Degree of extension of $mathbb{Q}(S)$ with $S={ sqrt3,sqrt5}$












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First time in abstract algebra, and right now I have some difficult to understand this example from one of my books.



How can I calculate the grade of extension of $mathbb{Q}(S)$ with $S={ sqrt3,sqrt5}$? If I extend $mathbb{Q})$ with $sqrt 3$ I have that $mathbb{Q}(sqrt 3)$ minimum polynomial is $x^2-3$, but then what?



Thanks all






abstract-algebra







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    Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
    $endgroup$
    – Jack D'Aurizio
    Dec 10 '18 at 12:43






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    $begingroup$
    You will also need $sqrt{15}$ to complete the field, so degree is 4.
    $endgroup$
    – orion
    Dec 10 '18 at 12:43












  • $begingroup$
    Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
    $endgroup$
    – Alessar
    Dec 10 '18 at 14:22
















0












$begingroup$


First time in abstract algebra, and right now I have some difficult to understand this example from one of my books.



How can I calculate the grade of extension of $mathbb{Q}(S)$ with $S={ sqrt3,sqrt5}$? If I extend $mathbb{Q})$ with $sqrt 3$ I have that $mathbb{Q}(sqrt 3)$ minimum polynomial is $x^2-3$, but then what?



Thanks all






abstract-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
    $endgroup$
    – Jack D'Aurizio
    Dec 10 '18 at 12:43






  • 1




    $begingroup$
    You will also need $sqrt{15}$ to complete the field, so degree is 4.
    $endgroup$
    – orion
    Dec 10 '18 at 12:43












  • $begingroup$
    Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
    $endgroup$
    – Alessar
    Dec 10 '18 at 14:22














0












0








0





$begingroup$


First time in abstract algebra, and right now I have some difficult to understand this example from one of my books.



How can I calculate the grade of extension of $mathbb{Q}(S)$ with $S={ sqrt3,sqrt5}$? If I extend $mathbb{Q})$ with $sqrt 3$ I have that $mathbb{Q}(sqrt 3)$ minimum polynomial is $x^2-3$, but then what?



Thanks all






abstract-algebra







share|cite|improve this question











$endgroup$




First time in abstract algebra, and right now I have some difficult to understand this example from one of my books.



How can I calculate the grade of extension of $mathbb{Q}(S)$ with $S={ sqrt3,sqrt5}$? If I extend $mathbb{Q})$ with $sqrt 3$ I have that $mathbb{Q}(sqrt 3)$ minimum polynomial is $x^2-3$, but then what?



Thanks all






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 11:03







Alessar

















asked Dec 10 '18 at 12:28









AlessarAlessar

313115




313115








  • 1




    $begingroup$
    Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
    $endgroup$
    – Jack D'Aurizio
    Dec 10 '18 at 12:43






  • 1




    $begingroup$
    You will also need $sqrt{15}$ to complete the field, so degree is 4.
    $endgroup$
    – orion
    Dec 10 '18 at 12:43












  • $begingroup$
    Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
    $endgroup$
    – Alessar
    Dec 10 '18 at 14:22














  • 1




    $begingroup$
    Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
    $endgroup$
    – Jack D'Aurizio
    Dec 10 '18 at 12:43






  • 1




    $begingroup$
    You will also need $sqrt{15}$ to complete the field, so degree is 4.
    $endgroup$
    – orion
    Dec 10 '18 at 12:43












  • $begingroup$
    Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
    $endgroup$
    – Alessar
    Dec 10 '18 at 14:22








1




1




$begingroup$
Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:43




$begingroup$
Grade$mapsto$degree. The degree of $mathbb{Q}(sqrt{3},sqrt{5})$ is at most $4=2times 2$, and it is exactly four since $mathbb{Q}(sqrt{3},sqrt{5})$ contains $mathbb{Q}(sqrt{3}+sqrt{5})$ and the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$ is biquadratic.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:43




1




1




$begingroup$
You will also need $sqrt{15}$ to complete the field, so degree is 4.
$endgroup$
– orion
Dec 10 '18 at 12:43






$begingroup$
You will also need $sqrt{15}$ to complete the field, so degree is 4.
$endgroup$
– orion
Dec 10 '18 at 12:43














$begingroup$
Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
$endgroup$
– Alessar
Dec 10 '18 at 14:22




$begingroup$
Ok guys, but why we need to include $sqrt 15$? This is in order to complete the field, not for the degree right?
$endgroup$
– Alessar
Dec 10 '18 at 14:22










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If you know how to find the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$, use @Jack D'Aurizio's comment. Otherwise prove that $x^2-5$ is the minimal polynomial of $sqrt{5}$ over $mathbb{Q}[sqrt{3}]$.






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    $begingroup$

    If you know how to find the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$, use @Jack D'Aurizio's comment. Otherwise prove that $x^2-5$ is the minimal polynomial of $sqrt{5}$ over $mathbb{Q}[sqrt{3}]$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you know how to find the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$, use @Jack D'Aurizio's comment. Otherwise prove that $x^2-5$ is the minimal polynomial of $sqrt{5}$ over $mathbb{Q}[sqrt{3}]$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you know how to find the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$, use @Jack D'Aurizio's comment. Otherwise prove that $x^2-5$ is the minimal polynomial of $sqrt{5}$ over $mathbb{Q}[sqrt{3}]$.






        share|cite|improve this answer









        $endgroup$



        If you know how to find the minimal polynomial of $sqrt{3}+sqrt{5}$ over $mathbb{Q}$, use @Jack D'Aurizio's comment. Otherwise prove that $x^2-5$ is the minimal polynomial of $sqrt{5}$ over $mathbb{Q}[sqrt{3}]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 13:19









        StockfishStockfish

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