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
          2






          active

          oldest

          votes


















          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











          $endgroup$





















            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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047204%2fquestions-on-a-proof-that-mathbb-f-2x-x2-1-and-mathbb-f-2x-x2-ar%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown
























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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












                $begingroup$

                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

                          4,7941513




                          4,7941513






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047204%2fquestions-on-a-proof-that-mathbb-f-2x-x2-1-and-mathbb-f-2x-x2-ar%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Probability when a professor distributes a quiz and homework assignment to a class of n students.

                              Image
                              0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                              Read more

                              Aardman Animations

                              Image
                              Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                              Read more

                              Are they similar matrix

                              Image
                              0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                              Read more