Isomorphic ring $mathbb{Q}[x]/(x^3+x) cong mathbb{Q} times mathbb{Q}[x]/(x^2+1)$












0












$begingroup$


I want to prove an isomorphism of the form $mathbb{Q}[x]/(x^3+x) cong mathbb{Q} times mathbb{Q}[x]/(x^2+1)$. I want to use the Chinese Remainder Theorem



$mathbb{Q}[x]/(x^3+x) = mathbb{Q}[x]/(x(x^2+1))$



So I choose $I = (x), J = (x^2 + 1)$



What I have to show first is that $I$, $J$ are coprime. (How can I do this? Because it feels trivial)



Secondly I have to show that $mathbb{Q}[x]/(x) = mathbb{Q}$. My question is: How can I prove this?






abstract-algebra ring-theory ideals ring-isomorphism







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$endgroup$

















    0












    $begingroup$


    I want to prove an isomorphism of the form $mathbb{Q}[x]/(x^3+x) cong mathbb{Q} times mathbb{Q}[x]/(x^2+1)$. I want to use the Chinese Remainder Theorem



    $mathbb{Q}[x]/(x^3+x) = mathbb{Q}[x]/(x(x^2+1))$



    So I choose $I = (x), J = (x^2 + 1)$



    What I have to show first is that $I$, $J$ are coprime. (How can I do this? Because it feels trivial)



    Secondly I have to show that $mathbb{Q}[x]/(x) = mathbb{Q}$. My question is: How can I prove this?






    abstract-algebra ring-theory ideals ring-isomorphism







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I want to prove an isomorphism of the form $mathbb{Q}[x]/(x^3+x) cong mathbb{Q} times mathbb{Q}[x]/(x^2+1)$. I want to use the Chinese Remainder Theorem



      $mathbb{Q}[x]/(x^3+x) = mathbb{Q}[x]/(x(x^2+1))$



      So I choose $I = (x), J = (x^2 + 1)$



      What I have to show first is that $I$, $J$ are coprime. (How can I do this? Because it feels trivial)



      Secondly I have to show that $mathbb{Q}[x]/(x) = mathbb{Q}$. My question is: How can I prove this?






      abstract-algebra ring-theory ideals ring-isomorphism







      share|cite|improve this question











      $endgroup$




      I want to prove an isomorphism of the form $mathbb{Q}[x]/(x^3+x) cong mathbb{Q} times mathbb{Q}[x]/(x^2+1)$. I want to use the Chinese Remainder Theorem



      $mathbb{Q}[x]/(x^3+x) = mathbb{Q}[x]/(x(x^2+1))$



      So I choose $I = (x), J = (x^2 + 1)$



      What I have to show first is that $I$, $J$ are coprime. (How can I do this? Because it feels trivial)



      Secondly I have to show that $mathbb{Q}[x]/(x) = mathbb{Q}$. My question is: How can I prove this?






      abstract-algebra ring-theory ideals ring-isomorphism




      abstract-algebra ring-theory ideals ring-isomorphism






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '18 at 12:50







      user593746

















      asked Nov 21 '18 at 11:29









      HansHans

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      537






















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          $begingroup$

          To find an explicit isomorphism, we note that



          $$1=(-x)cdot x+1cdot(x^2+1).$$

          This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, we can write
          $$p(x)=big(-xp(x)big)cdot x+p(x)cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$



          Therefore, a good isomorphism $varphi:Bbb{Q}/big(x(x^2+1)big)to big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)$ is given by
          $$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
          The inverse of $varphi$ is $psi: big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big)$ given by
          $$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
          Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. So, you can rewrite $varphi$ and $psi$ as $$Phi:Bbb{Q}/big(x(x^2+1)big)to mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)$$ and $$Psi:mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big),$$ which are given by
          $$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
          and
          $$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            To find an explicit isomorphism, we note that



            $$1=(-x)cdot x+1cdot(x^2+1).$$

            This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, we can write
            $$p(x)=big(-xp(x)big)cdot x+p(x)cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$



            Therefore, a good isomorphism $varphi:Bbb{Q}/big(x(x^2+1)big)to big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)$ is given by
            $$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
            The inverse of $varphi$ is $psi: big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big)$ given by
            $$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
            Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. So, you can rewrite $varphi$ and $psi$ as $$Phi:Bbb{Q}/big(x(x^2+1)big)to mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)$$ and $$Psi:mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big),$$ which are given by
            $$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
            and
            $$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              To find an explicit isomorphism, we note that



              $$1=(-x)cdot x+1cdot(x^2+1).$$

              This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, we can write
              $$p(x)=big(-xp(x)big)cdot x+p(x)cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$



              Therefore, a good isomorphism $varphi:Bbb{Q}/big(x(x^2+1)big)to big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)$ is given by
              $$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
              The inverse of $varphi$ is $psi: big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big)$ given by
              $$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
              Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. So, you can rewrite $varphi$ and $psi$ as $$Phi:Bbb{Q}/big(x(x^2+1)big)to mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)$$ and $$Psi:mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big),$$ which are given by
              $$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
              and
              $$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                To find an explicit isomorphism, we note that



                $$1=(-x)cdot x+1cdot(x^2+1).$$

                This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, we can write
                $$p(x)=big(-xp(x)big)cdot x+p(x)cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$



                Therefore, a good isomorphism $varphi:Bbb{Q}/big(x(x^2+1)big)to big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)$ is given by
                $$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
                The inverse of $varphi$ is $psi: big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big)$ given by
                $$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
                Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. So, you can rewrite $varphi$ and $psi$ as $$Phi:Bbb{Q}/big(x(x^2+1)big)to mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)$$ and $$Psi:mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big),$$ which are given by
                $$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
                and
                $$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$






                share|cite|improve this answer











                $endgroup$



                To find an explicit isomorphism, we note that



                $$1=(-x)cdot x+1cdot(x^2+1).$$

                This proves also that $I=(x)$ and $J=(x^2+1)$ are coprime ideals. So, for all $p(x)inmathbb{Q}[x]$, we can write
                $$p(x)=big(-xp(x)big)cdot x+p(x)cdot (x^2+1)=(x^2+1)p(x)-x^2p(x).$$



                Therefore, a good isomorphism $varphi:Bbb{Q}/big(x(x^2+1)big)to big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)$ is given by
                $$varphibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(x)operatorname{mod} x,p(x)operatorname{mod} (x^2+1)big).$$
                The inverse of $varphi$ is $psi: big(mathbb{Q}[x]/(x)big)oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big)$ given by
                $$psibig(a(x)operatorname{mod} x,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)a(x)-x^2b(x) Big)operatorname{mod} x(x^2+1).$$
                Note that there exists an isomorphism $mathbb{Q}[x]/(x)tomathbb{Q}$ sending $big(f(x)operatorname{mod} xbig)mapsto f(0)$. So, you can rewrite $varphi$ and $psi$ as $$Phi:Bbb{Q}/big(x(x^2+1)big)to mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)$$ and $$Psi:mathbb{Q}oplus big(mathbb{Q}/(x^2+1)big)to Bbb{Q}/big(x(x^2+1)big),$$ which are given by
                $$Phibig(p(x)operatorname{mod}x(x^2+1)big)=big(p(0),p(x)operatorname{mod} (x^2+1)big)$$
                and
                $$Psibig(t,b(x)operatorname{mod}(x^2+1)big)=Big((x^2+1)t-x^2b(x) Big)operatorname{mod} x(x^2+1).$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 23:00









                amWhy

                192k28225439




                192k28225439










                answered Nov 21 '18 at 11:48







                user593746





































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