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I wonder if $mathrm{Gal}(E/F) leq S_{d_1} times S_{d_2} times S_{d_3} times cdots times S_{d_n} $ or not

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2

$begingroup$

Let $F$ be a field and $f(x)$ in $F[x]$ satisfy $f(x)=f_1(x)f_2(x)cdots f_n(x)$, where $f_i(x)$ is irreducible in $F[x]$.

My opinion is:

If $E$ is splitting field of $f(x)$ and we denote degree of $f_i(x)$ by $d_i$, then $mathrm{Gal}(E/F) leq S_{d_1} times S_{d_2} times S_{d_3} times cdots times S_{d_n} $ ($mathrm{Gal}(E/F)$ is Galois group of $f(x)$ and $S_{n}$ is permutation group of $n$)

My Question: Is this collect?

If not, is there similar theorem of this opinion? what condition do I need?

For example, if $f(x)$ is separable, then this is true?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sure this is correct $Gal(E/F)$ is a group of permutation of the roots of $f$ and $f_i(alpha_j) = 0, f_i in F[x]$ implies $forall sigmain Gal(E/F),f_i(sigma(alpha_j)) = sigma(f_i(alpha_j))=0$ so $Gal(E/F)$ sends the roots of each $f_i$ to themselves.
  $endgroup$
  – reuns
  2 days ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $F$ be a field and $f(x)$ in $F[x]$ satisfy $f(x)=f_1(x)f_2(x)cdots f_n(x)$, where $f_i(x)$ is irreducible in $F[x]$.

My opinion is:

If $E$ is splitting field of $f(x)$ and we denote degree of $f_i(x)$ by $d_i$, then $mathrm{Gal}(E/F) leq S_{d_1} times S_{d_2} times S_{d_3} times cdots times S_{d_n} $ ($mathrm{Gal}(E/F)$ is Galois group of $f(x)$ and $S_{n}$ is permutation group of $n$)

My Question: Is this collect?

If not, is there similar theorem of this opinion? what condition do I need?

For example, if $f(x)$ is separable, then this is true?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sure this is correct $Gal(E/F)$ is a group of permutation of the roots of $f$ and $f_i(alpha_j) = 0, f_i in F[x]$ implies $forall sigmain Gal(E/F),f_i(sigma(alpha_j)) = sigma(f_i(alpha_j))=0$ so $Gal(E/F)$ sends the roots of each $f_i$ to themselves.
  $endgroup$
  – reuns
  2 days ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $F$ be a field and $f(x)$ in $F[x]$ satisfy $f(x)=f_1(x)f_2(x)cdots f_n(x)$, where $f_i(x)$ is irreducible in $F[x]$.

My opinion is:

If $E$ is splitting field of $f(x)$ and we denote degree of $f_i(x)$ by $d_i$, then $mathrm{Gal}(E/F) leq S_{d_1} times S_{d_2} times S_{d_3} times cdots times S_{d_n} $ ($mathrm{Gal}(E/F)$ is Galois group of $f(x)$ and $S_{n}$ is permutation group of $n$)

My Question: Is this collect?

If not, is there similar theorem of this opinion? what condition do I need?

For example, if $f(x)$ is separable, then this is true?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

$endgroup$

share|cite|improve this question

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

share|cite|improve this question

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

edited 2 days ago

user26857

39.3k124183

edited 2 days ago

user26857

39.3k124183

asked 2 days ago

hewhew

496

asked 2 days ago

hewhew

496