Extension Degree of Fields Composite
$begingroup$
Let $K/F$ be a field extension, $L/F$ and $M/F$ finite subextensions of $K/F$ and $LM$ the composite of $L$ and $M$. I'm trying to prove that $[LM:F] = [L:F][M:F]$ implies the trivial intersection $Lcap M=F$, and that the converse holds when $[L:F]=2$ or $[M:F]=2$.
In order to prove the trivial intersection of the fields, I want to show that $[Lcap M:F]=1$. From the hypothesis, we can obtain
$$[LM:L] = [M:F] = [M:Lcap M][Lcap M:F],$$
and
$$[LM:M] = [L:F] = [L:Lcap M][Lcap M:F],$$
but I don't if that is usefull for what I'm trying to prove. And I don't know how to prove the converse part. I would appreciate any help you can give me.
abstract-algebra field-theory extension-field
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
$endgroup$
add a comment |
$begingroup$
Let $K/F$ be a field extension, $L/F$ and $M/F$ finite subextensions of $K/F$ and $LM$ the composite of $L$ and $M$. I'm trying to prove that $[LM:F] = [L:F][M:F]$ implies the trivial intersection $Lcap M=F$, and that the converse holds when $[L:F]=2$ or $[M:F]=2$.
In order to prove the trivial intersection of the fields, I want to show that $[Lcap M:F]=1$. From the hypothesis, we can obtain
$$[LM:L] = [M:F] = [M:Lcap M][Lcap M:F],$$
and
$$[LM:M] = [L:F] = [L:Lcap M][Lcap M:F],$$
but I don't if that is usefull for what I'm trying to prove. And I don't know how to prove the converse part. I would appreciate any help you can give me.
abstract-algebra field-theory extension-field
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
$endgroup$
add a comment |
$begingroup$
Let $K/F$ be a field extension, $L/F$ and $M/F$ finite subextensions of $K/F$ and $LM$ the composite of $L$ and $M$. I'm trying to prove that $[LM:F] = [L:F][M:F]$ implies the trivial intersection $Lcap M=F$, and that the converse holds when $[L:F]=2$ or $[M:F]=2$.
In order to prove the trivial intersection of the fields, I want to show that $[Lcap M:F]=1$. From the hypothesis, we can obtain
$$[LM:L] = [M:F] = [M:Lcap M][Lcap M:F],$$
and
$$[LM:M] = [L:F] = [L:Lcap M][Lcap M:F],$$
but I don't if that is usefull for what I'm trying to prove. And I don't know how to prove the converse part. I would appreciate any help you can give me.
abstract-algebra field-theory extension-field
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
$endgroup$
Let $K/F$ be a field extension, $L/F$ and $M/F$ finite subextensions of $K/F$ and $LM$ the composite of $L$ and $M$. I'm trying to prove that $[LM:F] = [L:F][M:F]$ implies the trivial intersection $Lcap M=F$, and that the converse holds when $[L:F]=2$ or $[M:F]=2$.
In order to prove the trivial intersection of the fields, I want to show that $[Lcap M:F]=1$. From the hypothesis, we can obtain
$$[LM:L] = [M:F] = [M:Lcap M][Lcap M:F],$$
and
$$[LM:M] = [L:F] = [L:Lcap M][Lcap M:F],$$
but I don't if that is usefull for what I'm trying to prove. And I don't know how to prove the converse part. I would appreciate any help you can give me.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
asked Dec 28 '18 at 22:46
Juan David SamboníJuan David Samboní
917
917
add a comment |
add a comment |
1 Answer
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$begingroup$
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$require{AMScd}begin{CD}LM @<<< L \ @AAA @AAA\M @<<<Lcap M @<<< Fend{CD}$$
Now we have
$$[LM:L]=[M:F]=[M:Lcap M][Lcap M:F],$$
as you've already noted.
Note now that $[LM:L]le [M:Lcap M]$, so that
$$[M:Lcap M][Lcap M:F]le [M:Lcap M],$$
or
$$[Lcap M:F]le 1,$$
which of course implies that $Lcap M=F$.
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
$endgroup$
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$require{AMScd}begin{CD}LM @<<< L \ @AAA @AAA\M @<<<Lcap M @<<< Fend{CD}$$
Now we have
$$[LM:L]=[M:F]=[M:Lcap M][Lcap M:F],$$
as you've already noted.
Note now that $[LM:L]le [M:Lcap M]$, so that
$$[M:Lcap M][Lcap M:F]le [M:Lcap M],$$
or
$$[Lcap M:F]le 1,$$
which of course implies that $Lcap M=F$.
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
$endgroup$
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
add a comment |
$begingroup$
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$require{AMScd}begin{CD}LM @<<< L \ @AAA @AAA\M @<<<Lcap M @<<< Fend{CD}$$
Now we have
$$[LM:L]=[M:F]=[M:Lcap M][Lcap M:F],$$
as you've already noted.
Note now that $[LM:L]le [M:Lcap M]$, so that
$$[M:Lcap M][Lcap M:F]le [M:Lcap M],$$
or
$$[Lcap M:F]le 1,$$
which of course implies that $Lcap M=F$.
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
$endgroup$
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
add a comment |
$begingroup$
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$require{AMScd}begin{CD}LM @<<< L \ @AAA @AAA\M @<<<Lcap M @<<< Fend{CD}$$
Now we have
$$[LM:L]=[M:F]=[M:Lcap M][Lcap M:F],$$
as you've already noted.
Note now that $[LM:L]le [M:Lcap M]$, so that
$$[M:Lcap M][Lcap M:F]le [M:Lcap M],$$
or
$$[Lcap M:F]le 1,$$
which of course implies that $Lcap M=F$.
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
$endgroup$
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$require{AMScd}begin{CD}LM @<<< L \ @AAA @AAA\M @<<<Lcap M @<<< Fend{CD}$$
Now we have
$$[LM:L]=[M:F]=[M:Lcap M][Lcap M:F],$$
as you've already noted.
Note now that $[LM:L]le [M:Lcap M]$, so that
$$[M:Lcap M][Lcap M:F]le [M:Lcap M],$$
or
$$[Lcap M:F]le 1,$$
which of course implies that $Lcap M=F$.
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
answered Dec 29 '18 at 2:58
jgonjgon
14.7k22042
14.7k22042
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
add a comment |
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
$begingroup$
Thank you very much! That was very helpful.
$endgroup$
– Juan David Samboní
Dec 29 '18 at 21:15
add a comment |
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