Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic?












4












$begingroup$


Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?






abstract-algebra ring-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are your thoughts on this matter?
    $endgroup$
    – MJD
    Dec 3 '18 at 4:59






  • 2




    $begingroup$
    What happens when $R$ is finite?
    $endgroup$
    – D_S
    Dec 3 '18 at 5:04










  • $begingroup$
    What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
    $endgroup$
    – AlexanderJ93
    Dec 3 '18 at 5:13










  • $begingroup$
    @D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
    $endgroup$
    – JM23
    Dec 3 '18 at 5:42






  • 1




    $begingroup$
    When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
    $endgroup$
    – bof
    Dec 3 '18 at 5:56
















4












$begingroup$


Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?






abstract-algebra ring-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are your thoughts on this matter?
    $endgroup$
    – MJD
    Dec 3 '18 at 4:59






  • 2




    $begingroup$
    What happens when $R$ is finite?
    $endgroup$
    – D_S
    Dec 3 '18 at 5:04










  • $begingroup$
    What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
    $endgroup$
    – AlexanderJ93
    Dec 3 '18 at 5:13










  • $begingroup$
    @D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
    $endgroup$
    – JM23
    Dec 3 '18 at 5:42






  • 1




    $begingroup$
    When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
    $endgroup$
    – bof
    Dec 3 '18 at 5:56














4












4








4


2



$begingroup$


Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?






abstract-algebra ring-isomorphism







share|cite|improve this question











$endgroup$




Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?






abstract-algebra ring-isomorphism




abstract-algebra ring-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 5:46







JM23

















asked Dec 3 '18 at 4:54









JM23JM23

212




212












  • $begingroup$
    What are your thoughts on this matter?
    $endgroup$
    – MJD
    Dec 3 '18 at 4:59






  • 2




    $begingroup$
    What happens when $R$ is finite?
    $endgroup$
    – D_S
    Dec 3 '18 at 5:04










  • $begingroup$
    What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
    $endgroup$
    – AlexanderJ93
    Dec 3 '18 at 5:13










  • $begingroup$
    @D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
    $endgroup$
    – JM23
    Dec 3 '18 at 5:42






  • 1




    $begingroup$
    When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
    $endgroup$
    – bof
    Dec 3 '18 at 5:56


















  • $begingroup$
    What are your thoughts on this matter?
    $endgroup$
    – MJD
    Dec 3 '18 at 4:59






  • 2




    $begingroup$
    What happens when $R$ is finite?
    $endgroup$
    – D_S
    Dec 3 '18 at 5:04










  • $begingroup$
    What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
    $endgroup$
    – AlexanderJ93
    Dec 3 '18 at 5:13










  • $begingroup$
    @D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
    $endgroup$
    – JM23
    Dec 3 '18 at 5:42






  • 1




    $begingroup$
    When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
    $endgroup$
    – bof
    Dec 3 '18 at 5:56
















$begingroup$
What are your thoughts on this matter?
$endgroup$
– MJD
Dec 3 '18 at 4:59




$begingroup$
What are your thoughts on this matter?
$endgroup$
– MJD
Dec 3 '18 at 4:59




2




2




$begingroup$
What happens when $R$ is finite?
$endgroup$
– D_S
Dec 3 '18 at 5:04




$begingroup$
What happens when $R$ is finite?
$endgroup$
– D_S
Dec 3 '18 at 5:04












$begingroup$
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
$endgroup$
– AlexanderJ93
Dec 3 '18 at 5:13




$begingroup$
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
$endgroup$
– AlexanderJ93
Dec 3 '18 at 5:13












$begingroup$
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
$endgroup$
– JM23
Dec 3 '18 at 5:42




$begingroup$
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
$endgroup$
– JM23
Dec 3 '18 at 5:42




1




1




$begingroup$
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
$endgroup$
– bof
Dec 3 '18 at 5:56




$begingroup$
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
$endgroup$
– bof
Dec 3 '18 at 5:56










2 Answers
2






active

oldest

votes


















4












$begingroup$

Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
    $endgroup$
    – JM23
    Dec 3 '18 at 7:59












  • $begingroup$
    @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 8:20



















1












$begingroup$

HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...






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%2f3023651%2fare-the-rings-mathbbr2-and-mathbbr3-isomorphic%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









    4












    $begingroup$

    Consider idempotents in each ring, that is solutions of $e^2=e$. In your
    first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
      $endgroup$
      – JM23
      Dec 3 '18 at 7:59












    • $begingroup$
      @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
      $endgroup$
      – Robert Lewis
      Dec 3 '18 at 8:20
















    4












    $begingroup$

    Consider idempotents in each ring, that is solutions of $e^2=e$. In your
    first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
      $endgroup$
      – JM23
      Dec 3 '18 at 7:59












    • $begingroup$
      @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
      $endgroup$
      – Robert Lewis
      Dec 3 '18 at 8:20














    4












    4








    4





    $begingroup$

    Consider idempotents in each ring, that is solutions of $e^2=e$. In your
    first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?






    share|cite|improve this answer









    $endgroup$



    Consider idempotents in each ring, that is solutions of $e^2=e$. In your
    first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 7:01









    Lord Shark the UnknownLord Shark the Unknown

    102k1160132




    102k1160132








    • 1




      $begingroup$
      For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
      $endgroup$
      – JM23
      Dec 3 '18 at 7:59












    • $begingroup$
      @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
      $endgroup$
      – Robert Lewis
      Dec 3 '18 at 8:20














    • 1




      $begingroup$
      For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
      $endgroup$
      – JM23
      Dec 3 '18 at 7:59












    • $begingroup$
      @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
      $endgroup$
      – Robert Lewis
      Dec 3 '18 at 8:20








    1




    1




    $begingroup$
    For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
    $endgroup$
    – JM23
    Dec 3 '18 at 7:59






    $begingroup$
    For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
    $endgroup$
    – JM23
    Dec 3 '18 at 7:59














    $begingroup$
    @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 8:20




    $begingroup$
    @JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 8:20











    1












    $begingroup$

    HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...






        share|cite|improve this answer









        $endgroup$



        HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 5:02









        mathematics2x2lifemathematics2x2life

        8,06121738




        8,06121738






























            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%2f3023651%2fare-the-rings-mathbbr2-and-mathbbr3-isomorphic%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more