Splitting Field of a Polynomial in $mathbb{Q}$












1












$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







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$endgroup$












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02
















1












$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02














1












1








1


1



$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







share|cite|improve this question











$endgroup$




We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field




abstract-algebra splitting-field






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 10:14









Bernard

120k740115




120k740115










asked Jan 16 at 9:55









KingDingelingKingDingeling

1497




1497












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02


















  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02
















$begingroup$
The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
$endgroup$
– Wojowu
Jan 16 at 9:59




$begingroup$
The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
$endgroup$
– Wojowu
Jan 16 at 9:59












$begingroup$
Do you mean $X^4-2X^2+2$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 10:01




$begingroup$
Do you mean $X^4-2X^2+2$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 10:01












$begingroup$
Guys... I'm sorry. I edited both my mistakes. Thank you.
$endgroup$
– KingDingeling
Jan 16 at 10:02




$begingroup$
Guys... I'm sorry. I edited both my mistakes. Thank you.
$endgroup$
– KingDingeling
Jan 16 at 10:02










1 Answer
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$begingroup$


Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
100% sure about $sqrt{1-i}$.




Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






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    $begingroup$


    Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
    100% sure about $sqrt{1-i}$.




    Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



    Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$


      Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
      100% sure about $sqrt{1-i}$.




      Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



      Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$


        Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
        100% sure about $sqrt{1-i}$.




        Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



        Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






        share|cite|improve this answer









        $endgroup$




        Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
        100% sure about $sqrt{1-i}$.




        Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



        Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 10:49









        Thomas ShelbyThomas Shelby

        2,877421




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