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Let $F$ be a field. Is it true that if $[F(sqrtD) : F] = 2$, then $D in F$?

Splitting field that isn't a Galois extensionField Extensions and Number of IsomorphismsQuadratic Extensions over Field of Characteristic $neq 2$.Help justifying that $mathbb Q(sqrt[3]2)$ is not a splitting field over $mathbb Q$.Degree of the extension $Bbb Q(sqrt2, sqrt3)/Bbb Q(sqrt2)$ is at most 2Field Extension With Cube Root of 7Let $n geq 3$ and let $p$ be prime. Show that $sqrt[n]p$ is not contained in a cyclotomic extension of $mathbbQ$Showing the minimal polynomial for an element in an extension field is the same as the minimal polynomial of a linear transformation.$Bbb Q(sqrt 2)$ and $Bbb Q(sqrt 3)$ are not isomorphicDegree of the extension $mathbbQ(sqrt3 + 2sqrt2)$.

1

1

$begingroup$

In Dummit & Foote, problem 14.2.17(c), the authors hand us a quadratic extension of the form $F(sqrt D)$. Now, while I am pretty sure you need that $D in F$ to do this particular problem, I can't help but wonder if assuming this is even necessary. In other words, if the degree of the extension $F(sqrt D)/F$ is $2$, then must $D$ belong to $F$?

My thoughts so far are as follows: If $D notin F$, then $[F(D) : F] > 1$. This puts $sqrt D in F(D)$, otherwise
$$
4 le [F(sqrt D) : F(D)][F(D):F] = [F(sqrt D): F] = 2.
$$
This tells us that $F(D) = F(sqrt D)$. My next observation was that if $sqrt D$ has minimal polynomial $x^2 + ax + b$, we can write $sqrt D$ in terms of $D$ and elements of $F$ and use this to find the minimal polynomial of $D$ over $F$:
$$
sqrt D = -(D + b)/a, m_D, F(x) = x^2 + (2ab - a)x + b^2.
$$
Notice here that $a neq 0$, or else $D in F$!

abstract-algebra field-theory galois-theory extension-field

share|cite|improve this question

asked 2 days ago

fauxefoxfauxefox

11410

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused. If $D not in F$ then what does $F(sqrtD )$ mean?
  $endgroup$
  – Ethan Bolker
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Ethan You don't know what $,Bbb Q(sqrt1+sqrt 3),$ means?
  $endgroup$
  – Bill Dubuque
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BillDubuque Now I'm no longer confused. Thanks.
  $endgroup$
  – Ethan Bolker
  2 days ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

In Dummit & Foote, problem 14.2.17(c), the authors hand us a quadratic extension of the form $F(sqrt D)$. Now, while I am pretty sure you need that $D in F$ to do this particular problem, I can't help but wonder if assuming this is even necessary. In other words, if the degree of the extension $F(sqrt D)/F$ is $2$, then must $D$ belong to $F$?

My thoughts so far are as follows: If $D notin F$, then $[F(D) : F] > 1$. This puts $sqrt D in F(D)$, otherwise
$$
4 le [F(sqrt D) : F(D)][F(D):F] = [F(sqrt D): F] = 2.
$$
This tells us that $F(D) = F(sqrt D)$. My next observation was that if $sqrt D$ has minimal polynomial $x^2 + ax + b$, we can write $sqrt D$ in terms of $D$ and elements of $F$ and use this to find the minimal polynomial of $D$ over $F$:
$$
sqrt D = -(D + b)/a, m_D, F(x) = x^2 + (2ab - a)x + b^2.
$$
Notice here that $a neq 0$, or else $D in F$!

abstract-algebra field-theory galois-theory extension-field

share|cite|improve this question

asked 2 days ago

fauxefoxfauxefox

11410

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused. If $D not in F$ then what does $F(sqrtD )$ mean?
  $endgroup$
  – Ethan Bolker
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Ethan You don't know what $,Bbb Q(sqrt1+sqrt 3),$ means?
  $endgroup$
  – Bill Dubuque
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BillDubuque Now I'm no longer confused. Thanks.
  $endgroup$
  – Ethan Bolker
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

In Dummit & Foote, problem 14.2.17(c), the authors hand us a quadratic extension of the form $F(sqrt D)$. Now, while I am pretty sure you need that $D in F$ to do this particular problem, I can't help but wonder if assuming this is even necessary. In other words, if the degree of the extension $F(sqrt D)/F$ is $2$, then must $D$ belong to $F$?

My thoughts so far are as follows: If $D notin F$, then $[F(D) : F] > 1$. This puts $sqrt D in F(D)$, otherwise
$$
4 le [F(sqrt D) : F(D)][F(D):F] = [F(sqrt D): F] = 2.
$$
This tells us that $F(D) = F(sqrt D)$. My next observation was that if $sqrt D$ has minimal polynomial $x^2 + ax + b$, we can write $sqrt D$ in terms of $D$ and elements of $F$ and use this to find the minimal polynomial of $D$ over $F$:
$$
sqrt D = -(D + b)/a, m_D, F(x) = x^2 + (2ab - a)x + b^2.
$$
Notice here that $a neq 0$, or else $D in F$!

abstract-algebra field-theory galois-theory extension-field

share|cite|improve this question

asked 2 days ago

fauxefoxfauxefox

11410

$endgroup$

share|cite|improve this question

asked 2 days ago

fauxefoxfauxefox

11410

share|cite|improve this question

asked 2 days ago

fauxefoxfauxefox

11410

asked 2 days ago

fauxefoxfauxefox

11410

asked 2 days ago

fauxefoxfauxefox

11410

2 Answers
2

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7

$begingroup$

This is false. Consider the case of $F=BbbQ$, $D=(1+sqrt2)^2=3+2sqrt2.$

We have $sqrtD=pm(1+sqrt2)$, so $BbbQ(sqrt D)=BbbQ(D)=BbbQ(sqrt2)$.

share|cite|improve this answer

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

$endgroup$

add a comment | 

3

$begingroup$

If $[F(sqrt D):F]=2$ the extension is Galois with group $1,sigma$ and
$$
sigma(sqrt D)=a+bsqrt D,qquad a,bin F.
$$
Since $sigma^2=1$ we must have
$sqrt D=sigma^2(sqrt D)=a+ab+b^2sqrt D$, i.e.
$$
a(b+1)=0qquadtextandqquad b^2=1.
$$
If $a=0$ and $b=-1$ it readily follows that $Din F$.

On the other hand, if $aneq0$ and $b=-1$ we have
$$
Fnirm N(sqrt D)=sqrt Dcdotsigma(sqrt D)=asqrt D-D
$$
from which $D=rm N(sqrt D)-asqrt Dnotin F$. Jyrki Lahtonen's answer shows an instance of the latter situation happening.

share|cite|improve this answer

edited 21 hours ago

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $operatornamechar F = 2$, then $a+bsqrt Dmapsto a-bsqrt D$ is the trivial automorphism.
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! @JyrkiLahtonen is right; I mentioned near the end of my question that in the minimal polynomial of $sqrt D$, $a neq 0$.
  $endgroup$
  – fauxefox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, I fixed my answer.
  $endgroup$
  – Andrea Mori
  21 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

This is false. Consider the case of $F=BbbQ$, $D=(1+sqrt2)^2=3+2sqrt2.$

We have $sqrtD=pm(1+sqrt2)$, so $BbbQ(sqrt D)=BbbQ(D)=BbbQ(sqrt2)$.

share|cite|improve this answer

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

$endgroup$

add a comment | 

7

$begingroup$

This is false. Consider the case of $F=BbbQ$, $D=(1+sqrt2)^2=3+2sqrt2.$

We have $sqrtD=pm(1+sqrt2)$, so $BbbQ(sqrt D)=BbbQ(D)=BbbQ(sqrt2)$.

share|cite|improve this answer

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

$endgroup$

add a comment | 

$begingroup$

This is false. Consider the case of $F=BbbQ$, $D=(1+sqrt2)^2=3+2sqrt2.$

We have $sqrtD=pm(1+sqrt2)$, so $BbbQ(sqrt D)=BbbQ(D)=BbbQ(sqrt2)$.

share|cite|improve this answer

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

$endgroup$

share|cite|improve this answer

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

answered yesterday

Jyrki LahtonenJyrki Lahtonen

110k13171381

3

$begingroup$

If $[F(sqrt D):F]=2$ the extension is Galois with group $1,sigma$ and
$$
sigma(sqrt D)=a+bsqrt D,qquad a,bin F.
$$
Since $sigma^2=1$ we must have
$sqrt D=sigma^2(sqrt D)=a+ab+b^2sqrt D$, i.e.
$$
a(b+1)=0qquadtextandqquad b^2=1.
$$
If $a=0$ and $b=-1$ it readily follows that $Din F$.

On the other hand, if $aneq0$ and $b=-1$ we have
$$
Fnirm N(sqrt D)=sqrt Dcdotsigma(sqrt D)=asqrt D-D
$$
from which $D=rm N(sqrt D)-asqrt Dnotin F$. Jyrki Lahtonen's answer shows an instance of the latter situation happening.

share|cite|improve this answer

edited 21 hours ago

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $operatornamechar F = 2$, then $a+bsqrt Dmapsto a-bsqrt D$ is the trivial automorphism.
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! @JyrkiLahtonen is right; I mentioned near the end of my question that in the minimal polynomial of $sqrt D$, $a neq 0$.
  $endgroup$
  – fauxefox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, I fixed my answer.
  $endgroup$
  – Andrea Mori
  21 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

If $[F(sqrt D):F]=2$ the extension is Galois with group $1,sigma$ and
$$
sigma(sqrt D)=a+bsqrt D,qquad a,bin F.
$$
Since $sigma^2=1$ we must have
$sqrt D=sigma^2(sqrt D)=a+ab+b^2sqrt D$, i.e.
$$
a(b+1)=0qquadtextandqquad b^2=1.
$$
If $a=0$ and $b=-1$ it readily follows that $Din F$.

On the other hand, if $aneq0$ and $b=-1$ we have
$$
Fnirm N(sqrt D)=sqrt Dcdotsigma(sqrt D)=asqrt D-D
$$
from which $D=rm N(sqrt D)-asqrt Dnotin F$. Jyrki Lahtonen's answer shows an instance of the latter situation happening.

share|cite|improve this answer

edited 21 hours ago

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $operatornamechar F = 2$, then $a+bsqrt Dmapsto a-bsqrt D$ is the trivial automorphism.
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! @JyrkiLahtonen is right; I mentioned near the end of my question that in the minimal polynomial of $sqrt D$, $a neq 0$.
  $endgroup$
  – fauxefox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, I fixed my answer.
  $endgroup$
  – Andrea Mori
  21 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $[F(sqrt D):F]=2$ the extension is Galois with group $1,sigma$ and
$$
sigma(sqrt D)=a+bsqrt D,qquad a,bin F.
$$
Since $sigma^2=1$ we must have
$sqrt D=sigma^2(sqrt D)=a+ab+b^2sqrt D$, i.e.
$$
a(b+1)=0qquadtextandqquad b^2=1.
$$
If $a=0$ and $b=-1$ it readily follows that $Din F$.

On the other hand, if $aneq0$ and $b=-1$ we have
$$
Fnirm N(sqrt D)=sqrt Dcdotsigma(sqrt D)=asqrt D-D
$$
from which $D=rm N(sqrt D)-asqrt Dnotin F$. Jyrki Lahtonen's answer shows an instance of the latter situation happening.

share|cite|improve this answer

edited 21 hours ago

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

$endgroup$

share|cite|improve this answer

edited 21 hours ago

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466

answered 2 days ago

Andrea MoriAndrea Mori

20.1k13466