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Intermediate Galois subfields of $ mathbbQ(sqrt[n]alpha) $

0

$begingroup$

Let $ alpha > 0 $ in $ mathbbQ $ and let $ K = mathbbQ(sqrt[n]alpha) $ of degree $ n $ over $ mathbbQ $. Determine all nontrivial subfields of $ K $ that are Galois over $ mathbbQ $. The only such subfield that I could find is $ mathbbQ(sqrtalpha) $; how do I find all such subfields or prove that they don't exist?

Next let $ L $ be the Galois closure of $ K $. Determine $ [L : mathbbQ] $. I know that if $ alpha = 1 $, then $ K = mathbbQ(zeta_n) $ where $ zeta_n $ is a primitive $ n $th root of unity. So $ L cong (mathbbZ / n mathbbZ)^times $, so $ [L : mathbbQ] = phi(n) $. But how do I approach with the case when $ alpha neq 1 $?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related : math.stackexchange.com/questions/348621/radical-extension
  $endgroup$
  – Thomas Shelby
  Mar 19 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The field $BbbQ$ already has infinitely many elements, so the tag finite-fields was inappropriate.
  $endgroup$
  – Jyrki Lahtonen
  Mar 20 at 3:37
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $ alpha > 0 $ in $ mathbbQ $ and let $ K = mathbbQ(sqrt[n]alpha) $ of degree $ n $ over $ mathbbQ $. Determine all nontrivial subfields of $ K $ that are Galois over $ mathbbQ $. The only such subfield that I could find is $ mathbbQ(sqrtalpha) $; how do I find all such subfields or prove that they don't exist?

Next let $ L $ be the Galois closure of $ K $. Determine $ [L : mathbbQ] $. I know that if $ alpha = 1 $, then $ K = mathbbQ(zeta_n) $ where $ zeta_n $ is a primitive $ n $th root of unity. So $ L cong (mathbbZ / n mathbbZ)^times $, so $ [L : mathbbQ] = phi(n) $. But how do I approach with the case when $ alpha neq 1 $?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related : math.stackexchange.com/questions/348621/radical-extension
  $endgroup$
  – Thomas Shelby
  Mar 19 at 21:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The field $BbbQ$ already has infinitely many elements, so the tag finite-fields was inappropriate.
  $endgroup$
  – Jyrki Lahtonen
  Mar 20 at 3:37
  

  
  
  
  

add a comment | 

$begingroup$

Let $ alpha > 0 $ in $ mathbbQ $ and let $ K = mathbbQ(sqrt[n]alpha) $ of degree $ n $ over $ mathbbQ $. Determine all nontrivial subfields of $ K $ that are Galois over $ mathbbQ $. The only such subfield that I could find is $ mathbbQ(sqrtalpha) $; how do I find all such subfields or prove that they don't exist?

Next let $ L $ be the Galois closure of $ K $. Determine $ [L : mathbbQ] $. I know that if $ alpha = 1 $, then $ K = mathbbQ(zeta_n) $ where $ zeta_n $ is a primitive $ n $th root of unity. So $ L cong (mathbbZ / n mathbbZ)^times $, so $ [L : mathbbQ] = phi(n) $. But how do I approach with the case when $ alpha neq 1 $?

abstract-algebra field-theory galois-theory

share|cite|improve this question

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

$endgroup$

share|cite|improve this question

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

share|cite|improve this question

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

edited Mar 20 at 3:36

Jyrki Lahtonen

110k13172387

asked Mar 19 at 21:37

MaddieMaddie

11

asked Mar 19 at 21:37

MaddieMaddie

11