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Show that degree of $mathbbQ(sqrt1+sqrt3):mathbbQ=4$

2

$begingroup$

I am solving an ex.: Find basis of $mathbbQ(sqrt1+sqrt3)$ and degree of $mathbbQ(sqrt1+sqrt3):mathbbQ$. SO first of all I have showed that $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$. Also I think that degree in this exercise we find with Short Tower Law. So, degree of $mathbbQ(sqrt3):mathbbQ$ is $2$. How to show that degree of $mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)$ is also $2$? And what is the basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$?

field-theory galois-theory extension-field

share|cite|improve this question

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $K$ be a field. $alpha$ is a root of some irreducible polynomial of degree $n$ with coefficients in $K$ if and only if $1, alpha, dots, alpha^n-1$ is a $K$-basis for $K(alpha)$. If you understand this result, then you will be able to answer both of your questions.
  $endgroup$
  – Gunnar Sveinsson
  Mar 25 at 8:30
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I am solving an ex.: Find basis of $mathbbQ(sqrt1+sqrt3)$ and degree of $mathbbQ(sqrt1+sqrt3):mathbbQ$. SO first of all I have showed that $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$. Also I think that degree in this exercise we find with Short Tower Law. So, degree of $mathbbQ(sqrt3):mathbbQ$ is $2$. How to show that degree of $mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)$ is also $2$? And what is the basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$?

field-theory galois-theory extension-field

share|cite|improve this question

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $K$ be a field. $alpha$ is a root of some irreducible polynomial of degree $n$ with coefficients in $K$ if and only if $1, alpha, dots, alpha^n-1$ is a $K$-basis for $K(alpha)$. If you understand this result, then you will be able to answer both of your questions.
  $endgroup$
  – Gunnar Sveinsson
  Mar 25 at 8:30
  
  

  
  
  
  

add a comment | 

$begingroup$

I am solving an ex.: Find basis of $mathbbQ(sqrt1+sqrt3)$ and degree of $mathbbQ(sqrt1+sqrt3):mathbbQ$. SO first of all I have showed that $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$. Also I think that degree in this exercise we find with Short Tower Law. So, degree of $mathbbQ(sqrt3):mathbbQ$ is $2$. How to show that degree of $mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)$ is also $2$? And what is the basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$?

field-theory galois-theory extension-field

share|cite|improve this question

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

$endgroup$

share|cite|improve this question

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

share|cite|improve this question

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

edited Mar 25 at 8:26

egreg

186k1486208

edited Mar 25 at 8:26

egreg

186k1486208

asked Mar 25 at 7:19

BambeilBambeil

356

asked Mar 25 at 7:19

BambeilBambeil

356

2 Answers
2

active

oldest

votes

0

$begingroup$

Since $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$, it follows that
$$
[mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)]=2
$$
because clearly $sqrt1+sqrt3$ is a root of $x^2-(1+sqrt3)inmathbbQ(sqrt3)[x]$.

Now, by general theory, you know that a basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$ is given by
$$
1,quadsqrt3,quadsqrt1+sqrt3,quadsqrt3sqrt1+sqrt3
$$

share|cite|improve this answer

answered Mar 25 at 8:30

egregegreg

186k1486208

$endgroup$

add a comment | 

3

$begingroup$

$$alpha =sqrt1+sqrt 3implies alpha ^2=1+sqrt 3implies alpha ^4-2alpha ^2-2=0,$$
this polynomial is irreducible by Eisenstein criterion. Therefore, $$[mathbb Q(alpha ):mathbb Q]=4.$$

share|cite|improve this answer

answered Mar 25 at 7:24

user657324user657324

59510

$endgroup$

add a comment | 

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0

$begingroup$

Since $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$, it follows that
$$
[mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)]=2
$$
because clearly $sqrt1+sqrt3$ is a root of $x^2-(1+sqrt3)inmathbbQ(sqrt3)[x]$.

Now, by general theory, you know that a basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$ is given by
$$
1,quadsqrt3,quadsqrt1+sqrt3,quadsqrt3sqrt1+sqrt3
$$

share|cite|improve this answer

answered Mar 25 at 8:30

egregegreg

186k1486208

$endgroup$

add a comment | 

0

$begingroup$

Since $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$, it follows that
$$
[mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)]=2
$$
because clearly $sqrt1+sqrt3$ is a root of $x^2-(1+sqrt3)inmathbbQ(sqrt3)[x]$.

Now, by general theory, you know that a basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$ is given by
$$
1,quadsqrt3,quadsqrt1+sqrt3,quadsqrt3sqrt1+sqrt3
$$

share|cite|improve this answer

answered Mar 25 at 8:30

egregegreg

186k1486208

$endgroup$

add a comment | 

$begingroup$

Since $1+sqrt3$ is not a square in $mathbbQ(sqrt3)$, it follows that
$$
[mathbbQ(sqrt1+sqrt3):mathbbQ(sqrt3)]=2
$$
because clearly $sqrt1+sqrt3$ is a root of $x^2-(1+sqrt3)inmathbbQ(sqrt3)[x]$.

Now, by general theory, you know that a basis of $mathbbQ(sqrt1+sqrt3)$ over $mathbbQ$ is given by
$$
1,quadsqrt3,quadsqrt1+sqrt3,quadsqrt3sqrt1+sqrt3
$$

share|cite|improve this answer

answered Mar 25 at 8:30

egregegreg

186k1486208

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:30

egregegreg

186k1486208

answered Mar 25 at 8:30

egregegreg

186k1486208

answered Mar 25 at 8:30

egregegreg

186k1486208

3

$begingroup$

$$alpha =sqrt1+sqrt 3implies alpha ^2=1+sqrt 3implies alpha ^4-2alpha ^2-2=0,$$
this polynomial is irreducible by Eisenstein criterion. Therefore, $$[mathbb Q(alpha ):mathbb Q]=4.$$

share|cite|improve this answer

answered Mar 25 at 7:24

user657324user657324

59510

$endgroup$

add a comment | 

3

$begingroup$

$$alpha =sqrt1+sqrt 3implies alpha ^2=1+sqrt 3implies alpha ^4-2alpha ^2-2=0,$$
this polynomial is irreducible by Eisenstein criterion. Therefore, $$[mathbb Q(alpha ):mathbb Q]=4.$$

share|cite|improve this answer

answered Mar 25 at 7:24

user657324user657324

59510

$endgroup$

add a comment | 

$begingroup$

$$alpha =sqrt1+sqrt 3implies alpha ^2=1+sqrt 3implies alpha ^4-2alpha ^2-2=0,$$
this polynomial is irreducible by Eisenstein criterion. Therefore, $$[mathbb Q(alpha ):mathbb Q]=4.$$

share|cite|improve this answer

answered Mar 25 at 7:24

user657324user657324

59510

$endgroup$

share|cite|improve this answer

answered Mar 25 at 7:24

user657324user657324

59510

answered Mar 25 at 7:24

user657324user657324

59510

answered Mar 25 at 7:24

user657324user657324

59510