Prove $sqrt14 inmathbbQ(sqrt2+ sqrt7)$. [on hold]

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Can't get my head around the question relating to field theory.



Prove $sqrt14 in mathbbQ( sqrt2 + sqrt7 )$.



which leads to



Compute $[mathbbQ( sqrt2 + sqrt7 ) : mathbbQ( sqrt14 )]$



Trying to read ahead for a head start on next years work, Thanks




field-theory




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put on hold as off-topic by amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 4




    Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
    – spiralstotheleft
    2 days ago






  • 1




    You should use mathjax for your equation formatting.
    – Rumpelstiltskin
    2 days ago






  • 1




    @spiralstotheleft It would be great if you post this as an answer :)
    – TheSimpliFire
    2 days ago














up vote
1
down vote

favorite












Can't get my head around the question relating to field theory.



Prove $sqrt14 in mathbbQ( sqrt2 + sqrt7 )$.



which leads to



Compute $[mathbbQ( sqrt2 + sqrt7 ) : mathbbQ( sqrt14 )]$



Trying to read ahead for a head start on next years work, Thanks




field-theory




share|cite|improve this question













put on hold as off-topic by amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 4




    Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
    – spiralstotheleft
    2 days ago






  • 1




    You should use mathjax for your equation formatting.
    – Rumpelstiltskin
    2 days ago






  • 1




    @spiralstotheleft It would be great if you post this as an answer :)
    – TheSimpliFire
    2 days ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Can't get my head around the question relating to field theory.



Prove $sqrt14 in mathbbQ( sqrt2 + sqrt7 )$.



which leads to



Compute $[mathbbQ( sqrt2 + sqrt7 ) : mathbbQ( sqrt14 )]$



Trying to read ahead for a head start on next years work, Thanks




field-theory




share|cite|improve this question













Can't get my head around the question relating to field theory.



Prove $sqrt14 in mathbbQ( sqrt2 + sqrt7 )$.



which leads to



Compute $[mathbbQ( sqrt2 + sqrt7 ) : mathbbQ( sqrt14 )]$



Trying to read ahead for a head start on next years work, Thanks





field-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Rumpelstiltskin

1,386315




1,386315









asked 2 days ago









Roman

112




112




put on hold as off-topic by amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, John Ma, José Carlos Santos, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 4




    Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
    – spiralstotheleft
    2 days ago






  • 1




    You should use mathjax for your equation formatting.
    – Rumpelstiltskin
    2 days ago






  • 1




    @spiralstotheleft It would be great if you post this as an answer :)
    – TheSimpliFire
    2 days ago












  • 4




    Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
    – spiralstotheleft
    2 days ago






  • 1




    You should use mathjax for your equation formatting.
    – Rumpelstiltskin
    2 days ago






  • 1




    @spiralstotheleft It would be great if you post this as an answer :)
    – TheSimpliFire
    2 days ago







4




4




Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
– spiralstotheleft
2 days ago




Elements of $mathbbQ(sqrt2+sqrt7)$ are polynomials of $r=sqrt2+sqrt7$ with rational coefficients. You just need to see if one of them gives you $sqrt14$. Since rational linear combinations of $1,r,r^2,r^3,...$ generate all polynomials in $r$, then you only need to compute powers of $r$ and see if a linear combination of them is $sqrt17$. As soon as you compute $r^2=2+2sqrt14+7$ you see that a linear combination of $1$ and $r^2$ is $sqrt14$.
– spiralstotheleft
2 days ago




1




1




You should use mathjax for your equation formatting.
– Rumpelstiltskin
2 days ago




You should use mathjax for your equation formatting.
– Rumpelstiltskin
2 days ago




1




1




@spiralstotheleft It would be great if you post this as an answer :)
– TheSimpliFire
2 days ago




@spiralstotheleft It would be great if you post this as an answer :)
– TheSimpliFire
2 days ago










1 Answer
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We claim that
$$
mathbbQ(sqrt2+sqrt7)=mathbbQ(sqrt2, sqrt7)
$$
from which the desired result would follow. Since $sqrt2+sqrt7in mathbbQ(sqrt2, sqrt7)$, it follows that $mathbbQ(sqrt2+sqrt7)subsetmathbbQ(sqrt2, sqrt7)$. Further, note that
$$
-5(sqrt2+sqrt7)^-1=sqrt2-sqrt7inmathbbQ(sqrt2+sqrt7)
$$
Thus $sqrt2, sqrt7in mathbbQ(sqrt2+sqrt7)$, from which the reverse containment follows.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    We claim that
    $$
    mathbbQ(sqrt2+sqrt7)=mathbbQ(sqrt2, sqrt7)
    $$
    from which the desired result would follow. Since $sqrt2+sqrt7in mathbbQ(sqrt2, sqrt7)$, it follows that $mathbbQ(sqrt2+sqrt7)subsetmathbbQ(sqrt2, sqrt7)$. Further, note that
    $$
    -5(sqrt2+sqrt7)^-1=sqrt2-sqrt7inmathbbQ(sqrt2+sqrt7)
    $$
    Thus $sqrt2, sqrt7in mathbbQ(sqrt2+sqrt7)$, from which the reverse containment follows.






    share|cite|improve this answer

























      up vote
      0
      down vote













      We claim that
      $$
      mathbbQ(sqrt2+sqrt7)=mathbbQ(sqrt2, sqrt7)
      $$
      from which the desired result would follow. Since $sqrt2+sqrt7in mathbbQ(sqrt2, sqrt7)$, it follows that $mathbbQ(sqrt2+sqrt7)subsetmathbbQ(sqrt2, sqrt7)$. Further, note that
      $$
      -5(sqrt2+sqrt7)^-1=sqrt2-sqrt7inmathbbQ(sqrt2+sqrt7)
      $$
      Thus $sqrt2, sqrt7in mathbbQ(sqrt2+sqrt7)$, from which the reverse containment follows.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        We claim that
        $$
        mathbbQ(sqrt2+sqrt7)=mathbbQ(sqrt2, sqrt7)
        $$
        from which the desired result would follow. Since $sqrt2+sqrt7in mathbbQ(sqrt2, sqrt7)$, it follows that $mathbbQ(sqrt2+sqrt7)subsetmathbbQ(sqrt2, sqrt7)$. Further, note that
        $$
        -5(sqrt2+sqrt7)^-1=sqrt2-sqrt7inmathbbQ(sqrt2+sqrt7)
        $$
        Thus $sqrt2, sqrt7in mathbbQ(sqrt2+sqrt7)$, from which the reverse containment follows.






        share|cite|improve this answer













        We claim that
        $$
        mathbbQ(sqrt2+sqrt7)=mathbbQ(sqrt2, sqrt7)
        $$
        from which the desired result would follow. Since $sqrt2+sqrt7in mathbbQ(sqrt2, sqrt7)$, it follows that $mathbbQ(sqrt2+sqrt7)subsetmathbbQ(sqrt2, sqrt7)$. Further, note that
        $$
        -5(sqrt2+sqrt7)^-1=sqrt2-sqrt7inmathbbQ(sqrt2+sqrt7)
        $$
        Thus $sqrt2, sqrt7in mathbbQ(sqrt2+sqrt7)$, from which the reverse containment follows.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        Foobaz John

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        17.9k41244












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