If $L_1$, $L_2$ are intermediate fields of $E/K$, then $[L_1 L_2 : K] = [L_1 : K] [L_2 : K] implies L_1 cap...
$begingroup$
Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.
field-theory extension-field
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
$endgroup$
add a comment |
$begingroup$
Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.
field-theory extension-field
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
$endgroup$
add a comment |
$begingroup$
Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.
field-theory extension-field
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
$endgroup$
Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.
field-theory extension-field
field-theory extension-field
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
edited Dec 14 '18 at 8:14
Brahadeesh
6,36442363
6,36442363
asked Sep 30 '16 at 14:25
F.KF.K
589415
asked Sep 30 '16 at 14:25
F.KF.K
589415
asked Sep 30 '16 at 14:25
F.KF.K
589415
589415
add a comment |
add a comment |
1 Answer
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$begingroup$
One can proceed as follows:
If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
$$
L_1L_2 = text{span}_{L_2}(S)
$$
This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)Once you have that, it shows that
$$
[L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
$$
(This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
$$
|B|=1
$$
and so $L_1cap L_2 = K$.
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One can proceed as follows:
If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
$$
L_1L_2 = text{span}_{L_2}(S)
$$
This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)Once you have that, it shows that
$$
[L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
$$
(This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
$$
|B|=1
$$
and so $L_1cap L_2 = K$.
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
$endgroup$
add a comment |
$begingroup$
One can proceed as follows:
If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
$$
L_1L_2 = text{span}_{L_2}(S)
$$
This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)Once you have that, it shows that
$$
[L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
$$
(This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
$$
|B|=1
$$
and so $L_1cap L_2 = K$.
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
$endgroup$
add a comment |
$begingroup$
One can proceed as follows:
If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
$$
L_1L_2 = text{span}_{L_2}(S)
$$
This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)Once you have that, it shows that
$$
[L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
$$
(This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
$$
|B|=1
$$
and so $L_1cap L_2 = K$.
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
$endgroup$
One can proceed as follows:
If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
$$
L_1L_2 = text{span}_{L_2}(S)
$$
This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)Once you have that, it shows that
$$
[L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
$$
(This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
$$
|B|=1
$$
and so $L_1cap L_2 = K$.
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
answered Oct 5 '16 at 6:27
Prahlad VaidyanathanPrahlad Vaidyanathan
26.5k12152
26.5k12152
add a comment |
add a comment |
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