Questions on a proof that $mathbb F_2[x]/(x^2-1)$ and $mathbb F_2[x]/(x^2)$ are isomorphic.












2












$begingroup$


This has been proven more generally for a field of characteristic 2 in another question.



Here is the proof from brianbi.ca $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2) =
mathbb F_2[x]/(x^2 + 2x + 1) = mathbb F_2[x]/(x^2 - 1)$$




  1. Why is $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2)$$?


-We can use the correspondence theorem by the surjective homomorphism $varphi:mathbb F_2[x] to mathbb F_2[x+1], varphi(p(x))=p(x+1)$, so the ideal $(x^2)$ corresponds to the ideal $(x + 1)^2$. Is this correct?





  • Is injectivity irrelevant?




    1. Why is $$mathbb F_2[x+1]/((x + 1)^2) = mathbb F_2[x]/(x^2 + 2x + 1)$$




?




  • $mathbb F_2[x+1] cong mathbb F_2[x]$ by the same homomorphism $varphi$ because $varphi$ is injective too. Is this correct? I think this proves only $cong$ and not necessarily $=$.


  • Do we actually have $mathbb F_2[x+1] = mathbb F_2[x]$? If so, how? In general for a ring $R$ and $a,b in R$, when is $R[x-a] = R[x-b]$? When is $R[x-a] cong R[x-b]$?







abstract-algebra number-theory field-theory modular-arithmetic ideals







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    This has been proven more generally for a field of characteristic 2 in another question.



    Here is the proof from brianbi.ca $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2) =
    mathbb F_2[x]/(x^2 + 2x + 1) = mathbb F_2[x]/(x^2 - 1)$$




    1. Why is $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2)$$?


    -We can use the correspondence theorem by the surjective homomorphism $varphi:mathbb F_2[x] to mathbb F_2[x+1], varphi(p(x))=p(x+1)$, so the ideal $(x^2)$ corresponds to the ideal $(x + 1)^2$. Is this correct?





    • Is injectivity irrelevant?




      1. Why is $$mathbb F_2[x+1]/((x + 1)^2) = mathbb F_2[x]/(x^2 + 2x + 1)$$




    ?




    • $mathbb F_2[x+1] cong mathbb F_2[x]$ by the same homomorphism $varphi$ because $varphi$ is injective too. Is this correct? I think this proves only $cong$ and not necessarily $=$.


    • Do we actually have $mathbb F_2[x+1] = mathbb F_2[x]$? If so, how? In general for a ring $R$ and $a,b in R$, when is $R[x-a] = R[x-b]$? When is $R[x-a] cong R[x-b]$?







    abstract-algebra number-theory field-theory modular-arithmetic ideals







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      This has been proven more generally for a field of characteristic 2 in another question.



      Here is the proof from brianbi.ca $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2) =
      mathbb F_2[x]/(x^2 + 2x + 1) = mathbb F_2[x]/(x^2 - 1)$$




      1. Why is $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2)$$?


      -We can use the correspondence theorem by the surjective homomorphism $varphi:mathbb F_2[x] to mathbb F_2[x+1], varphi(p(x))=p(x+1)$, so the ideal $(x^2)$ corresponds to the ideal $(x + 1)^2$. Is this correct?





      • Is injectivity irrelevant?




        1. Why is $$mathbb F_2[x+1]/((x + 1)^2) = mathbb F_2[x]/(x^2 + 2x + 1)$$




      ?




      • $mathbb F_2[x+1] cong mathbb F_2[x]$ by the same homomorphism $varphi$ because $varphi$ is injective too. Is this correct? I think this proves only $cong$ and not necessarily $=$.


      • Do we actually have $mathbb F_2[x+1] = mathbb F_2[x]$? If so, how? In general for a ring $R$ and $a,b in R$, when is $R[x-a] = R[x-b]$? When is $R[x-a] cong R[x-b]$?







      abstract-algebra number-theory field-theory modular-arithmetic ideals







      share|cite|improve this question











      $endgroup$




      This has been proven more generally for a field of characteristic 2 in another question.



      Here is the proof from brianbi.ca $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2) =
      mathbb F_2[x]/(x^2 + 2x + 1) = mathbb F_2[x]/(x^2 - 1)$$




      1. Why is $$mathbb F_2[x]/(x^2) cong mathbb F_2[x+1]/((x + 1)^2)$$?


      -We can use the correspondence theorem by the surjective homomorphism $varphi:mathbb F_2[x] to mathbb F_2[x+1], varphi(p(x))=p(x+1)$, so the ideal $(x^2)$ corresponds to the ideal $(x + 1)^2$. Is this correct?





      • Is injectivity irrelevant?




        1. Why is $$mathbb F_2[x+1]/((x + 1)^2) = mathbb F_2[x]/(x^2 + 2x + 1)$$




      ?




      • $mathbb F_2[x+1] cong mathbb F_2[x]$ by the same homomorphism $varphi$ because $varphi$ is injective too. Is this correct? I think this proves only $cong$ and not necessarily $=$.


      • Do we actually have $mathbb F_2[x+1] = mathbb F_2[x]$? If so, how? In general for a ring $R$ and $a,b in R$, when is $R[x-a] = R[x-b]$? When is $R[x-a] cong R[x-b]$?







      abstract-algebra number-theory field-theory modular-arithmetic ideals




      abstract-algebra number-theory field-theory modular-arithmetic ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 6:37

























      asked Dec 20 '18 at 5:45







      user198044





























          2 Answers
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          $begingroup$

          I seems to me that in fact $mathbb{F}_2[x+1] = mathbb{F}_2[x]$, because we have $x+1 in mathbb{F}_2[x]$, and also in characteristic 2 we have $2=0$, hence $x = x+2 = (x+1) + 1 in mathbb{F}_2[x+1]$.



          More generally polynomials in $x$ with coefficients in any field $mathbb{F}$ are the same as polynomials in $x+1$ with coefficients in $mathbb{F}$, since we can always write $x = (x+1) - 1$, so $p(x) = p((x+1)-1) = q(x+1)$ for some polynomial $q$, and we can pass from $q(x+1)$ to $p(x)$ by writing $q(x+1)$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            The quotient rings ${Bbb F}_2[x]/langle x^2+1rangle$ and ${Bbb F}_2[x]/langle x^2rangle$ are isomorphic as ${Bbb F}_2$-vector spaces.



            In view of multiplication, it is obvious that a ring isomorphism $phi:{Bbb F}_2[x]/langle x^2ranglerightarrow {Bbb F}_2[x]/langle x^2+1rangle$ maps $x$ to $x+1$.



            Indeed, check that in ${Bbb F}_2[x]/langle x^2rangle$:
            $$xcdot x = 0,quad xcdot(x+1)=x^2+x=x, quad(x+1)(x+1)=x^2+1=1,$$
            while in ${Bbb F}_2[x]/langle x^2+1rangle$:
            $$xcdot x = x^2=1,quad xcdot(x+1)=x^2+x=x+1,quad (x+1)(x+1)=0.$$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
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              $begingroup$

              I seems to me that in fact $mathbb{F}_2[x+1] = mathbb{F}_2[x]$, because we have $x+1 in mathbb{F}_2[x]$, and also in characteristic 2 we have $2=0$, hence $x = x+2 = (x+1) + 1 in mathbb{F}_2[x+1]$.



              More generally polynomials in $x$ with coefficients in any field $mathbb{F}$ are the same as polynomials in $x+1$ with coefficients in $mathbb{F}$, since we can always write $x = (x+1) - 1$, so $p(x) = p((x+1)-1) = q(x+1)$ for some polynomial $q$, and we can pass from $q(x+1)$ to $p(x)$ by writing $q(x+1)$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                I seems to me that in fact $mathbb{F}_2[x+1] = mathbb{F}_2[x]$, because we have $x+1 in mathbb{F}_2[x]$, and also in characteristic 2 we have $2=0$, hence $x = x+2 = (x+1) + 1 in mathbb{F}_2[x+1]$.



                More generally polynomials in $x$ with coefficients in any field $mathbb{F}$ are the same as polynomials in $x+1$ with coefficients in $mathbb{F}$, since we can always write $x = (x+1) - 1$, so $p(x) = p((x+1)-1) = q(x+1)$ for some polynomial $q$, and we can pass from $q(x+1)$ to $p(x)$ by writing $q(x+1)$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  I seems to me that in fact $mathbb{F}_2[x+1] = mathbb{F}_2[x]$, because we have $x+1 in mathbb{F}_2[x]$, and also in characteristic 2 we have $2=0$, hence $x = x+2 = (x+1) + 1 in mathbb{F}_2[x+1]$.



                  More generally polynomials in $x$ with coefficients in any field $mathbb{F}$ are the same as polynomials in $x+1$ with coefficients in $mathbb{F}$, since we can always write $x = (x+1) - 1$, so $p(x) = p((x+1)-1) = q(x+1)$ for some polynomial $q$, and we can pass from $q(x+1)$ to $p(x)$ by writing $q(x+1)$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.






                  share|cite|improve this answer











                  $endgroup$



                  I seems to me that in fact $mathbb{F}_2[x+1] = mathbb{F}_2[x]$, because we have $x+1 in mathbb{F}_2[x]$, and also in characteristic 2 we have $2=0$, hence $x = x+2 = (x+1) + 1 in mathbb{F}_2[x+1]$.



                  More generally polynomials in $x$ with coefficients in any field $mathbb{F}$ are the same as polynomials in $x+1$ with coefficients in $mathbb{F}$, since we can always write $x = (x+1) - 1$, so $p(x) = p((x+1)-1) = q(x+1)$ for some polynomial $q$, and we can pass from $q(x+1)$ to $p(x)$ by writing $q(x+1)$ and expanding terms. This reasoning applies to commutative rings with multiplicative identity 1 in general too.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 6:22

























                  answered Dec 20 '18 at 6:12









                  mlerma54mlerma54

                  1,177148




                  1,177148























                      0












                      $begingroup$

                      The quotient rings ${Bbb F}_2[x]/langle x^2+1rangle$ and ${Bbb F}_2[x]/langle x^2rangle$ are isomorphic as ${Bbb F}_2$-vector spaces.



                      In view of multiplication, it is obvious that a ring isomorphism $phi:{Bbb F}_2[x]/langle x^2ranglerightarrow {Bbb F}_2[x]/langle x^2+1rangle$ maps $x$ to $x+1$.



                      Indeed, check that in ${Bbb F}_2[x]/langle x^2rangle$:
                      $$xcdot x = 0,quad xcdot(x+1)=x^2+x=x, quad(x+1)(x+1)=x^2+1=1,$$
                      while in ${Bbb F}_2[x]/langle x^2+1rangle$:
                      $$xcdot x = x^2=1,quad xcdot(x+1)=x^2+x=x+1,quad (x+1)(x+1)=0.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The quotient rings ${Bbb F}_2[x]/langle x^2+1rangle$ and ${Bbb F}_2[x]/langle x^2rangle$ are isomorphic as ${Bbb F}_2$-vector spaces.



                        In view of multiplication, it is obvious that a ring isomorphism $phi:{Bbb F}_2[x]/langle x^2ranglerightarrow {Bbb F}_2[x]/langle x^2+1rangle$ maps $x$ to $x+1$.



                        Indeed, check that in ${Bbb F}_2[x]/langle x^2rangle$:
                        $$xcdot x = 0,quad xcdot(x+1)=x^2+x=x, quad(x+1)(x+1)=x^2+1=1,$$
                        while in ${Bbb F}_2[x]/langle x^2+1rangle$:
                        $$xcdot x = x^2=1,quad xcdot(x+1)=x^2+x=x+1,quad (x+1)(x+1)=0.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The quotient rings ${Bbb F}_2[x]/langle x^2+1rangle$ and ${Bbb F}_2[x]/langle x^2rangle$ are isomorphic as ${Bbb F}_2$-vector spaces.



                          In view of multiplication, it is obvious that a ring isomorphism $phi:{Bbb F}_2[x]/langle x^2ranglerightarrow {Bbb F}_2[x]/langle x^2+1rangle$ maps $x$ to $x+1$.



                          Indeed, check that in ${Bbb F}_2[x]/langle x^2rangle$:
                          $$xcdot x = 0,quad xcdot(x+1)=x^2+x=x, quad(x+1)(x+1)=x^2+1=1,$$
                          while in ${Bbb F}_2[x]/langle x^2+1rangle$:
                          $$xcdot x = x^2=1,quad xcdot(x+1)=x^2+x=x+1,quad (x+1)(x+1)=0.$$






                          share|cite|improve this answer









                          $endgroup$



                          The quotient rings ${Bbb F}_2[x]/langle x^2+1rangle$ and ${Bbb F}_2[x]/langle x^2rangle$ are isomorphic as ${Bbb F}_2$-vector spaces.



                          In view of multiplication, it is obvious that a ring isomorphism $phi:{Bbb F}_2[x]/langle x^2ranglerightarrow {Bbb F}_2[x]/langle x^2+1rangle$ maps $x$ to $x+1$.



                          Indeed, check that in ${Bbb F}_2[x]/langle x^2rangle$:
                          $$xcdot x = 0,quad xcdot(x+1)=x^2+x=x, quad(x+1)(x+1)=x^2+1=1,$$
                          while in ${Bbb F}_2[x]/langle x^2+1rangle$:
                          $$xcdot x = x^2=1,quad xcdot(x+1)=x^2+x=x+1,quad (x+1)(x+1)=0.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 20 '18 at 8:54









                          WuestenfuxWuestenfux

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