If $g,h in L[X]$ prime divisors of f irreducible, than it exists $phi in Aut(L/K)$ with $phi(g) = h$.
$begingroup$
Let f $in$ K[X] be irreducible and L/K a normal field extension. If g,h $in$ L[X] normalized prime divisor of f in L[X], so it exist a automorphism $tau in$ Aut(L/K) with $tau$(g)=h.
I got so far:
Set f = gt = hs with t,s $in$ L[X]. Than
- g divides hs
- h divides gt
and because of f irreducible follows with definition that
- g or t $in L[X]^x=L^x$
- h or s $in L[X]^x=L^x$
Since g,h prime, so irreducible, so $notin L[X]^x=L^x$, it must be t,s $in L^x$ and so we can set g = $h cdot frac{s}{t} := h cdot b$ with $b in L^x$.
Any comments or ideas how to go on?
abstract-algebra extension-field automorphism-group
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
$endgroup$
add a comment |
$begingroup$
Let f $in$ K[X] be irreducible and L/K a normal field extension. If g,h $in$ L[X] normalized prime divisor of f in L[X], so it exist a automorphism $tau in$ Aut(L/K) with $tau$(g)=h.
I got so far:
Set f = gt = hs with t,s $in$ L[X]. Than
- g divides hs
- h divides gt
and because of f irreducible follows with definition that
- g or t $in L[X]^x=L^x$
- h or s $in L[X]^x=L^x$
Since g,h prime, so irreducible, so $notin L[X]^x=L^x$, it must be t,s $in L^x$ and so we can set g = $h cdot frac{s}{t} := h cdot b$ with $b in L^x$.
Any comments or ideas how to go on?
abstract-algebra extension-field automorphism-group
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
$endgroup$
$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51
add a comment |
$begingroup$
Let f $in$ K[X] be irreducible and L/K a normal field extension. If g,h $in$ L[X] normalized prime divisor of f in L[X], so it exist a automorphism $tau in$ Aut(L/K) with $tau$(g)=h.
I got so far:
Set f = gt = hs with t,s $in$ L[X]. Than
- g divides hs
- h divides gt
and because of f irreducible follows with definition that
- g or t $in L[X]^x=L^x$
- h or s $in L[X]^x=L^x$
Since g,h prime, so irreducible, so $notin L[X]^x=L^x$, it must be t,s $in L^x$ and so we can set g = $h cdot frac{s}{t} := h cdot b$ with $b in L^x$.
Any comments or ideas how to go on?
abstract-algebra extension-field automorphism-group
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
$endgroup$
Let f $in$ K[X] be irreducible and L/K a normal field extension. If g,h $in$ L[X] normalized prime divisor of f in L[X], so it exist a automorphism $tau in$ Aut(L/K) with $tau$(g)=h.
I got so far:
Set f = gt = hs with t,s $in$ L[X]. Than
- g divides hs
- h divides gt
and because of f irreducible follows with definition that
- g or t $in L[X]^x=L^x$
- h or s $in L[X]^x=L^x$
Since g,h prime, so irreducible, so $notin L[X]^x=L^x$, it must be t,s $in L^x$ and so we can set g = $h cdot frac{s}{t} := h cdot b$ with $b in L^x$.
Any comments or ideas how to go on?
abstract-algebra extension-field automorphism-group
abstract-algebra extension-field automorphism-group
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
asked Dec 6 '18 at 13:40
Zorro_CZorro_C
82
82
$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51
add a comment |
$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51
$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51
$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51
add a comment |
1 Answer
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$begingroup$
You can write $f=Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $Pi_{hin Aut(L/K), {f_1}^hneq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You can write $f=Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $Pi_{hin Aut(L/K), {f_1}^hneq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
$begingroup$
You can write $f=Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $Pi_{hin Aut(L/K), {f_1}^hneq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
add a comment |
$begingroup$
You can write $f=Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $Pi_{hin Aut(L/K), {f_1}^hneq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
You can write $f=Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $Pi_{hin Aut(L/K), {f_1}^hneq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
answered Dec 6 '18 at 14:07
Tsemo AristideTsemo Aristide
56.9k11444
56.9k11444
add a comment |
add a comment |
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$begingroup$
What is your $x$ ? To prove everything let $alpha, beta$ some roots of $g,h$. Do you know a field morphism acting on $alpha,beta$ ?
$endgroup$
– reuns
Dec 6 '18 at 13:51