find the maximal ideal of the ring ?.












2














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







share|cite




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    43 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    41 mins ago






  • 1




    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 $mathbf R$.
    – Bernard
    37 mins ago






  • 2




    @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
    – IAmNoOne
    25 mins ago


















2














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







share|cite




















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    43 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    41 mins ago






  • 1




    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 $mathbf R$.
    – Bernard
    37 mins ago






  • 2




    @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
    – IAmNoOne
    25 mins ago
















2












2








2







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







share|cite















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




abstract-algebra






share|cite















share|cite













share|cite




share|cite








edited 19 mins ago









Key Flex

7,54441232




7,54441232










asked 45 mins ago









jasmine

1,582416




1,582416








  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    43 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    41 mins ago






  • 1




    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 $mathbf R$.
    – Bernard
    37 mins ago






  • 2




    @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
    – IAmNoOne
    25 mins ago
















  • 3




    No: $(x^2)$ is not maximal.
    – Bernard
    43 mins ago










  • @Bernard im not getting can u elaborate this ?
    – jasmine
    41 mins ago






  • 1




    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 $mathbf R$.
    – Bernard
    37 mins ago






  • 2




    @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
    – IAmNoOne
    25 mins ago










3




3




No: $(x^2)$ is not maximal.
– Bernard
43 mins ago




No: $(x^2)$ is not maximal.
– Bernard
43 mins ago












@Bernard im not getting can u elaborate this ?
– jasmine
41 mins ago




@Bernard im not getting can u elaborate this ?
– jasmine
41 mins ago




1




1




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 $mathbf R$.
– Bernard
37 mins ago




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 $mathbf R$.
– Bernard
37 mins ago




2




2




@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
– IAmNoOne
25 mins ago






@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
– IAmNoOne
25 mins ago












2 Answers
2






active

oldest

votes


















3














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





























    2














    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.






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









      3














      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


























        3














        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
























          3












          3








          3






          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












          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 23 mins ago









          user544921

          607




          607























              2














              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.






              share|cite|improve this answer


























                2














                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.






                share|cite|improve this answer
























                  2












                  2








                  2






                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 mins ago









                  Key Flex

                  7,54441232




                  7,54441232






























                      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%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

                      flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                      Image
                      0 I recently upgraded an Ubuntu 16.04 virtual machine to 18.04. I regularly use this VM to run apt-mirror at school on their high-speed internet, then rsync these files to my rack-mount server at home for updating multiple computers at home without wasting the tiny amount of bandwidth I have there. Since updating I can't get apt-mirror to run on my VM. It keeps giving the error: flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror line 214 immediately followed by: apt-mirror is already running, exiting at /usr/bin/apt-mirror line 217 apt-mirror is NOT running, and does not have a cron job. I've tried rebooting, and even looked for the old-style lockfile to delete, but I guess that method is no longer used. Is there some way to force unlock this so I can run apt-mirror? UPDATE: So... I f
                      Read more

                      Mangá

                      Image
                        Nota:  Não confundir com Manga. Mangás e Animes Mangá Mangaká • Tankōbon Film comic La nouvelle manga Mangá original em inglês Anime História Animação influenciada por animes • Filler • OVA • Seiyū Públicos Kodomo • Shōjo • Shōnen Josei • Seinen Gêneros Mahō shōjo • Mecha • Harém Ecchi • Hentai ( Yaoi • Yuri Futanari • Shotacon • Lolicon Toddlercon ) Termos Bishōjo • Chibi • Fan service Gaiden • Kawaii • Kaitō • Light novel Mahō shōnen • Moe (antropomorfismo • Kemonomimi Nekomimi ) • Super deformed Omake • Tsundere • Yonkoma Listas Mangás • Animes • Ecchi • Mangaka • Estúdios • Revistas Fãs ( otaku ) Anime Music Video Convenções • Cosplay Dōjinshi • Fansub • Fandub • Fujoshi OS-tan • Scanlation Portal de Anime e mangá v d e O mangá (português brasileiro) ou manga (português europeu) [ 1 ] (em japonês: 漫画 , manga ? , lit. “história em quadrinhos”) , é a
                      Read more

                      Eduardo VII do Reino Unido

                      Image
                      Eduardo VII Rei do Reino Unido da Grã-Bretanha e Irlanda e dos Domínios Britânicos de Além-mar Imperador da Índia Rei do Reino Unido e Imperador da Índia Reinado 22 de janeiro de 1901 a 6 de maio de 1910 Coroação 9 de agosto de 1902 Delhi Durbar 1 de janeiro de 1903 Predecessora Vitória Sucessor Jorge V   Esposa Alexandra da Dinamarca Descendência Alberto Vitor, Duque de Clarence e Avondale Jorge V do Reino Unido Luísa, Princesa Real Vitória de Gales Maud de Gales Alexandre João de Gales Casa Saxe-Coburgo-Gota Nome completo Alberto Eduardo Nascimento 9 de novembro de 1841   Palácio de Buckingham, Londres, Reino Unido Morte 6 de maio de 1910 (68 anos)   Palácio de Buckingham, Londres, Reino Unido Enterro Capela de São Jorge, Windsor, Berkshire, Reino Unido 20 de maio de 1910 Pai Alberto de Saxe-Coburgo-Gota Mãe Vitória do Reino Unido Assinatura
                      Read more