Mixing
Degree of composite field
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Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]leqslant[E:K]$ and $[EF:K]leqslant[E:K]cdot[F:K]$.
If we assume $Ecap F=K$, then will we get "="?
abstract-algebra field-theory
asked Jul 28 at 5:00
Lao-tzu
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Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]leqslant[E:K]$ and $[EF:K]leqslant[E:K]cdot[F:K]$.
If we assume $Ecap F=K$, then will we get "="?
abstract-algebra field-theory
asked Jul 28 at 5:00
Lao-tzu
1,004923
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]leqslant[E:K]$ and $[EF:K]leqslant[E:K]cdot[F:K]$.
If we assume $Ecap F=K$, then will we get "="?
abstract-algebra field-theory
asked Jul 28 at 5:00
Lao-tzu
1,004923
Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]leqslant[E:K]$ and $[EF:K]leqslant[E:K]cdot[F:K]$.
If we assume $Ecap F=K$, then will we get "="?
abstract-algebra field-theory
asked Jul 28 at 5:00
Lao-tzu
1,004923
asked Jul 28 at 5:00
Lao-tzu
1,004923
asked Jul 28 at 5:00
Lao-tzu
1,004923
asked Jul 28 at 5:00
Lao-tzu
1,004923
1,004923
add a comment |Â
add a comment |Â
1 Answer
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No. Trivial intersection does not guarantee that $[EF:K]=[E:K]cdot [F:K]$.
A standard counterexample of this is $K=BbbQ$, $E=BbbQ(root3of2)$,
$F=BbbQ(e^2pi i/3root3of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[Ecap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $Ecap F=K$.
But, the compositum $EF$ is the splitting field of $p(x)$ over $BbbQ$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3cdot3$.
The condition that guarantees equality, $[EF:K]=[E:K]cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.
Pete L. Clark has better material on this theme.
- See his question, and
- Notes on field theory. You want Chapter 12 in particular.
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No. Trivial intersection does not guarantee that $[EF:K]=[E:K]cdot [F:K]$.
A standard counterexample of this is $K=BbbQ$, $E=BbbQ(root3of2)$,
$F=BbbQ(e^2pi i/3root3of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[Ecap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $Ecap F=K$.
But, the compositum $EF$ is the splitting field of $p(x)$ over $BbbQ$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3cdot3$.
The condition that guarantees equality, $[EF:K]=[E:K]cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.
Pete L. Clark has better material on this theme.
- See his question, and
- Notes on field theory. You want Chapter 12 in particular.
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
add a comment |Â
up vote
3
down vote
accepted
No. Trivial intersection does not guarantee that $[EF:K]=[E:K]cdot [F:K]$.
A standard counterexample of this is $K=BbbQ$, $E=BbbQ(root3of2)$,
$F=BbbQ(e^2pi i/3root3of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[Ecap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $Ecap F=K$.
But, the compositum $EF$ is the splitting field of $p(x)$ over $BbbQ$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3cdot3$.
The condition that guarantees equality, $[EF:K]=[E:K]cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.
Pete L. Clark has better material on this theme.
- See his question, and
- Notes on field theory. You want Chapter 12 in particular.
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No. Trivial intersection does not guarantee that $[EF:K]=[E:K]cdot [F:K]$.
A standard counterexample of this is $K=BbbQ$, $E=BbbQ(root3of2)$,
$F=BbbQ(e^2pi i/3root3of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[Ecap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $Ecap F=K$.
But, the compositum $EF$ is the splitting field of $p(x)$ over $BbbQ$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3cdot3$.
The condition that guarantees equality, $[EF:K]=[E:K]cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.
Pete L. Clark has better material on this theme.
- See his question, and
- Notes on field theory. You want Chapter 12 in particular.
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
No. Trivial intersection does not guarantee that $[EF:K]=[E:K]cdot [F:K]$.
A standard counterexample of this is $K=BbbQ$, $E=BbbQ(root3of2)$,
$F=BbbQ(e^2pi i/3root3of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[Ecap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $Ecap F=K$.
But, the compositum $EF$ is the splitting field of $p(x)$ over $BbbQ$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3cdot3$.
The condition that guarantees equality, $[EF:K]=[E:K]cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.
Pete L. Clark has better material on this theme.
- See his question, and
- Notes on field theory. You want Chapter 12 in particular.
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
edited Jul 28 at 7:04
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
answered Jul 28 at 6:57
Jyrki Lahtonen
105k12161355
105k12161355
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
add a comment |Â
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
1
1
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks a lot, very good ref. I almost reach the notion of linearly disjoint extensions. I just remark that, in the last result (Corollary 8) of your linked notes, it's enough to assume only L/K to be fiô°œnite Galois, in fact in J. Milne's "Fields and Galois Theory" (version 4.40) Corollary 3.19, the author gives a more general formula.
– Lao-tzu
Jul 28 at 7:11
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
Thanks @Lao-tzu. I should have checked out Milne's notes as well, a great find! Well, writing up those notes was also an exercise for myself :-)
– Jyrki Lahtonen
Jul 28 at 7:15
add a comment |Â
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