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









$endgroup$













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    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






























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