find the maximal ideal of the ring ?.












1












$begingroup$


find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No: $(x^2)$ is not maximal.
    $endgroup$
    – Bernard
    Jan 6 at 2:13










  • $begingroup$
    @Bernard im not getting can u elaborate this ?
    $endgroup$
    – jasmine
    Jan 6 at 2:15






  • 1




    $begingroup$
    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    $endgroup$
    – Bernard
    Jan 6 at 2:19






  • 2




    $begingroup$
    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    $endgroup$
    – IAmNoOne
    Jan 6 at 2:31


















1












$begingroup$


find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No: $(x^2)$ is not maximal.
    $endgroup$
    – Bernard
    Jan 6 at 2:13










  • $begingroup$
    @Bernard im not getting can u elaborate this ?
    $endgroup$
    – jasmine
    Jan 6 at 2:15






  • 1




    $begingroup$
    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    $endgroup$
    – Bernard
    Jan 6 at 2:19






  • 2




    $begingroup$
    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    $endgroup$
    – IAmNoOne
    Jan 6 at 2:31
















1












1








1


1



$begingroup$


find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$



My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$



Is its correct ?



any hints/solution will be appreciated






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 3:28









J. W. Tanner

51010




51010










asked Jan 6 at 2:11









jasminejasmine

1,689416




1,689416








  • 3




    $begingroup$
    No: $(x^2)$ is not maximal.
    $endgroup$
    – Bernard
    Jan 6 at 2:13










  • $begingroup$
    @Bernard im not getting can u elaborate this ?
    $endgroup$
    – jasmine
    Jan 6 at 2:15






  • 1




    $begingroup$
    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    $endgroup$
    – Bernard
    Jan 6 at 2:19






  • 2




    $begingroup$
    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    $endgroup$
    – IAmNoOne
    Jan 6 at 2:31
















  • 3




    $begingroup$
    No: $(x^2)$ is not maximal.
    $endgroup$
    – Bernard
    Jan 6 at 2:13










  • $begingroup$
    @Bernard im not getting can u elaborate this ?
    $endgroup$
    – jasmine
    Jan 6 at 2:15






  • 1




    $begingroup$
    The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
    $endgroup$
    – Bernard
    Jan 6 at 2:19






  • 2




    $begingroup$
    @jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
    $endgroup$
    – IAmNoOne
    Jan 6 at 2:31










3




3




$begingroup$
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13




$begingroup$
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13












$begingroup$
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15




$begingroup$
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15




1




1




$begingroup$
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19




$begingroup$
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19




2




2




$begingroup$
@jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31






$begingroup$
@jasmine, $x^2 = x(x - 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31












4 Answers
4






active

oldest

votes


















3












$begingroup$

Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






    share|cite|improve this answer











    $endgroup$





















      3












      $begingroup$

      Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



      Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).



      Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6xequiv -6x$. Then we get $x$, then $6$, then $1$.






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






        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%2f3063406%2ffind-the-maximal-ideal-of-the-ring%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.






              share|cite|improve this answer









              $endgroup$



              Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 6 at 2:33









              user544921user544921

              12717




              12717























                  3












                  $begingroup$

                  There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.






                      share|cite|improve this answer











                      $endgroup$



                      There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 6 at 3:00

























                      answered Jan 6 at 2:35









                      Key FlexKey Flex

                      7,83961232




                      7,83961232























                          3












                          $begingroup$

                          Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                          Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).



                          Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6xequiv -6x$. Then we get $x$, then $6$, then $1$.






                          share|cite|improve this answer











                          $endgroup$


















                            3












                            $begingroup$

                            Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                            Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).



                            Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6xequiv -6x$. Then we get $x$, then $6$, then $1$.






                            share|cite|improve this answer











                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                              Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).



                              Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6xequiv -6x$. Then we get $x$, then $6$, then $1$.






                              share|cite|improve this answer











                              $endgroup$



                              Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.



                              Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).



                              Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6xequiv -6x$. Then we get $x$, then $6$, then $1$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 6 at 4:23

























                              answered Jan 6 at 3:25









                              Chris CusterChris Custer

                              11.6k3824




                              11.6k3824























                                  1












                                  $begingroup$

                                  Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 6 at 7:54









                                      Es.RoEs.Ro

                                      83228




                                      83228






























                                          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%2f3063406%2ffind-the-maximal-ideal-of-the-ring%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