Strategies for finding $[mathbb{Q} (sqrt{2} + sqrt{3}) : mathbb{Q} ]$












4














I need to find the degree of the extension $mathbb{Q}(sqrt{2} + sqrt{3})$ over $mathbb{Q}$. I don't quite know how to do it, nor can I exhibit any polynomial with root $sqrt{2} + sqrt{3}$, but I think it has to have at least degree $4$. I tried to work with $mathbb{Q}(sqrt{2} + sqrt{3})$ as a subspace of $mathbb{Q}(sqrt{2}, sqrt{3})$ over $mathbb{Q}$, but I also don't know if that is the case.






field-theory extension-field irreducible-polynomials







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  • $x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
    – Seth
    Nov 22 '18 at 2:57


















4














I need to find the degree of the extension $mathbb{Q}(sqrt{2} + sqrt{3})$ over $mathbb{Q}$. I don't quite know how to do it, nor can I exhibit any polynomial with root $sqrt{2} + sqrt{3}$, but I think it has to have at least degree $4$. I tried to work with $mathbb{Q}(sqrt{2} + sqrt{3})$ as a subspace of $mathbb{Q}(sqrt{2}, sqrt{3})$ over $mathbb{Q}$, but I also don't know if that is the case.






field-theory extension-field irreducible-polynomials







share|cite|improve this question






















  • $x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
    – Seth
    Nov 22 '18 at 2:57
















4












4








4







I need to find the degree of the extension $mathbb{Q}(sqrt{2} + sqrt{3})$ over $mathbb{Q}$. I don't quite know how to do it, nor can I exhibit any polynomial with root $sqrt{2} + sqrt{3}$, but I think it has to have at least degree $4$. I tried to work with $mathbb{Q}(sqrt{2} + sqrt{3})$ as a subspace of $mathbb{Q}(sqrt{2}, sqrt{3})$ over $mathbb{Q}$, but I also don't know if that is the case.






field-theory extension-field irreducible-polynomials







share|cite|improve this question













I need to find the degree of the extension $mathbb{Q}(sqrt{2} + sqrt{3})$ over $mathbb{Q}$. I don't quite know how to do it, nor can I exhibit any polynomial with root $sqrt{2} + sqrt{3}$, but I think it has to have at least degree $4$. I tried to work with $mathbb{Q}(sqrt{2} + sqrt{3})$ as a subspace of $mathbb{Q}(sqrt{2}, sqrt{3})$ over $mathbb{Q}$, but I also don't know if that is the case.






field-theory extension-field irreducible-polynomials




field-theory extension-field irreducible-polynomials






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asked Nov 22 '18 at 2:43









Nuntractatuses Amável

61812




61812












  • $x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
    – Seth
    Nov 22 '18 at 2:57




















  • $x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
    – Seth
    Nov 22 '18 at 2:57


















$x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
– Seth
Nov 22 '18 at 2:57






$x^4-10x^2+1$ is the smallest monic polynomial that has $sqrt{2}+sqrt{3}$ as a root
– Seth
Nov 22 '18 at 2:57












4 Answers
4






active

oldest

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2














As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
$1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
We have
$$alpha^2=5+2sqrt6$$
and
$$alpha^3=11sqrt2+9sqrt3.$$
Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
and $sqrt3=-frac12(alpha^3-11alpha)$
are all elements of $Bbb Q(alpha)$. Therefore $Bbb Q(alpha)=Bbb Q(sqrt2,sqrt3)$.



To show that $|Bbb Q(sqrt2,sqrt3):Bbb Q|=4$, we need $sqrt3notin
Bbb Q(sqrt2)$. To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
Bbb Q$ and get a contradiction from $(a+bsqrt2)^2=3$.






share|cite|improve this answer





























    3














    $frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.



    Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.






    share|cite|improve this answer





























      1














      Hint
      begin{align*}
      x &= sqrt{2}+sqrt{3}\
      x - sqrt{2}&=sqrt{3}\
      (x - sqrt{2})^2&=3\
      x^2-1&=2xsqrt{2}\
      (x^2-1)^2&=8x^2\
      x^4-10x^2+1&=0.
      end{align*}






      share|cite|improve this answer























      • second last line should be $(x^2-1)^2=8x^2$
        – Seth
        Nov 22 '18 at 2:59










      • @Seth Thanks. I have fixed the typo.
        – Anurag A
        Nov 22 '18 at 3:01





















      0














      Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. Then $frac{(w^2-5)^2}{4}=6$ is in $mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2-1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
        It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
        equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
        $1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
        We have
        $$alpha^2=5+2sqrt6$$
        and
        $$alpha^3=11sqrt2+9sqrt3.$$
        Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
        and $sqrt3=-frac12(alpha^3-11alpha)$
        are all elements of $Bbb Q(alpha)$. Therefore $Bbb Q(alpha)=Bbb Q(sqrt2,sqrt3)$.



        To show that $|Bbb Q(sqrt2,sqrt3):Bbb Q|=4$, we need $sqrt3notin
        Bbb Q(sqrt2)$        . To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
        Bbb Q$         and get a contradiction from $(a+bsqrt2)^2=3$.






        share|cite|improve this answer


























          2














          As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
          It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
          equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
          $1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
          We have
          $$alpha^2=5+2sqrt6$$
          and
          $$alpha^3=11sqrt2+9sqrt3.$$
          Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
          and $sqrt3=-frac12(alpha^3-11alpha)$
          are all elements of $Bbb Q(alpha)$. Therefore $Bbb Q(alpha)=Bbb Q(sqrt2,sqrt3)$.



          To show that $|Bbb Q(sqrt2,sqrt3):Bbb Q|=4$, we need $sqrt3notin
          Bbb Q(sqrt2)$          . To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
          Bbb Q$           and get a contradiction from $(a+bsqrt2)^2=3$.






          share|cite|improve this answer
























            2












            2








            2






            As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
            It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
            equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
            $1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
            We have
            $$alpha^2=5+2sqrt6$$
            and
            $$alpha^3=11sqrt2+9sqrt3.$$
            Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
            and $sqrt3=-frac12(alpha^3-11alpha)$
            are all elements of $Bbb Q(alpha)$. Therefore $Bbb Q(alpha)=Bbb Q(sqrt2,sqrt3)$.



            To show that $|Bbb Q(sqrt2,sqrt3):Bbb Q|=4$, we need $sqrt3notin
            Bbb Q(sqrt2)$            . To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
            Bbb Q$             and get a contradiction from $(a+bsqrt2)^2=3$.






            share|cite|improve this answer












            As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
            It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
            equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
            $1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
            We have
            $$alpha^2=5+2sqrt6$$
            and
            $$alpha^3=11sqrt2+9sqrt3.$$
            Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
            and $sqrt3=-frac12(alpha^3-11alpha)$
            are all elements of $Bbb Q(alpha)$. Therefore $Bbb Q(alpha)=Bbb Q(sqrt2,sqrt3)$.



            To show that $|Bbb Q(sqrt2,sqrt3):Bbb Q|=4$, we need $sqrt3notin
            Bbb Q(sqrt2)$            . To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
            Bbb Q$             and get a contradiction from $(a+bsqrt2)^2=3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 '18 at 2:59









            Lord Shark the Unknown

            102k958132




            102k958132























                3














                $frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.



                Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.






                share|cite|improve this answer


























                  3














                  $frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.



                  Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.






                  share|cite|improve this answer
























                    3












                    3








                    3






                    $frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.



                    Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.






                    share|cite|improve this answer












                    $frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.



                    Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 '18 at 2:58









                    Chris Custer

                    10.9k3824




                    10.9k3824























                        1














                        Hint
                        begin{align*}
                        x &= sqrt{2}+sqrt{3}\
                        x - sqrt{2}&=sqrt{3}\
                        (x - sqrt{2})^2&=3\
                        x^2-1&=2xsqrt{2}\
                        (x^2-1)^2&=8x^2\
                        x^4-10x^2+1&=0.
                        end{align*}






                        share|cite|improve this answer























                        • second last line should be $(x^2-1)^2=8x^2$
                          – Seth
                          Nov 22 '18 at 2:59










                        • @Seth Thanks. I have fixed the typo.
                          – Anurag A
                          Nov 22 '18 at 3:01


















                        1














                        Hint
                        begin{align*}
                        x &= sqrt{2}+sqrt{3}\
                        x - sqrt{2}&=sqrt{3}\
                        (x - sqrt{2})^2&=3\
                        x^2-1&=2xsqrt{2}\
                        (x^2-1)^2&=8x^2\
                        x^4-10x^2+1&=0.
                        end{align*}






                        share|cite|improve this answer























                        • second last line should be $(x^2-1)^2=8x^2$
                          – Seth
                          Nov 22 '18 at 2:59










                        • @Seth Thanks. I have fixed the typo.
                          – Anurag A
                          Nov 22 '18 at 3:01
















                        1












                        1








                        1






                        Hint
                        begin{align*}
                        x &= sqrt{2}+sqrt{3}\
                        x - sqrt{2}&=sqrt{3}\
                        (x - sqrt{2})^2&=3\
                        x^2-1&=2xsqrt{2}\
                        (x^2-1)^2&=8x^2\
                        x^4-10x^2+1&=0.
                        end{align*}






                        share|cite|improve this answer














                        Hint
                        begin{align*}
                        x &= sqrt{2}+sqrt{3}\
                        x - sqrt{2}&=sqrt{3}\
                        (x - sqrt{2})^2&=3\
                        x^2-1&=2xsqrt{2}\
                        (x^2-1)^2&=8x^2\
                        x^4-10x^2+1&=0.
                        end{align*}







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Nov 22 '18 at 3:00

























                        answered Nov 22 '18 at 2:56









                        Anurag A

                        25.6k12249




                        25.6k12249












                        • second last line should be $(x^2-1)^2=8x^2$
                          – Seth
                          Nov 22 '18 at 2:59










                        • @Seth Thanks. I have fixed the typo.
                          – Anurag A
                          Nov 22 '18 at 3:01




















                        • second last line should be $(x^2-1)^2=8x^2$
                          – Seth
                          Nov 22 '18 at 2:59










                        • @Seth Thanks. I have fixed the typo.
                          – Anurag A
                          Nov 22 '18 at 3:01


















                        second last line should be $(x^2-1)^2=8x^2$
                        – Seth
                        Nov 22 '18 at 2:59




                        second last line should be $(x^2-1)^2=8x^2$
                        – Seth
                        Nov 22 '18 at 2:59












                        @Seth Thanks. I have fixed the typo.
                        – Anurag A
                        Nov 22 '18 at 3:01






                        @Seth Thanks. I have fixed the typo.
                        – Anurag A
                        Nov 22 '18 at 3:01













                        0














                        Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. Then $frac{(w^2-5)^2}{4}=6$ is in $mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2-1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.






                        share|cite|improve this answer


























                          0














                          Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. Then $frac{(w^2-5)^2}{4}=6$ is in $mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2-1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. Then $frac{(w^2-5)^2}{4}=6$ is in $mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2-1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.






                            share|cite|improve this answer












                            Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. Then $frac{(w^2-5)^2}{4}=6$ is in $mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2-1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 '18 at 3:00









                            Dante Grevino

                            94319




                            94319






























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