Finding elements of $mathbb{Z}_3[x] /langle x^2+2rangle$ and writing the multiplication table












0














I want to list elements of $mathbb{Z}_3[x] /langle x^2+2rangle$ and write a multiplication table. Here is my attempt to finding the elements:



Let $A =langle x^2+2rangle= {(x^2+2)f(x): f(x) in mathbb{Z}_3[x]}$ and $mathbb{Z}_3[x] / A = {f(x) + A: f(x) in mathbb{Z}_3[x]}$ by definition.



Let $f(x) in mathbb{Z}_3[x]$. By division algorithm, $f(x) = (x^2+2)q(x) + a + bx$ for some $q(x) in mathbb{Z}_3[x]$ and $a,b in mathbb{Z}_3$. Hence, $f(x) + A = (x^2+2)q(x) + a + bx + A = a + bx + (x^2+2)q(x) + A$. Since $(x^2+2)q(x) in A$, $(x^2+2)q(x) + A = A$. Thus, $f(x) + A = a + bx + A$. So we have $mathbb{Z_3}[x] / A = {a + bx + A: a,b in mathbb{Z_3}}$. Hence, the elements are the following:




  1. $A$

  2. $1 + A$

  3. $x + A$

  4. $2 + A$

  5. $2x + A$

  6. $1 + x + A$

  7. $2 + x + A$

  8. $2 + 2x + A$

  9. $1 + 2x + A$


My question is:



1) Is this the right derivation?



2) How does multiplication table work in $mathbb{Z}_3[x] / A$? For example, if I have $(x+A)(1+2x+A) = x(1+2x) + A = x+2x^2 + A$ which is not in the same form as $ax+b+A$.






abstract-algebra polynomials ring-theory







share|cite|improve this question





























    0














    I want to list elements of $mathbb{Z}_3[x] /langle x^2+2rangle$ and write a multiplication table. Here is my attempt to finding the elements:



    Let $A =langle x^2+2rangle= {(x^2+2)f(x): f(x) in mathbb{Z}_3[x]}$ and $mathbb{Z}_3[x] / A = {f(x) + A: f(x) in mathbb{Z}_3[x]}$ by definition.



    Let $f(x) in mathbb{Z}_3[x]$. By division algorithm, $f(x) = (x^2+2)q(x) + a + bx$ for some $q(x) in mathbb{Z}_3[x]$ and $a,b in mathbb{Z}_3$. Hence, $f(x) + A = (x^2+2)q(x) + a + bx + A = a + bx + (x^2+2)q(x) + A$. Since $(x^2+2)q(x) in A$, $(x^2+2)q(x) + A = A$. Thus, $f(x) + A = a + bx + A$. So we have $mathbb{Z_3}[x] / A = {a + bx + A: a,b in mathbb{Z_3}}$. Hence, the elements are the following:




    1. $A$

    2. $1 + A$

    3. $x + A$

    4. $2 + A$

    5. $2x + A$

    6. $1 + x + A$

    7. $2 + x + A$

    8. $2 + 2x + A$

    9. $1 + 2x + A$


    My question is:



    1) Is this the right derivation?



    2) How does multiplication table work in $mathbb{Z}_3[x] / A$? For example, if I have $(x+A)(1+2x+A) = x(1+2x) + A = x+2x^2 + A$ which is not in the same form as $ax+b+A$.






    abstract-algebra polynomials ring-theory







    share|cite|improve this question



























      0












      0








      0







      I want to list elements of $mathbb{Z}_3[x] /langle x^2+2rangle$ and write a multiplication table. Here is my attempt to finding the elements:



      Let $A =langle x^2+2rangle= {(x^2+2)f(x): f(x) in mathbb{Z}_3[x]}$ and $mathbb{Z}_3[x] / A = {f(x) + A: f(x) in mathbb{Z}_3[x]}$ by definition.



      Let $f(x) in mathbb{Z}_3[x]$. By division algorithm, $f(x) = (x^2+2)q(x) + a + bx$ for some $q(x) in mathbb{Z}_3[x]$ and $a,b in mathbb{Z}_3$. Hence, $f(x) + A = (x^2+2)q(x) + a + bx + A = a + bx + (x^2+2)q(x) + A$. Since $(x^2+2)q(x) in A$, $(x^2+2)q(x) + A = A$. Thus, $f(x) + A = a + bx + A$. So we have $mathbb{Z_3}[x] / A = {a + bx + A: a,b in mathbb{Z_3}}$. Hence, the elements are the following:




      1. $A$

      2. $1 + A$

      3. $x + A$

      4. $2 + A$

      5. $2x + A$

      6. $1 + x + A$

      7. $2 + x + A$

      8. $2 + 2x + A$

      9. $1 + 2x + A$


      My question is:



      1) Is this the right derivation?



      2) How does multiplication table work in $mathbb{Z}_3[x] / A$? For example, if I have $(x+A)(1+2x+A) = x(1+2x) + A = x+2x^2 + A$ which is not in the same form as $ax+b+A$.






      abstract-algebra polynomials ring-theory







      share|cite|improve this question















      I want to list elements of $mathbb{Z}_3[x] /langle x^2+2rangle$ and write a multiplication table. Here is my attempt to finding the elements:



      Let $A =langle x^2+2rangle= {(x^2+2)f(x): f(x) in mathbb{Z}_3[x]}$ and $mathbb{Z}_3[x] / A = {f(x) + A: f(x) in mathbb{Z}_3[x]}$ by definition.



      Let $f(x) in mathbb{Z}_3[x]$. By division algorithm, $f(x) = (x^2+2)q(x) + a + bx$ for some $q(x) in mathbb{Z}_3[x]$ and $a,b in mathbb{Z}_3$. Hence, $f(x) + A = (x^2+2)q(x) + a + bx + A = a + bx + (x^2+2)q(x) + A$. Since $(x^2+2)q(x) in A$, $(x^2+2)q(x) + A = A$. Thus, $f(x) + A = a + bx + A$. So we have $mathbb{Z_3}[x] / A = {a + bx + A: a,b in mathbb{Z_3}}$. Hence, the elements are the following:




      1. $A$

      2. $1 + A$

      3. $x + A$

      4. $2 + A$

      5. $2x + A$

      6. $1 + x + A$

      7. $2 + x + A$

      8. $2 + 2x + A$

      9. $1 + 2x + A$


      My question is:



      1) Is this the right derivation?



      2) How does multiplication table work in $mathbb{Z}_3[x] / A$? For example, if I have $(x+A)(1+2x+A) = x(1+2x) + A = x+2x^2 + A$ which is not in the same form as $ax+b+A$.






      abstract-algebra polynomials ring-theory




      abstract-algebra polynomials ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 16:23









      José Carlos Santos

      150k22122223




      150k22122223










      asked Nov 20 '18 at 16:11









      Ted

      738




      738






















          2 Answers
          2






          active

          oldest

          votes


















          1















          1. Yes, that is the right derivation.

          2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $mathbb{Z}_3[x]$).






          share|cite|improve this answer





























            0














            Using mod notation, in the quotient ring we have $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ which implies every polynomial is congruent to one of degree $le 1,,$ because $,x^{large 2q+r}! = (color{#c00}{x^{large 2}})^{large q},x^{large r} equiv color{#c00}{1}^{large q},x^{large r}equiv x^{large r}, $ for $,rin {0,1}$



            Alternatively breaking into even+odd parts $,f(x) = g(color{#c00}{x^2}) + x, h(color{#c00}{x^2})$ $Rightarrow, f(x)equiv g(color{#c00}{1}) + x, h(color{#c00}{1})$



            Or we can apply Division with Remainder: $,f(x) = q(x) (color{#c00}{x^2}!-!color{#c00}1) + ax+b,Rightarrow, f(x)equiv ax+b$



            So every $f(x)$ is congruent to $,(f,bmod x^2!-!1)bmod 3,$ having degree $le 1$, and these linear reps $,f,g,$ are incongruent else $,x^2-1,$ divides a lower degree polynomial $,f - g notequiv 0pmod{!3}.,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $,3^2 = 9$ such linear reps $,ax+b$ corresponding to the $3$ choices for the coef's $,a,bbmod 3$. Your table correctly lists these $9$ reps.



            To multiply these linear normal-form reps compute the polynomial product then replace $,color{#c00}{x^2},$ by $,color{#c00}{1},$



            $$ (a_1x + a_0)(b_1 x + b_0), equiv, (a_0 b_1 + a_1 b_0), x + a_0 b_0 color{#c00}{+1},a_1 b_1$$



            while performing coefficient arithmetic $!bmod 3.,$ The coefficient arithmetic will be slightly simpler if we use $,-1,$ vs. $,2,$ as our rep for $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$






            share|cite|improve this answer























              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%2f3006512%2ffinding-elements-of-mathbbz-3x-langle-x22-rangle-and-writing-the-mult%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









              1















              1. Yes, that is the right derivation.

              2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $mathbb{Z}_3[x]$).






              share|cite|improve this answer


























                1















                1. Yes, that is the right derivation.

                2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $mathbb{Z}_3[x]$).






                share|cite|improve this answer
























                  1












                  1








                  1







                  1. Yes, that is the right derivation.

                  2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $mathbb{Z}_3[x]$).






                  share|cite|improve this answer













                  1. Yes, that is the right derivation.

                  2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $mathbb{Z}_3[x]$).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 20 '18 at 16:22









                  José Carlos Santos

                  150k22122223




                  150k22122223























                      0














                      Using mod notation, in the quotient ring we have $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ which implies every polynomial is congruent to one of degree $le 1,,$ because $,x^{large 2q+r}! = (color{#c00}{x^{large 2}})^{large q},x^{large r} equiv color{#c00}{1}^{large q},x^{large r}equiv x^{large r}, $ for $,rin {0,1}$



                      Alternatively breaking into even+odd parts $,f(x) = g(color{#c00}{x^2}) + x, h(color{#c00}{x^2})$ $Rightarrow, f(x)equiv g(color{#c00}{1}) + x, h(color{#c00}{1})$



                      Or we can apply Division with Remainder: $,f(x) = q(x) (color{#c00}{x^2}!-!color{#c00}1) + ax+b,Rightarrow, f(x)equiv ax+b$



                      So every $f(x)$ is congruent to $,(f,bmod x^2!-!1)bmod 3,$ having degree $le 1$, and these linear reps $,f,g,$ are incongruent else $,x^2-1,$ divides a lower degree polynomial $,f - g notequiv 0pmod{!3}.,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $,3^2 = 9$ such linear reps $,ax+b$ corresponding to the $3$ choices for the coef's $,a,bbmod 3$. Your table correctly lists these $9$ reps.



                      To multiply these linear normal-form reps compute the polynomial product then replace $,color{#c00}{x^2},$ by $,color{#c00}{1},$



                      $$ (a_1x + a_0)(b_1 x + b_0), equiv, (a_0 b_1 + a_1 b_0), x + a_0 b_0 color{#c00}{+1},a_1 b_1$$



                      while performing coefficient arithmetic $!bmod 3.,$ The coefficient arithmetic will be slightly simpler if we use $,-1,$ vs. $,2,$ as our rep for $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$






                      share|cite|improve this answer




























                        0














                        Using mod notation, in the quotient ring we have $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ which implies every polynomial is congruent to one of degree $le 1,,$ because $,x^{large 2q+r}! = (color{#c00}{x^{large 2}})^{large q},x^{large r} equiv color{#c00}{1}^{large q},x^{large r}equiv x^{large r}, $ for $,rin {0,1}$



                        Alternatively breaking into even+odd parts $,f(x) = g(color{#c00}{x^2}) + x, h(color{#c00}{x^2})$ $Rightarrow, f(x)equiv g(color{#c00}{1}) + x, h(color{#c00}{1})$



                        Or we can apply Division with Remainder: $,f(x) = q(x) (color{#c00}{x^2}!-!color{#c00}1) + ax+b,Rightarrow, f(x)equiv ax+b$



                        So every $f(x)$ is congruent to $,(f,bmod x^2!-!1)bmod 3,$ having degree $le 1$, and these linear reps $,f,g,$ are incongruent else $,x^2-1,$ divides a lower degree polynomial $,f - g notequiv 0pmod{!3}.,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $,3^2 = 9$ such linear reps $,ax+b$ corresponding to the $3$ choices for the coef's $,a,bbmod 3$. Your table correctly lists these $9$ reps.



                        To multiply these linear normal-form reps compute the polynomial product then replace $,color{#c00}{x^2},$ by $,color{#c00}{1},$



                        $$ (a_1x + a_0)(b_1 x + b_0), equiv, (a_0 b_1 + a_1 b_0), x + a_0 b_0 color{#c00}{+1},a_1 b_1$$



                        while performing coefficient arithmetic $!bmod 3.,$ The coefficient arithmetic will be slightly simpler if we use $,-1,$ vs. $,2,$ as our rep for $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$






                        share|cite|improve this answer


























                          0












                          0








                          0






                          Using mod notation, in the quotient ring we have $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ which implies every polynomial is congruent to one of degree $le 1,,$ because $,x^{large 2q+r}! = (color{#c00}{x^{large 2}})^{large q},x^{large r} equiv color{#c00}{1}^{large q},x^{large r}equiv x^{large r}, $ for $,rin {0,1}$



                          Alternatively breaking into even+odd parts $,f(x) = g(color{#c00}{x^2}) + x, h(color{#c00}{x^2})$ $Rightarrow, f(x)equiv g(color{#c00}{1}) + x, h(color{#c00}{1})$



                          Or we can apply Division with Remainder: $,f(x) = q(x) (color{#c00}{x^2}!-!color{#c00}1) + ax+b,Rightarrow, f(x)equiv ax+b$



                          So every $f(x)$ is congruent to $,(f,bmod x^2!-!1)bmod 3,$ having degree $le 1$, and these linear reps $,f,g,$ are incongruent else $,x^2-1,$ divides a lower degree polynomial $,f - g notequiv 0pmod{!3}.,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $,3^2 = 9$ such linear reps $,ax+b$ corresponding to the $3$ choices for the coef's $,a,bbmod 3$. Your table correctly lists these $9$ reps.



                          To multiply these linear normal-form reps compute the polynomial product then replace $,color{#c00}{x^2},$ by $,color{#c00}{1},$



                          $$ (a_1x + a_0)(b_1 x + b_0), equiv, (a_0 b_1 + a_1 b_0), x + a_0 b_0 color{#c00}{+1},a_1 b_1$$



                          while performing coefficient arithmetic $!bmod 3.,$ The coefficient arithmetic will be slightly simpler if we use $,-1,$ vs. $,2,$ as our rep for $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$






                          share|cite|improve this answer














                          Using mod notation, in the quotient ring we have $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ which implies every polynomial is congruent to one of degree $le 1,,$ because $,x^{large 2q+r}! = (color{#c00}{x^{large 2}})^{large q},x^{large r} equiv color{#c00}{1}^{large q},x^{large r}equiv x^{large r}, $ for $,rin {0,1}$



                          Alternatively breaking into even+odd parts $,f(x) = g(color{#c00}{x^2}) + x, h(color{#c00}{x^2})$ $Rightarrow, f(x)equiv g(color{#c00}{1}) + x, h(color{#c00}{1})$



                          Or we can apply Division with Remainder: $,f(x) = q(x) (color{#c00}{x^2}!-!color{#c00}1) + ax+b,Rightarrow, f(x)equiv ax+b$



                          So every $f(x)$ is congruent to $,(f,bmod x^2!-!1)bmod 3,$ having degree $le 1$, and these linear reps $,f,g,$ are incongruent else $,x^2-1,$ divides a lower degree polynomial $,f - g notequiv 0pmod{!3}.,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $,3^2 = 9$ such linear reps $,ax+b$ corresponding to the $3$ choices for the coef's $,a,bbmod 3$. Your table correctly lists these $9$ reps.



                          To multiply these linear normal-form reps compute the polynomial product then replace $,color{#c00}{x^2},$ by $,color{#c00}{1},$



                          $$ (a_1x + a_0)(b_1 x + b_0), equiv, (a_0 b_1 + a_1 b_0), x + a_0 b_0 color{#c00}{+1},a_1 b_1$$



                          while performing coefficient arithmetic $!bmod 3.,$ The coefficient arithmetic will be slightly simpler if we use $,-1,$ vs. $,2,$ as our rep for $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 20 '18 at 18:10

























                          answered Nov 20 '18 at 17:40









                          Bill Dubuque

                          208k29190628




                          208k29190628






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3006512%2ffinding-elements-of-mathbbz-3x-langle-x22-rangle-and-writing-the-mult%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

                              MongoDB - Not Authorized To Execute Command

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                              Read more

                              How to fix TextFormField cause rebuild widget in Flutter

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                              Read more

                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                              Image
                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                              Read more