Mixing
Extension fields of coprime degrees
$begingroup$
Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $gcd(m,n)=1$, then $[K(u,v):K]=mn$.
Once $[K(u,v):K]=[K(u,v):K(u)][K(u):K]=n[K(u,v):K(u)]$, I'm trying to show that $[K(u,v):K(u)]=m$.
abstract-algebra field-theory extension-field
edited Sep 24 '16 at 19:43
user26857
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
$endgroup$
add a comment |
$begingroup$
Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $gcd(m,n)=1$, then $[K(u,v):K]=mn$.
Once $[K(u,v):K]=[K(u,v):K(u)][K(u):K]=n[K(u,v):K(u)]$, I'm trying to show that $[K(u,v):K(u)]=m$.
abstract-algebra field-theory extension-field
edited Sep 24 '16 at 19:43
user26857
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
$endgroup$
3
$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37
add a comment |
$begingroup$
Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $gcd(m,n)=1$, then $[K(u,v):K]=mn$.
Once $[K(u,v):K]=[K(u,v):K(u)][K(u):K]=n[K(u,v):K(u)]$, I'm trying to show that $[K(u,v):K(u)]=m$.
abstract-algebra field-theory extension-field
edited Sep 24 '16 at 19:43
user26857
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
$endgroup$
Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $gcd(m,n)=1$, then $[K(u,v):K]=mn$.
Once $[K(u,v):K]=[K(u,v):K(u)][K(u):K]=n[K(u,v):K(u)]$, I'm trying to show that $[K(u,v):K(u)]=m$.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
edited Sep 24 '16 at 19:43
user26857
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
edited Sep 24 '16 at 19:43
user26857
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
edited Sep 24 '16 at 19:43
user26857
39.5k124284
edited Sep 24 '16 at 19:43
user26857
39.5k124284
edited Sep 24 '16 at 19:43
user26857
39.5k124284
39.5k124284
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
asked Aug 9 '15 at 20:35
Andre GomesAndre Gomes
920516
920516
3
$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37
add a comment |
3
$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37
3
3
$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37
$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K subset K(v)$ we have $s le n$.
We also have $rn=sm$ and if $gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
add a comment |
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$begingroup$
Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K subset K(v)$ we have $s le n$.
We also have $rn=sm$ and if $gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
add a comment |
$begingroup$
Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K subset K(v)$ we have $s le n$.
We also have $rn=sm$ and if $gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
add a comment |
$begingroup$
Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K subset K(v)$ we have $s le n$.
We also have $rn=sm$ and if $gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K subset K(v)$ we have $s le n$.
We also have $rn=sm$ and if $gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Aug 10 '15 at 8:19
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Aug 9 '15 at 20:59
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
add a comment |
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
$begingroup$
Why is $[K(u,v):K(v)] le [K(u):K]$?
$endgroup$
– Oliver G
Sep 24 '16 at 19:50
2
2
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
$begingroup$
Let $p in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$.
$endgroup$
– mathcounterexamples.net
Sep 24 '16 at 19:53
add a comment |
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$begingroup$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
$endgroup$
– anon
Aug 9 '15 at 20:37