Minimal polynomial of an element over a subfield when the degrees of extensions are coprime
$begingroup$
This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?
Suppose $Kleq Lleq L(alpha)$ are field extensions, and that
$|L(alpha):L|$ and $|L:K|$ are relatively prime. Show that the
minimal polynomial of $alpha$ over $L$ has its coefficients in $K$.
Let $|L(alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(alpha):K|=nt$ for some integer $t$. Let $|L(alpha):K(alpha)|=s$.
Now by the tower law, we have $nts = mn implies ts =m$.
If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$?
I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.
It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?
abstract-algebra field-theory galois-theory extension-field minimal-polynomials
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
$endgroup$
add a comment |
$begingroup$
This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?
Suppose $Kleq Lleq L(alpha)$ are field extensions, and that
$|L(alpha):L|$ and $|L:K|$ are relatively prime. Show that the
minimal polynomial of $alpha$ over $L$ has its coefficients in $K$.
Let $|L(alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(alpha):K|=nt$ for some integer $t$. Let $|L(alpha):K(alpha)|=s$.
Now by the tower law, we have $nts = mn implies ts =m$.
If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$?
I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.
It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?
abstract-algebra field-theory galois-theory extension-field minimal-polynomials
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
$endgroup$
$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53
add a comment |
$begingroup$
This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?
Suppose $Kleq Lleq L(alpha)$ are field extensions, and that
$|L(alpha):L|$ and $|L:K|$ are relatively prime. Show that the
minimal polynomial of $alpha$ over $L$ has its coefficients in $K$.
Let $|L(alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(alpha):K|=nt$ for some integer $t$. Let $|L(alpha):K(alpha)|=s$.
Now by the tower law, we have $nts = mn implies ts =m$.
If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$?
I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.
It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?
abstract-algebra field-theory galois-theory extension-field minimal-polynomials
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
$endgroup$
This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?
Suppose $Kleq Lleq L(alpha)$ are field extensions, and that
$|L(alpha):L|$ and $|L:K|$ are relatively prime. Show that the
minimal polynomial of $alpha$ over $L$ has its coefficients in $K$.
Let $|L(alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(alpha):K|=nt$ for some integer $t$. Let $|L(alpha):K(alpha)|=s$.
Now by the tower law, we have $nts = mn implies ts =m$.
If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$?
I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.
It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?
abstract-algebra field-theory galois-theory extension-field minimal-polynomials
abstract-algebra field-theory galois-theory extension-field minimal-polynomials
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
edited Jan 5 '17 at 10:09
Globe Theatre
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
asked Jan 5 '17 at 8:53
Globe TheatreGlobe Theatre
311210
311210
$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53
add a comment |
$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53
$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Try $K=mathbb{Q}$, $L=mathbb{Q}(sqrt{2})$, $alpha=sqrt[6]{2}$.
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
$endgroup$
add a comment |
$begingroup$
Another counterexample to the exercise as written is to take $alpha in L setminus K$. Then $|L(alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $alpha$ over $L$ is simply $x - alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(alpha):K|$ and not $|L(alpha):L|$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:
Suppose that $K subset L subset L(alpha)$ are field extensions, and that $|K(alpha):K|$ and $|L:K|$ are relatively prime, then $|L(alpha):L| = |K(alpha):K|$.
A proof would be:
Since $|K(alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(alpha):K|$, their product must divide $|L(alpha):K|$, implying $|L(alpha):K| geq |K(alpha):K| cdot |L:K|$.
But then, the general fact that $|K(alpha):K| geq |L(alpha):L|$ implies that $|K(alpha):K| cdot |L:K| geq |L(alpha):L| cdot |L:K| = |L(alpha):K|$.
So the inequalities collapse, forcing $|L(alpha):L| = |K(alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $mathbb Q$.
answered Dec 10 '18 at 19:04
user3339517user3339517
315
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
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votes
$begingroup$
Try $K=mathbb{Q}$, $L=mathbb{Q}(sqrt{2})$, $alpha=sqrt[6]{2}$.
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
$endgroup$
add a comment |
$begingroup$
Try $K=mathbb{Q}$, $L=mathbb{Q}(sqrt{2})$, $alpha=sqrt[6]{2}$.
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
$endgroup$
add a comment |
$begingroup$
Try $K=mathbb{Q}$, $L=mathbb{Q}(sqrt{2})$, $alpha=sqrt[6]{2}$.
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
$endgroup$
Try $K=mathbb{Q}$, $L=mathbb{Q}(sqrt{2})$, $alpha=sqrt[6]{2}$.
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
answered Jan 5 '17 at 12:57
Ievgen BondarenkoIevgen Bondarenko
33617
33617
add a comment |
add a comment |
$begingroup$
Another counterexample to the exercise as written is to take $alpha in L setminus K$. Then $|L(alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $alpha$ over $L$ is simply $x - alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(alpha):K|$ and not $|L(alpha):L|$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:
Suppose that $K subset L subset L(alpha)$ are field extensions, and that $|K(alpha):K|$ and $|L:K|$ are relatively prime, then $|L(alpha):L| = |K(alpha):K|$.
A proof would be:
Since $|K(alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(alpha):K|$, their product must divide $|L(alpha):K|$, implying $|L(alpha):K| geq |K(alpha):K| cdot |L:K|$.
But then, the general fact that $|K(alpha):K| geq |L(alpha):L|$ implies that $|K(alpha):K| cdot |L:K| geq |L(alpha):L| cdot |L:K| = |L(alpha):K|$.
So the inequalities collapse, forcing $|L(alpha):L| = |K(alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $mathbb Q$.
answered Dec 10 '18 at 19:04
user3339517user3339517
315
$endgroup$
add a comment |
$begingroup$
Another counterexample to the exercise as written is to take $alpha in L setminus K$. Then $|L(alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $alpha$ over $L$ is simply $x - alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(alpha):K|$ and not $|L(alpha):L|$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:
Suppose that $K subset L subset L(alpha)$ are field extensions, and that $|K(alpha):K|$ and $|L:K|$ are relatively prime, then $|L(alpha):L| = |K(alpha):K|$.
A proof would be:
Since $|K(alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(alpha):K|$, their product must divide $|L(alpha):K|$, implying $|L(alpha):K| geq |K(alpha):K| cdot |L:K|$.
But then, the general fact that $|K(alpha):K| geq |L(alpha):L|$ implies that $|K(alpha):K| cdot |L:K| geq |L(alpha):L| cdot |L:K| = |L(alpha):K|$.
So the inequalities collapse, forcing $|L(alpha):L| = |K(alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $mathbb Q$.
answered Dec 10 '18 at 19:04
user3339517user3339517
315
$endgroup$
add a comment |
$begingroup$
Another counterexample to the exercise as written is to take $alpha in L setminus K$. Then $|L(alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $alpha$ over $L$ is simply $x - alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(alpha):K|$ and not $|L(alpha):L|$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:
Suppose that $K subset L subset L(alpha)$ are field extensions, and that $|K(alpha):K|$ and $|L:K|$ are relatively prime, then $|L(alpha):L| = |K(alpha):K|$.
A proof would be:
Since $|K(alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(alpha):K|$, their product must divide $|L(alpha):K|$, implying $|L(alpha):K| geq |K(alpha):K| cdot |L:K|$.
But then, the general fact that $|K(alpha):K| geq |L(alpha):L|$ implies that $|K(alpha):K| cdot |L:K| geq |L(alpha):L| cdot |L:K| = |L(alpha):K|$.
So the inequalities collapse, forcing $|L(alpha):L| = |K(alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $mathbb Q$.
answered Dec 10 '18 at 19:04
user3339517user3339517
315
$endgroup$
Another counterexample to the exercise as written is to take $alpha in L setminus K$. Then $|L(alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $alpha$ over $L$ is simply $x - alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(alpha):K|$ and not $|L(alpha):L|$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:
Suppose that $K subset L subset L(alpha)$ are field extensions, and that $|K(alpha):K|$ and $|L:K|$ are relatively prime, then $|L(alpha):L| = |K(alpha):K|$.
A proof would be:
Since $|K(alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(alpha):K|$, their product must divide $|L(alpha):K|$, implying $|L(alpha):K| geq |K(alpha):K| cdot |L:K|$.
But then, the general fact that $|K(alpha):K| geq |L(alpha):L|$ implies that $|K(alpha):K| cdot |L:K| geq |L(alpha):L| cdot |L:K| = |L(alpha):K|$.
So the inequalities collapse, forcing $|L(alpha):L| = |K(alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $mathbb Q$.
answered Dec 10 '18 at 19:04
user3339517user3339517
315
edited Dec 11 '18 at 11:53
answered Dec 10 '18 at 19:04
user3339517user3339517
315
answered Dec 10 '18 at 19:04
user3339517user3339517
315
answered Dec 10 '18 at 19:04
user3339517user3339517
315
315
add a comment |
add a comment |
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$begingroup$
Did you mean Galois extensions ?
$endgroup$
– reuns
Jan 5 '17 at 13:14
$begingroup$
Not necessarily. Just finite extensions.
$endgroup$
– Globe Theatre
Jan 5 '17 at 13:24
$begingroup$
@Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question?
$endgroup$
– Marcel S
Aug 25 '17 at 11:23
$begingroup$
GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/…
$endgroup$
– Marcel S
Aug 25 '17 at 14:53