minimal polynomial of $1+alpha^2$ where $alpha$ is a root of $x^3-x-1$











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Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.



What is the minimal polynomial of $gamma$ over $mathbb Q$?






abstract-algebra roots minimal-polynomials







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    What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
    – lioness99a
    Apr 7 '17 at 12:00










  • Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
    – Magdiragdag
    Apr 8 '17 at 7:34















up vote
0
down vote

favorite












Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.



What is the minimal polynomial of $gamma$ over $mathbb Q$?






abstract-algebra roots minimal-polynomials







share|cite|improve this question




















  • 1




    What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
    – lioness99a
    Apr 7 '17 at 12:00










  • Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
    – Magdiragdag
    Apr 8 '17 at 7:34













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.



What is the minimal polynomial of $gamma$ over $mathbb Q$?






abstract-algebra roots minimal-polynomials







share|cite|improve this question















Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.



What is the minimal polynomial of $gamma$ over $mathbb Q$?






abstract-algebra roots minimal-polynomials




abstract-algebra roots minimal-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 8 '17 at 7:28









Marc van Leeuwen

86.2k5105218




86.2k5105218










asked Apr 7 '17 at 11:56









Lehostal11

214




214








  • 1




    What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
    – lioness99a
    Apr 7 '17 at 12:00










  • Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
    – Magdiragdag
    Apr 8 '17 at 7:34














  • 1




    What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
    – lioness99a
    Apr 7 '17 at 12:00










  • Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
    – Magdiragdag
    Apr 8 '17 at 7:34








1




1




What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00




What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00












Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34




Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34










4 Answers
4






active

oldest

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up vote
3
down vote













Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.



The minimal polynomial of this matrix is the minimal polynomial of $gamma$.






share|cite|improve this answer




























    up vote
    2
    down vote













    A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
    $$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
    $$
    The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
    The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.






    share|cite|improve this answer






























      up vote
      1
      down vote













      Hints:



      Observe that $;alpha^3=alpha+1;$ , and from here



      $$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$






      share|cite|improve this answer




























        up vote
        1
        down vote













        Hint:



        Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.



        The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.



        Some details:



        Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
        $$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
        This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.



        Example:
        $$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$






        share|cite|improve this answer























        • I try it this way. But it is still hard. Can you give me more hints?
          – Lehostal11
          Apr 7 '17 at 12:33










        • @Lehostal: I've added some details.
          – Bernard
          Apr 7 '17 at 13:10











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        4 Answers
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        4 Answers
        4






        active

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        active

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        active

        oldest

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        up vote
        3
        down vote













        Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.



        The minimal polynomial of this matrix is the minimal polynomial of $gamma$.






        share|cite|improve this answer

























          up vote
          3
          down vote













          Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.



          The minimal polynomial of this matrix is the minimal polynomial of $gamma$.






          share|cite|improve this answer























            up vote
            3
            down vote










            up vote
            3
            down vote









            Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.



            The minimal polynomial of this matrix is the minimal polynomial of $gamma$.






            share|cite|improve this answer












            Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.



            The minimal polynomial of this matrix is the minimal polynomial of $gamma$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 7 '17 at 13:25









            lhf

            162k9166385




            162k9166385






















                up vote
                2
                down vote













                A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
                $$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
                $$
                The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
                The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.






                share|cite|improve this answer



























                  up vote
                  2
                  down vote













                  A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
                  $$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
                  $$
                  The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
                  The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.






                  share|cite|improve this answer

























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
                    $$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
                    $$
                    The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
                    The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.






                    share|cite|improve this answer














                    A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
                    $$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
                    $$
                    The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
                    The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 22 at 10:25

























                    answered Apr 8 '17 at 10:07









                    Marc van Leeuwen

                    86.2k5105218




                    86.2k5105218






















                        up vote
                        1
                        down vote













                        Hints:



                        Observe that $;alpha^3=alpha+1;$ , and from here



                        $$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          Hints:



                          Observe that $;alpha^3=alpha+1;$ , and from here



                          $$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Hints:



                            Observe that $;alpha^3=alpha+1;$ , and from here



                            $$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$






                            share|cite|improve this answer












                            Hints:



                            Observe that $;alpha^3=alpha+1;$ , and from here



                            $$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 7 '17 at 12:06









                            DonAntonio

                            176k1491224




                            176k1491224






















                                up vote
                                1
                                down vote













                                Hint:



                                Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.



                                The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.



                                Some details:



                                Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
                                $$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
                                This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.



                                Example:
                                $$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$






                                share|cite|improve this answer























                                • I try it this way. But it is still hard. Can you give me more hints?
                                  – Lehostal11
                                  Apr 7 '17 at 12:33










                                • @Lehostal: I've added some details.
                                  – Bernard
                                  Apr 7 '17 at 13:10















                                up vote
                                1
                                down vote













                                Hint:



                                Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.



                                The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.



                                Some details:



                                Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
                                $$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
                                This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.



                                Example:
                                $$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$






                                share|cite|improve this answer























                                • I try it this way. But it is still hard. Can you give me more hints?
                                  – Lehostal11
                                  Apr 7 '17 at 12:33










                                • @Lehostal: I've added some details.
                                  – Bernard
                                  Apr 7 '17 at 13:10













                                up vote
                                1
                                down vote










                                up vote
                                1
                                down vote









                                Hint:



                                Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.



                                The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.



                                Some details:



                                Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
                                $$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
                                This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.



                                Example:
                                $$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$






                                share|cite|improve this answer














                                Hint:



                                Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.



                                The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.



                                Some details:



                                Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
                                $$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
                                This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.



                                Example:
                                $$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Apr 7 '17 at 13:10

























                                answered Apr 7 '17 at 12:06









                                Bernard

                                117k637109




                                117k637109












                                • I try it this way. But it is still hard. Can you give me more hints?
                                  – Lehostal11
                                  Apr 7 '17 at 12:33










                                • @Lehostal: I've added some details.
                                  – Bernard
                                  Apr 7 '17 at 13:10


















                                • I try it this way. But it is still hard. Can you give me more hints?
                                  – Lehostal11
                                  Apr 7 '17 at 12:33










                                • @Lehostal: I've added some details.
                                  – Bernard
                                  Apr 7 '17 at 13:10
















                                I try it this way. But it is still hard. Can you give me more hints?
                                – Lehostal11
                                Apr 7 '17 at 12:33




                                I try it this way. But it is still hard. Can you give me more hints?
                                – Lehostal11
                                Apr 7 '17 at 12:33












                                @Lehostal: I've added some details.
                                – Bernard
                                Apr 7 '17 at 13:10




                                @Lehostal: I've added some details.
                                – Bernard
                                Apr 7 '17 at 13:10


















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                                Mont Emei

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                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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