alpha)$ is the product of the bases.
$$B={1,sqrt{5},alpha,alpha sqrt{5}}$$
Is the basis and method correct?
galois-theory extension-field
galois-theory extension-field
asked Feb 3 at 10:26
orangeorange
729316
asked Feb 3 at 10:26
orangeorange
729316
asked Feb 3 at 10:26
orangeorange
729316
asked Feb 3 at 10:26
orangeorange
729316
asked Feb 3 at 10:26
orangeorange
729316
729316
$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44
add a comment |
$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44
$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44
add a comment |
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$begingroup$
See also math.stackexchange.com/questions/2446699/…
$endgroup$
– lhf
Feb 3 at 11:03
$begingroup$
is the basis correct? i wrote this in an exam I want to make sure @lhf
$endgroup$
– orange
Feb 6 at 6:44