Problem in Galois theory preliminaries












2












$begingroup$


If K/F is a field extension and $alpha in K$ is algebraic, then for any automorphism $sigma in operatorname{Aut}(K/F)$, $sigma a$ is a root of the minimal polynomial for $alpha$ over F.
I dont know how to proceed with it please help me!






field-theory galois-theory extension-field automorphism-group







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  • 2




    $begingroup$
    I think you mean algebraic over field $F$?
    $endgroup$
    – Shamim Akhtar
    27 mins ago










  • $begingroup$
    Yes i forget tl wrote that
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    24 mins ago
















2












$begingroup$


If K/F is a field extension and $alpha in K$ is algebraic, then for any automorphism $sigma in operatorname{Aut}(K/F)$, $sigma a$ is a root of the minimal polynomial for $alpha$ over F.
I dont know how to proceed with it please help me!






field-theory galois-theory extension-field automorphism-group







share|cite|improve this question









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Mr. BabaKGhalebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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  • 2




    $begingroup$
    I think you mean algebraic over field $F$?
    $endgroup$
    – Shamim Akhtar
    27 mins ago










  • $begingroup$
    Yes i forget tl wrote that
    $endgroup$
    – Mr. BabaKGhalebi
    24 mins ago














2












2








2





$begingroup$


If K/F is a field extension and $alpha in K$ is algebraic, then for any automorphism $sigma in operatorname{Aut}(K/F)$, $sigma a$ is a root of the minimal polynomial for $alpha$ over F.
I dont know how to proceed with it please help me!






field-theory galois-theory extension-field automorphism-group







share|cite|improve this question









New contributor




Mr. BabaKGhalebi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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If K/F is a field extension and $alpha in K$ is algebraic, then for any automorphism $sigma in operatorname{Aut}(K/F)$, $sigma a$ is a root of the minimal polynomial for $alpha$ over F.
I dont know how to proceed with it please help me!






field-theory galois-theory extension-field automorphism-group




field-theory galois-theory extension-field automorphism-group






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edited 36 mins ago









Bernard

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asked 38 mins ago









Mr. BabaKGhalebiMr. BabaKGhalebi

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  • 2




    $begingroup$
    I think you mean algebraic over field $F$?
    $endgroup$
    – Shamim Akhtar
    27 mins ago










  • $begingroup$
    Yes i forget tl wrote that
    $endgroup$
    – Mr. BabaKGhalebi
    24 mins ago














  • 2




    $begingroup$
    I think you mean algebraic over field $F$?
    $endgroup$
    – Shamim Akhtar
    27 mins ago










  • $begingroup$
    Yes i forget tl wrote that
    $endgroup$
    – Mr. BabaKGhalebi
    24 mins ago








2




2




$begingroup$
I think you mean algebraic over field $F$?
$endgroup$
– Shamim Akhtar
27 mins ago




$begingroup$
I think you mean algebraic over field $F$?
$endgroup$
– Shamim Akhtar
27 mins ago












$begingroup$
Yes i forget tl wrote that
$endgroup$
– Mr. BabaKGhalebi
24 mins ago




$begingroup$
Yes i forget tl wrote that
$endgroup$
– Mr. BabaKGhalebi
24 mins ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Suppose $alpha$ satisfies $alpha^n+a_{n-1}alpha^{n-1}+...+a_0=0$ where the coefficients belong to the field $F$. Then applying the automorphism, we find..
$sigma(alpha^n)+sigma (a_{n-1}alpha^{n-1})+...+sigma (a_0)=sigma0$. Now since $sigma$ is a multiplicative homomorphism, we get $(sigma(alpha))^n+sigma (a_{n-1}) (sigma(alpha)^{n-1})+...+sigma (a_0)=0$. Now this automorphism $alpha in text{Aut}(K/F)$ fixes all elements of the field $F$, so $sigma(a_i)=a_i forall i=0,1,...,n-1$. So, this becomes $(sigmaalpha)^n+a_{n-1}(sigmaalpha)^{n-1}+...+a_0=0$ implying that $sigma alpha$ is also a root of this polynomial.






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  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago



















2












$begingroup$

Very simple: if $;p(X)= a_0+a_1X+dots+a_nX^n$ is this minimal polynomial, then
begin{align}
p(sigmaalpha)&=a_0+a_1sigmaalpha+dots+a_n(sigmaalpha)^n\
&=sigma(a_0)+sigma(a_1)sigmaalpha+dots+sigma(a_n)(sigmaalpha)^n&text{since $sigma$ is a $F$-homomorphism}\
&=sigma(a_0+a_1alpha+dots+a_nalpha^n)=p(0)=0.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago



















1












$begingroup$

HINT



For $sigma in Aut(K/F)$ it is $sigma(a)=a forall ain F$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Suppose $alpha$ satisfies $alpha^n+a_{n-1}alpha^{n-1}+...+a_0=0$ where the coefficients belong to the field $F$. Then applying the automorphism, we find..
$sigma(alpha^n)+sigma (a_{n-1}alpha^{n-1})+...+sigma (a_0)=sigma0$. Now since $sigma$ is a multiplicative homomorphism, we get $(sigma(alpha))^n+sigma (a_{n-1}) (sigma(alpha)^{n-1})+...+sigma (a_0)=0$. Now this automorphism $alpha in text{Aut}(K/F)$ fixes all elements of the field $F$, so $sigma(a_i)=a_i forall i=0,1,...,n-1$. So, this becomes $(sigmaalpha)^n+a_{n-1}(sigmaalpha)^{n-1}+...+a_0=0$ implying that $sigma alpha$ is also a root of this polynomial.






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$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago
















2












$begingroup$

Suppose $alpha$ satisfies $alpha^n+a_{n-1}alpha^{n-1}+...+a_0=0$ where the coefficients belong to the field $F$. Then applying the automorphism, we find..
$sigma(alpha^n)+sigma (a_{n-1}alpha^{n-1})+...+sigma (a_0)=sigma0$. Now since $sigma$ is a multiplicative homomorphism, we get $(sigma(alpha))^n+sigma (a_{n-1}) (sigma(alpha)^{n-1})+...+sigma (a_0)=0$. Now this automorphism $alpha in text{Aut}(K/F)$ fixes all elements of the field $F$, so $sigma(a_i)=a_i forall i=0,1,...,n-1$. So, this becomes $(sigmaalpha)^n+a_{n-1}(sigmaalpha)^{n-1}+...+a_0=0$ implying that $sigma alpha$ is also a root of this polynomial.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago














2












2








2





$begingroup$

Suppose $alpha$ satisfies $alpha^n+a_{n-1}alpha^{n-1}+...+a_0=0$ where the coefficients belong to the field $F$. Then applying the automorphism, we find..
$sigma(alpha^n)+sigma (a_{n-1}alpha^{n-1})+...+sigma (a_0)=sigma0$. Now since $sigma$ is a multiplicative homomorphism, we get $(sigma(alpha))^n+sigma (a_{n-1}) (sigma(alpha)^{n-1})+...+sigma (a_0)=0$. Now this automorphism $alpha in text{Aut}(K/F)$ fixes all elements of the field $F$, so $sigma(a_i)=a_i forall i=0,1,...,n-1$. So, this becomes $(sigmaalpha)^n+a_{n-1}(sigmaalpha)^{n-1}+...+a_0=0$ implying that $sigma alpha$ is also a root of this polynomial.






share|cite|improve this answer









$endgroup$



Suppose $alpha$ satisfies $alpha^n+a_{n-1}alpha^{n-1}+...+a_0=0$ where the coefficients belong to the field $F$. Then applying the automorphism, we find..
$sigma(alpha^n)+sigma (a_{n-1}alpha^{n-1})+...+sigma (a_0)=sigma0$. Now since $sigma$ is a multiplicative homomorphism, we get $(sigma(alpha))^n+sigma (a_{n-1}) (sigma(alpha)^{n-1})+...+sigma (a_0)=0$. Now this automorphism $alpha in text{Aut}(K/F)$ fixes all elements of the field $F$, so $sigma(a_i)=a_i forall i=0,1,...,n-1$. So, this becomes $(sigmaalpha)^n+a_{n-1}(sigmaalpha)^{n-1}+...+a_0=0$ implying that $sigma alpha$ is also a root of this polynomial.







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answered 28 mins ago









Shamim AkhtarShamim Akhtar

57819




57819












  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago


















  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago
















$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago




$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago











2












$begingroup$

Very simple: if $;p(X)= a_0+a_1X+dots+a_nX^n$ is this minimal polynomial, then
begin{align}
p(sigmaalpha)&=a_0+a_1sigmaalpha+dots+a_n(sigmaalpha)^n\
&=sigma(a_0)+sigma(a_1)sigmaalpha+dots+sigma(a_n)(sigmaalpha)^n&text{since $sigma$ is a $F$-homomorphism}\
&=sigma(a_0+a_1alpha+dots+a_nalpha^n)=p(0)=0.
end{align}






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$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago
















2












$begingroup$

Very simple: if $;p(X)= a_0+a_1X+dots+a_nX^n$ is this minimal polynomial, then
begin{align}
p(sigmaalpha)&=a_0+a_1sigmaalpha+dots+a_n(sigmaalpha)^n\
&=sigma(a_0)+sigma(a_1)sigmaalpha+dots+sigma(a_n)(sigmaalpha)^n&text{since $sigma$ is a $F$-homomorphism}\
&=sigma(a_0+a_1alpha+dots+a_nalpha^n)=p(0)=0.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago














2












2








2





$begingroup$

Very simple: if $;p(X)= a_0+a_1X+dots+a_nX^n$ is this minimal polynomial, then
begin{align}
p(sigmaalpha)&=a_0+a_1sigmaalpha+dots+a_n(sigmaalpha)^n\
&=sigma(a_0)+sigma(a_1)sigmaalpha+dots+sigma(a_n)(sigmaalpha)^n&text{since $sigma$ is a $F$-homomorphism}\
&=sigma(a_0+a_1alpha+dots+a_nalpha^n)=p(0)=0.
end{align}






share|cite|improve this answer









$endgroup$



Very simple: if $;p(X)= a_0+a_1X+dots+a_nX^n$ is this minimal polynomial, then
begin{align}
p(sigmaalpha)&=a_0+a_1sigmaalpha+dots+a_n(sigmaalpha)^n\
&=sigma(a_0)+sigma(a_1)sigmaalpha+dots+sigma(a_n)(sigmaalpha)^n&text{since $sigma$ is a $F$-homomorphism}\
&=sigma(a_0+a_1alpha+dots+a_nalpha^n)=p(0)=0.
end{align}







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share|cite|improve this answer



share|cite|improve this answer










answered 27 mins ago









BernardBernard

125k743119




125k743119












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    – Mr. BabaKGhalebi
    22 mins ago


















  • $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago
















$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago




$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago











1












$begingroup$

HINT



For $sigma in Aut(K/F)$ it is $sigma(a)=a forall ain F$






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$endgroup$









  • 1




    $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago
















1












$begingroup$

HINT



For $sigma in Aut(K/F)$ it is $sigma(a)=a forall ain F$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much
    $endgroup$
    – Mr. BabaKGhalebi
    22 mins ago














1












1








1





$begingroup$

HINT



For $sigma in Aut(K/F)$ it is $sigma(a)=a forall ain F$






share|cite|improve this answer









$endgroup$



HINT



For $sigma in Aut(K/F)$ it is $sigma(a)=a forall ain F$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 28 mins ago









giannispapavgiannispapav

2,040325




2,040325








  • 1




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    22 mins ago














  • 1




    $begingroup$
    Thank you very much
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    – Mr. BabaKGhalebi
    22 mins ago








1




1




$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago




$begingroup$
Thank you very much
$endgroup$
– Mr. BabaKGhalebi
22 mins ago










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