Show that a ring of fractions and a quotient ring are isomorphic [duplicate]












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This question already has an answer here:




  • Localisation is isomorphic to a quotient of polynomial ring

    5 answers




For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$



I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?






abstract-algebra ring-theory







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marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














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    Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
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    – Anurag A
    Dec 6 '18 at 22:07












  • $begingroup$
    Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
    $endgroup$
    – André 3000
    Dec 6 '18 at 23:31
















1












$begingroup$



This question already has an answer here:




  • Localisation is isomorphic to a quotient of polynomial ring

    5 answers




For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$



I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$



marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:07












  • $begingroup$
    Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
    $endgroup$
    – André 3000
    Dec 6 '18 at 23:31














1












1








1





$begingroup$



This question already has an answer here:




  • Localisation is isomorphic to a quotient of polynomial ring

    5 answers




For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$



I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Localisation is isomorphic to a quotient of polynomial ring

    5 answers




For a commutative ring $R$ with $1neq 0$ and a nonzerodivisor $r in R$, let $S$ be the set
$S={r^nmid nin mathbb{Z}, ngeq 0}$ and denote $S^{-1}R=Rleft[frac{1}{r}right]$.
Prove that there is a ring isomorphism $$Rleft[frac{1}{r}right]cong frac{R[x]}{(rx -1)}.$$



I'm thinking maybe I can find a homomorphism from $R[x]$ to $Rleft[frac{1}{r}right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?





This question already has an answer here:




  • Localisation is isomorphic to a quotient of polynomial ring

    5 answers







abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 6 '18 at 22:15









Bernard

120k740116




120k740116










asked Dec 6 '18 at 21:58









WesleyWesley

518313




518313




marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by André 3000, darij grinberg, user10354138, Brahadeesh, KReiser Dec 7 '18 at 8:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:07












  • $begingroup$
    Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
    $endgroup$
    – André 3000
    Dec 6 '18 at 23:31














  • 1




    $begingroup$
    Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:07












  • $begingroup$
    Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
    $endgroup$
    – André 3000
    Dec 6 '18 at 23:31








1




1




$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07






$begingroup$
Yes, evaluation homomorphism can do the trick $phi(p(x))=pleft(frac{1}{r}right)$.
$endgroup$
– Anurag A
Dec 6 '18 at 22:07














$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31




$begingroup$
Please search the site before asking a new question. This has been asked many, many times before: 1, 2, 3, 4 to name a few.
$endgroup$
– André 3000
Dec 6 '18 at 23:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division



$$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$



Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.



Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
    $$ R[x]xrightarrow{phi} S^{-1}R.$$
    Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why is the map clearly a surjection?
      $endgroup$
      – Wesley
      Dec 6 '18 at 22:18










    • $begingroup$
      Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
      $endgroup$
      – Wesley
      Dec 6 '18 at 22:26










    • $begingroup$
      Yes that is one way to see it.
      $endgroup$
      – Antonios-Alexandros Robotis
      Dec 6 '18 at 22:50










    • $begingroup$
      It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
      $endgroup$
      – Wesley
      Dec 6 '18 at 23:14






    • 1




      $begingroup$
      @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
      $endgroup$
      – Bill Dubuque
      Dec 7 '18 at 0:02


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division



    $$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$



    Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.



    Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division



      $$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$



      Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.



      Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division



        $$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$



        Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.



        Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.






        share|cite|improve this answer











        $endgroup$



        Tricky is proving the kernel $K = (rx!-!1).,$ A simple way: if $fin K$ then by nonmonic division



        $$ r^n f(x) = (rx!-!1),q(x) + r', {rm for} r'in R, nin Bbb N$$



        Evaluating at $, x = 1/r,$ shows $,r'! = 0,$ so $,rx!-!1mid r^n f,Rightarrow,rx!-!1mid f,,$ by $,(rx!-!1,r) = (1);,$ more explicitly $,rx!-!1mid rg,Rightarrow, rx!-!1mid g = x(rg)-(rx!-!1)g$.



        Remark $ $ See this answer for another proof and further discussion. If you already know basic properties of localizations then see also the linked dupe for ways to exploit these proeprties.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 0:22

























        answered Dec 6 '18 at 23:59









        Bill DubuqueBill Dubuque

        210k29192643




        210k29192643























            0












            $begingroup$

            You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
            $$ R[x]xrightarrow{phi} S^{-1}R.$$
            Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why is the map clearly a surjection?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:18










            • $begingroup$
              Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:26










            • $begingroup$
              Yes that is one way to see it.
              $endgroup$
              – Antonios-Alexandros Robotis
              Dec 6 '18 at 22:50










            • $begingroup$
              It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
              $endgroup$
              – Wesley
              Dec 6 '18 at 23:14






            • 1




              $begingroup$
              @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 7 '18 at 0:02
















            0












            $begingroup$

            You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
            $$ R[x]xrightarrow{phi} S^{-1}R.$$
            Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why is the map clearly a surjection?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:18










            • $begingroup$
              Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:26










            • $begingroup$
              Yes that is one way to see it.
              $endgroup$
              – Antonios-Alexandros Robotis
              Dec 6 '18 at 22:50










            • $begingroup$
              It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
              $endgroup$
              – Wesley
              Dec 6 '18 at 23:14






            • 1




              $begingroup$
              @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 7 '18 at 0:02














            0












            0








            0





            $begingroup$

            You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
            $$ R[x]xrightarrow{phi} S^{-1}R.$$
            Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.






            share|cite|improve this answer











            $endgroup$



            You should note that $frac{1}{r}$ has the property $frac{1}{r}cdot r=1$. That is, $frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $sin R$ to itself for all $s$. Send $xmapsto frac{1}{r}$. This is clearly a surjective homomorphism
            $$ R[x]xrightarrow{phi} S^{-1}R.$$
            Calculate $kerphi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 6 '18 at 22:37

























            answered Dec 6 '18 at 22:08









            Antonios-Alexandros RobotisAntonios-Alexandros Robotis

            10.2k41641




            10.2k41641












            • $begingroup$
              Why is the map clearly a surjection?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:18










            • $begingroup$
              Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:26










            • $begingroup$
              Yes that is one way to see it.
              $endgroup$
              – Antonios-Alexandros Robotis
              Dec 6 '18 at 22:50










            • $begingroup$
              It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
              $endgroup$
              – Wesley
              Dec 6 '18 at 23:14






            • 1




              $begingroup$
              @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 7 '18 at 0:02


















            • $begingroup$
              Why is the map clearly a surjection?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:18










            • $begingroup$
              Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
              $endgroup$
              – Wesley
              Dec 6 '18 at 22:26










            • $begingroup$
              Yes that is one way to see it.
              $endgroup$
              – Antonios-Alexandros Robotis
              Dec 6 '18 at 22:50










            • $begingroup$
              It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
              $endgroup$
              – Wesley
              Dec 6 '18 at 23:14






            • 1




              $begingroup$
              @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
              $endgroup$
              – Bill Dubuque
              Dec 7 '18 at 0:02
















            $begingroup$
            Why is the map clearly a surjection?
            $endgroup$
            – Wesley
            Dec 6 '18 at 22:18




            $begingroup$
            Why is the map clearly a surjection?
            $endgroup$
            – Wesley
            Dec 6 '18 at 22:18












            $begingroup$
            Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
            $endgroup$
            – Wesley
            Dec 6 '18 at 22:26




            $begingroup$
            Let $ain R$ and $r^kin S$. Is it surjective because $phi(ax^k) = frac{a}{r^k}$, and the latter term is a general term of the ring of fractions?
            $endgroup$
            – Wesley
            Dec 6 '18 at 22:26












            $begingroup$
            Yes that is one way to see it.
            $endgroup$
            – Antonios-Alexandros Robotis
            Dec 6 '18 at 22:50




            $begingroup$
            Yes that is one way to see it.
            $endgroup$
            – Antonios-Alexandros Robotis
            Dec 6 '18 at 22:50












            $begingroup$
            It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
            $endgroup$
            – Wesley
            Dec 6 '18 at 23:14




            $begingroup$
            It's clear that $(rx-1) subseteq ker(phi)$, but how would I show that $ker(phi) subseteq (rx-1)$? If I could use the division algorithm I could see how, but I can't see how to do it without the division algorithm.
            $endgroup$
            – Wesley
            Dec 6 '18 at 23:14




            1




            1




            $begingroup$
            @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
            $endgroup$
            – Bill Dubuque
            Dec 7 '18 at 0:02




            $begingroup$
            @Wesley One simple way is to use the nonmonic division algorithm - see my answer.
            $endgroup$
            – Bill Dubuque
            Dec 7 '18 at 0:02



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