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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












    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













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      egreg

      166k1180189




      166k1180189










      asked 2 days ago









      blue boy

      1,095513




      1,095513




















          2 Answers
          2






          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
          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




















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911884%2fto-prove-a-set-is-a-subring-but-not-an-ideal%23new-answer', 'question_page');

            );

            Post as a guest

















            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



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911884%2fto-prove-a-set-is-a-subring-but-not-an-ideal%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    Long meetings (6-7 hours a day): Being “babysat” by supervisor

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
                    Read more

                    Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
                    Read more

                    Confectionery

                    Image
                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
                    Read more