about flat $R$-modules with different conditions [closed]











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Need to know if the following statements are true or false:




  1. Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.


  2. Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.


  3. Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.


  4. $R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?











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closed as off-topic by user26857, rschwieb, user302797, max_zorn, Brahadeesh Nov 23 at 6:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, user302797, max_zorn, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.

















    up vote
    -2
    down vote

    favorite












    Need to know if the following statements are true or false:




    1. Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.


    2. Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.


    3. Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.


    4. $R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?











    share|cite|improve this question















    closed as off-topic by user26857, rschwieb, user302797, max_zorn, Brahadeesh Nov 23 at 6:28


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, user302797, max_zorn, Brahadeesh

    If this question can be reworded to fit the rules in the help center, please edit the question.















      up vote
      -2
      down vote

      favorite









      up vote
      -2
      down vote

      favorite











      Need to know if the following statements are true or false:




      1. Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.


      2. Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.


      3. Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.


      4. $R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?











      share|cite|improve this question















      Need to know if the following statements are true or false:




      1. Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.


      2. Let $I,J subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.


      3. Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/left<sum^{infty}_{i=2} x^iright>$ is always flat $R$-module.


      4. $R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?








      abstract-algebra commutative-algebra modules flatness






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      edited Nov 22 at 19:43









      user26857

      39.2k123882




      39.2k123882










      asked Nov 22 at 13:37









      idriskameni

      608




      608




      closed as off-topic by user26857, rschwieb, user302797, max_zorn, Brahadeesh Nov 23 at 6:28


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, user302797, max_zorn, Brahadeesh

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by user26857, rschwieb, user302797, max_zorn, Brahadeesh Nov 23 at 6:28


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, rschwieb, user302797, max_zorn, Brahadeesh

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






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          oldest

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          up vote
          1
          down vote



          accepted










          I assume you are working over commutative rings.




          1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.


          2. This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.


          3. This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.


          4. This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.







          share|cite|improve this answer






























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            I assume you are working over commutative rings.




            1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.


            2. This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.


            3. This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.


            4. This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.







            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              I assume you are working over commutative rings.




              1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.


              2. This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.


              3. This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.


              4. This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.







              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                I assume you are working over commutative rings.




                1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.


                2. This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.


                3. This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.


                4. This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.







                share|cite|improve this answer














                I assume you are working over commutative rings.




                1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $Ato Bto C$ are ring homomorphisms with $Ato B$ and $Bto C$ are flat, then so is $Ato C$. Since $S^{-1}Rto S^{-1}[x,y,z]$ is flat, you are done.


                2. This is false. Take $R=mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.


                3. This is correct. $f=sum_{n=2}^{infty} x^n=x^2u$ where $u=sum_{n=0}^{infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.


                4. This is false. For example, let $R=mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.








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                share|cite|improve this answer








                edited Nov 22 at 21:37









                user26857

                39.2k123882




                39.2k123882










                answered Nov 22 at 18:23









                Mohan

                11.5k1817




                11.5k1817















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