To prove a set is a subring but not an ideal.

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Problem



Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.



Silly doubt



Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?



If that part is clear, then this problem can be solved.






abstract-algebra ring-theory







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    up vote
    2
    down vote

    favorite












    Problem



    Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.



    Silly doubt



    Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?



    If that part is clear, then this problem can be solved.






    abstract-algebra ring-theory







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Problem



      Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.



      Silly doubt



      Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?



      If that part is clear, then this problem can be solved.






      abstract-algebra ring-theory







      share|cite|improve this question















      Problem



      Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.



      Silly doubt



      Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?



      If that part is clear, then this problem can be solved.






      abstract-algebra ring-theory




      abstract-algebra ring-theory






      share|cite|improve this question















      share|cite|improve this question













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      edited 2 days ago









      egreg

      166k1180189




      166k1180189










      asked 2 days ago









      blue boy

      1,095513




      1,095513




















          2 Answers
          2






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          up vote
          6
          down vote



          accepted










          The operations are addition and multiplication of functions, i.e.
          $$
          (f+g)(x)=f(x)+g(x) \
          (fg)(x)=f(x)g(x).
          $$






          share|cite|improve this answer




















          • Thanks for answering. So if nothing is given we assume this ?
            – blue boy
            2 days ago










          • @blueboy Yes, you do.
            – Janik
            2 days ago

















          up vote
          3
          down vote













          The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            6
            down vote



            accepted










            The operations are addition and multiplication of functions, i.e.
            $$
            (f+g)(x)=f(x)+g(x) \
            (fg)(x)=f(x)g(x).
            $$






            share|cite|improve this answer




















            • Thanks for answering. So if nothing is given we assume this ?
              – blue boy
              2 days ago










            • @blueboy Yes, you do.
              – Janik
              2 days ago














            up vote
            6
            down vote



            accepted










            The operations are addition and multiplication of functions, i.e.
            $$
            (f+g)(x)=f(x)+g(x) \
            (fg)(x)=f(x)g(x).
            $$






            share|cite|improve this answer




















            • Thanks for answering. So if nothing is given we assume this ?
              – blue boy
              2 days ago










            • @blueboy Yes, you do.
              – Janik
              2 days ago












            up vote
            6
            down vote



            accepted







            up vote
            6
            down vote



            accepted






            The operations are addition and multiplication of functions, i.e.
            $$
            (f+g)(x)=f(x)+g(x) \
            (fg)(x)=f(x)g(x).
            $$






            share|cite|improve this answer












            The operations are addition and multiplication of functions, i.e.
            $$
            (f+g)(x)=f(x)+g(x) \
            (fg)(x)=f(x)g(x).
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Janik

            1,4352418




            1,4352418











            • Thanks for answering. So if nothing is given we assume this ?
              – blue boy
              2 days ago










            • @blueboy Yes, you do.
              – Janik
              2 days ago
















            • Thanks for answering. So if nothing is given we assume this ?
              – blue boy
              2 days ago










            • @blueboy Yes, you do.
              – Janik
              2 days ago















            Thanks for answering. So if nothing is given we assume this ?
            – blue boy
            2 days ago




            Thanks for answering. So if nothing is given we assume this ?
            – blue boy
            2 days ago












            @blueboy Yes, you do.
            – Janik
            2 days ago




            @blueboy Yes, you do.
            – Janik
            2 days ago










            up vote
            3
            down vote













            The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.






            share|cite|improve this answer
























              up vote
              3
              down vote













              The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.






                share|cite|improve this answer












                The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Fra

                532211




                532211



























                     

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