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Is this how to find a polynomial with a given splitting field?

Determining whether or not an extension is a splitting fieldSplitting field for polynomialsplitting field of a polynomial over a finite fieldFind the splitting field without the explicit rootsFinding one primitive element for splitting field of polynomial over $mathbb{Q}$Finding minimal polynomial and splitting fieldCalculating the splitting field of this polynomialFinding the splitting field of a polynomial.Confusion about the splitting field of $x^3-7$Find polynomial given splitting field

0

$begingroup$

Suppose we want to find a polynomial whose splitting field is $Bbb Q(sqrt{2}, sqrt{-3})$ over $Bbb Q$. Then the following is how you'd do it right ?

We want to adjoin the roots $alpha=sqrt{2}, beta=sqrt{-3}$. so we could just say $(x^2-2)(x^2+3)=0$ and expanding the brackets gives the polynomial .

abstract-algebra ring-theory splitting-field

share|cite|improve this question

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's exactly how I would do it.
  $endgroup$
  – Lord Shark the Unknown
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose we want to find a polynomial whose splitting field is $Bbb Q(sqrt{2}, sqrt{-3})$ over $Bbb Q$. Then the following is how you'd do it right ?

We want to adjoin the roots $alpha=sqrt{2}, beta=sqrt{-3}$. so we could just say $(x^2-2)(x^2+3)=0$ and expanding the brackets gives the polynomial .

abstract-algebra ring-theory splitting-field

share|cite|improve this question

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that's exactly how I would do it.
  $endgroup$
  – Lord Shark the Unknown
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we want to find a polynomial whose splitting field is $Bbb Q(sqrt{2}, sqrt{-3})$ over $Bbb Q$. Then the following is how you'd do it right ?

We want to adjoin the roots $alpha=sqrt{2}, beta=sqrt{-3}$. so we could just say $(x^2-2)(x^2+3)=0$ and expanding the brackets gives the polynomial .

abstract-algebra ring-theory splitting-field

share|cite|improve this question

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

$endgroup$

share|cite|improve this question

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

share|cite|improve this question

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

asked 2 days ago

can'tcauchycan'tcauchy

1,016417

asked 2 days ago

can'tcauchycan'tcauchy

1,016417