Prove that $ I + J = R iff b + J $ is invertible in $ R / J$, where $ I = langle b rangle$












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$begingroup$


Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$



If it is a duplicate, I apologize, but I've already looked up for it.
I was able to show this:
$ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.






abstract-algebra proof-verification ring-theory ideals







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$endgroup$

















    1












    $begingroup$


    Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$



    If it is a duplicate, I apologize, but I've already looked up for it.
    I was able to show this:
    $ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.






    abstract-algebra proof-verification ring-theory ideals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$



      If it is a duplicate, I apologize, but I've already looked up for it.
      I was able to show this:
      $ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.






      abstract-algebra proof-verification ring-theory ideals







      share|cite|improve this question











      $endgroup$




      Let $R$ be an integral domain and $I, J$ its ideals, where $I = langle b rangle$, for a $ b in R$. Prove that $ I + J = R iff b + J $ is invertible in $ R / J$



      If it is a duplicate, I apologize, but I've already looked up for it.
      I was able to show this:
      $ I + J = R iff (forall r in R) r = i + j $, for some $ i in I, j in J iff (forall r in R) r = ba + j $, for some $ a in R, j in J$. Am I missing something or am I in a completely wrong direction? Thank you.






      abstract-algebra proof-verification ring-theory ideals




      abstract-algebra proof-verification ring-theory ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 16:21







      user593746

















      asked Sep 20 '18 at 11:45









      Nemanja BericNemanja Beric

      37018




      37018






















          1 Answer
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          3












          $begingroup$

          You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:01








          • 1




            $begingroup$
            @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
            $endgroup$
            – user593746
            Sep 20 '18 at 12:13










          • $begingroup$
            Right, right. I must not omit the "+ J"
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:14












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          1 Answer
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          active

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          3












          $begingroup$

          You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:01








          • 1




            $begingroup$
            @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
            $endgroup$
            – user593746
            Sep 20 '18 at 12:13










          • $begingroup$
            Right, right. I must not omit the "+ J"
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:14
















          3












          $begingroup$

          You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:01








          • 1




            $begingroup$
            @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
            $endgroup$
            – user593746
            Sep 20 '18 at 12:13










          • $begingroup$
            Right, right. I must not omit the "+ J"
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:14














          3












          3








          3





          $begingroup$

          You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?






          share|cite|improve this answer









          $endgroup$



          You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $ain R$ and $jin J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 20 '18 at 11:51







          user593746



















          • $begingroup$
            Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:01








          • 1




            $begingroup$
            @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
            $endgroup$
            – user593746
            Sep 20 '18 at 12:13










          • $begingroup$
            Right, right. I must not omit the "+ J"
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:14


















          • $begingroup$
            Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:01








          • 1




            $begingroup$
            @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
            $endgroup$
            – user593746
            Sep 20 '18 at 12:13










          • $begingroup$
            Right, right. I must not omit the "+ J"
            $endgroup$
            – Nemanja Beric
            Sep 20 '18 at 12:14
















          $begingroup$
          Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
          $endgroup$
          – Nemanja Beric
          Sep 20 '18 at 12:01






          $begingroup$
          Well, if ab + j = 1, for some a, that means that b is invertible in R / J, because R / J = r + J = {r + j : j in J}, so for j = 0 (which is in J), ab + j= 1 + 0, i.e. b is invertible. Am I right?
          $endgroup$
          – Nemanja Beric
          Sep 20 '18 at 12:01






          1




          1




          $begingroup$
          @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
          $endgroup$
          – user593746
          Sep 20 '18 at 12:13




          $begingroup$
          @NemanjaBeric I would write it in the following way: $ab+j=1$ implies $ab+J=1+J$. Thus, $(a+J)(b+J)=1+J$. That is, $a+Jin R/J$ is the inverse of $b+Jin R/J$.
          $endgroup$
          – user593746
          Sep 20 '18 at 12:13












          $begingroup$
          Right, right. I must not omit the "+ J"
          $endgroup$
          – Nemanja Beric
          Sep 20 '18 at 12:14




          $begingroup$
          Right, right. I must not omit the "+ J"
          $endgroup$
          – Nemanja Beric
          Sep 20 '18 at 12:14


















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