Equality of field extensions given Splitting field is $S_n$.

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Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$.

Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$.

Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$.

Attempt
$[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$.

$[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible.

$mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$.

Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$.

If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$.

How can I claim it is not possible?

Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$

galois-theory splitting-field

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asked Aug 6 at 23:40

Jo'

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up vote
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Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$.

Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$.

Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$.

Attempt
$[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$.

$[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible.

$mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$.

Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$.

If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$.

How can I claim it is not possible?

Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$

galois-theory splitting-field

share|cite|improve this question

asked Aug 6 at 23:40

Jo'

1469

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$.

Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$.

Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$.

Attempt
$[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$.

$[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible.

$mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$.

Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$.

If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$.

How can I claim it is not possible?

Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$

galois-theory splitting-field

share|cite|improve this question

asked Aug 6 at 23:40

Jo'

1469

Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$.

Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$.

Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$.

Attempt
$[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$.

$[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible.

$mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$.

Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$.

If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$.

How can I claim it is not possible?

Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$

galois-theory splitting-field

share|cite|improve this question

asked Aug 6 at 23:40

Jo'

1469

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 23:40

Jo'

1469

asked Aug 6 at 23:40

Jo'

1469

asked Aug 6 at 23:40

Jo'

1469

1469

add a comment | 

add a comment | 

1 Answer
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accepted

Note that $[L:mathbbQ(alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $fracf(x)x-alpha in Q(alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_n-1$.

As you've noted the Galois group of the extension $mathbbQ(alpha^4) subset L$ is $A_n$. Also as $mathbbQ(alpha^4) subset mathbbQ(alpha)$ we must have that the Galois group of $mathbbQ(alpha) subset L$ is contained in the Galois group of $mathbbQ(alpha^4) subset L$. In other words $S_n-1 le A_n$. However this is an obvious contradiction.

---

Another way is to consider the extension field $mathbbQ(alpha^2)$. With similar reasoning we get that $[mathbbQ(alpha^2):mathbbQ] = 2$ or $n$. If it's $2$, then we must have that $mathbbQ(alpha^2) = mathbbQ(alpha^4)$. However $mathbbQ(alpha^2) subseteq mathbbQ(alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - alpha^2 in mathbbQ(alpha^2)[x]$. As $A_n$, the Galois group of the extension $mathbbQ(alpha^2) subset L$ is a simple group for $n ge 5$ we must get that $mathbbQ(alpha^2) = mathbbQ(alpha)$.

If $[mathbbQ(alpha^2):mathbbQ] = n$, then we have that $mathbbQ(alpha) = mathbbQ(alpha^2)$. Similalry $mathbbQ(alpha^4) subseteq mathbbQ(alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.

Hence the proof.

share|cite|improve this answer

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Many thanks. @Stefan4024
  – Jo'
  Aug 7 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo' You're welcome
  – Stefan4024
  Aug 7 at 0:52
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Note that $[L:mathbbQ(alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $fracf(x)x-alpha in Q(alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_n-1$.

As you've noted the Galois group of the extension $mathbbQ(alpha^4) subset L$ is $A_n$. Also as $mathbbQ(alpha^4) subset mathbbQ(alpha)$ we must have that the Galois group of $mathbbQ(alpha) subset L$ is contained in the Galois group of $mathbbQ(alpha^4) subset L$. In other words $S_n-1 le A_n$. However this is an obvious contradiction.

---

Another way is to consider the extension field $mathbbQ(alpha^2)$. With similar reasoning we get that $[mathbbQ(alpha^2):mathbbQ] = 2$ or $n$. If it's $2$, then we must have that $mathbbQ(alpha^2) = mathbbQ(alpha^4)$. However $mathbbQ(alpha^2) subseteq mathbbQ(alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - alpha^2 in mathbbQ(alpha^2)[x]$. As $A_n$, the Galois group of the extension $mathbbQ(alpha^2) subset L$ is a simple group for $n ge 5$ we must get that $mathbbQ(alpha^2) = mathbbQ(alpha)$.

If $[mathbbQ(alpha^2):mathbbQ] = n$, then we have that $mathbbQ(alpha) = mathbbQ(alpha^2)$. Similalry $mathbbQ(alpha^4) subseteq mathbbQ(alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.

Hence the proof.

share|cite|improve this answer

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Many thanks. @Stefan4024
  – Jo'
  Aug 7 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo' You're welcome
  – Stefan4024
  Aug 7 at 0:52
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

Note that $[L:mathbbQ(alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $fracf(x)x-alpha in Q(alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_n-1$.

As you've noted the Galois group of the extension $mathbbQ(alpha^4) subset L$ is $A_n$. Also as $mathbbQ(alpha^4) subset mathbbQ(alpha)$ we must have that the Galois group of $mathbbQ(alpha) subset L$ is contained in the Galois group of $mathbbQ(alpha^4) subset L$. In other words $S_n-1 le A_n$. However this is an obvious contradiction.

---

Another way is to consider the extension field $mathbbQ(alpha^2)$. With similar reasoning we get that $[mathbbQ(alpha^2):mathbbQ] = 2$ or $n$. If it's $2$, then we must have that $mathbbQ(alpha^2) = mathbbQ(alpha^4)$. However $mathbbQ(alpha^2) subseteq mathbbQ(alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - alpha^2 in mathbbQ(alpha^2)[x]$. As $A_n$, the Galois group of the extension $mathbbQ(alpha^2) subset L$ is a simple group for $n ge 5$ we must get that $mathbbQ(alpha^2) = mathbbQ(alpha)$.

If $[mathbbQ(alpha^2):mathbbQ] = n$, then we have that $mathbbQ(alpha) = mathbbQ(alpha^2)$. Similalry $mathbbQ(alpha^4) subseteq mathbbQ(alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.

Hence the proof.

share|cite|improve this answer

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Many thanks. @Stefan4024
  – Jo'
  Aug 7 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo' You're welcome
  – Stefan4024
  Aug 7 at 0:52
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Note that $[L:mathbbQ(alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $fracf(x)x-alpha in Q(alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_n-1$.

As you've noted the Galois group of the extension $mathbbQ(alpha^4) subset L$ is $A_n$. Also as $mathbbQ(alpha^4) subset mathbbQ(alpha)$ we must have that the Galois group of $mathbbQ(alpha) subset L$ is contained in the Galois group of $mathbbQ(alpha^4) subset L$. In other words $S_n-1 le A_n$. However this is an obvious contradiction.

---

Another way is to consider the extension field $mathbbQ(alpha^2)$. With similar reasoning we get that $[mathbbQ(alpha^2):mathbbQ] = 2$ or $n$. If it's $2$, then we must have that $mathbbQ(alpha^2) = mathbbQ(alpha^4)$. However $mathbbQ(alpha^2) subseteq mathbbQ(alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - alpha^2 in mathbbQ(alpha^2)[x]$. As $A_n$, the Galois group of the extension $mathbbQ(alpha^2) subset L$ is a simple group for $n ge 5$ we must get that $mathbbQ(alpha^2) = mathbbQ(alpha)$.

If $[mathbbQ(alpha^2):mathbbQ] = n$, then we have that $mathbbQ(alpha) = mathbbQ(alpha^2)$. Similalry $mathbbQ(alpha^4) subseteq mathbbQ(alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.

Hence the proof.

share|cite|improve this answer

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

Note that $[L:mathbbQ(alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $fracf(x)x-alpha in Q(alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_n-1$.

As you've noted the Galois group of the extension $mathbbQ(alpha^4) subset L$ is $A_n$. Also as $mathbbQ(alpha^4) subset mathbbQ(alpha)$ we must have that the Galois group of $mathbbQ(alpha) subset L$ is contained in the Galois group of $mathbbQ(alpha^4) subset L$. In other words $S_n-1 le A_n$. However this is an obvious contradiction.

---

Another way is to consider the extension field $mathbbQ(alpha^2)$. With similar reasoning we get that $[mathbbQ(alpha^2):mathbbQ] = 2$ or $n$. If it's $2$, then we must have that $mathbbQ(alpha^2) = mathbbQ(alpha^4)$. However $mathbbQ(alpha^2) subseteq mathbbQ(alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - alpha^2 in mathbbQ(alpha^2)[x]$. As $A_n$, the Galois group of the extension $mathbbQ(alpha^2) subset L$ is a simple group for $n ge 5$ we must get that $mathbbQ(alpha^2) = mathbbQ(alpha)$.

If $[mathbbQ(alpha^2):mathbbQ] = n$, then we have that $mathbbQ(alpha) = mathbbQ(alpha^2)$. Similalry $mathbbQ(alpha^4) subseteq mathbbQ(alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.

Hence the proof.

share|cite|improve this answer

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

share|cite|improve this answer

share|cite|improve this answer

edited Aug 7 at 0:46

edited Aug 7 at 0:46

edited Aug 7 at 0:46

answered Aug 7 at 0:40

Stefan4024

28.5k53174

answered Aug 7 at 0:40

Stefan4024

28.5k53174

answered Aug 7 at 0:40

Stefan4024

28.5k53174

28.5k53174

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Many thanks. @Stefan4024
  – Jo'
  Aug 7 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo' You're welcome
  – Stefan4024
  Aug 7 at 0:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Many thanks. @Stefan4024
  – Jo'
  Aug 7 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jo' You're welcome
  – Stefan4024
  Aug 7 at 0:52
  

  
  
  
  

1

1

Many thanks. @Stefan4024
– Jo'
Aug 7 at 0:51

Many thanks. @Stefan4024
– Jo'
Aug 7 at 0:51

@Jo' You're welcome
– Stefan4024
Aug 7 at 0:52

@Jo' You're welcome
– Stefan4024
Aug 7 at 0:52

add a comment | 

 

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