What algebraic structure does the set of endomorphisms of a ring have?












2












$begingroup$


Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:





  1. $+$, defined by $(f+g)(x)=f(x)+g(x)$


  2. $cdot$, defined by $(fcdot g)(x)=f(g(x))$


  3. $*$, defiend by $(f*g)(x)=f(x)g(x)$


Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?



Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?






abstract-algebra ring-theory modules terminology ring-homomorphism







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_ring
    $endgroup$
    – Chris Culter
    6 hours ago






  • 1




    $begingroup$
    @ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
    $endgroup$
    – Keshav Srinivasan
    5 hours ago
















2












$begingroup$


Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:





  1. $+$, defined by $(f+g)(x)=f(x)+g(x)$


  2. $cdot$, defined by $(fcdot g)(x)=f(g(x))$


  3. $*$, defiend by $(f*g)(x)=f(x)g(x)$


Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?



Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?






abstract-algebra ring-theory modules terminology ring-homomorphism







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_ring
    $endgroup$
    – Chris Culter
    6 hours ago






  • 1




    $begingroup$
    @ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
    $endgroup$
    – Keshav Srinivasan
    5 hours ago














2












2








2





$begingroup$


Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:





  1. $+$, defined by $(f+g)(x)=f(x)+g(x)$


  2. $cdot$, defined by $(fcdot g)(x)=f(g(x))$


  3. $*$, defiend by $(f*g)(x)=f(x)g(x)$


Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?



Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?






abstract-algebra ring-theory modules terminology ring-homomorphism







share|cite|improve this question









$endgroup$




Let $R$ be a ring, and let $End(R)$ be the set of ring endomorphisms of $R$, i.e. the set of all ring homomorphisms form $R$ to $R$. Then we can define three binary operations on $End(R)$:





  1. $+$, defined by $(f+g)(x)=f(x)+g(x)$


  2. $cdot$, defined by $(fcdot g)(x)=f(g(x))$


  3. $*$, defiend by $(f*g)(x)=f(x)g(x)$


Now $(End(R),+,cdot)$ is a ring, being a subring of the endomorphism ring of the abelian group $(R,+)$. But my question is, what is the algebraic structure of $(End(R),+,cdot,*)$? Does this beast with three binary operations have a name?



Also, what algebraic structure does $(End(R),+,*)$ have? Is that also a ring?






abstract-algebra ring-theory modules terminology ring-homomorphism




abstract-algebra ring-theory modules terminology ring-homomorphism






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









Keshav SrinivasanKeshav Srinivasan

2,12811441




2,12811441








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_ring
    $endgroup$
    – Chris Culter
    6 hours ago






  • 1




    $begingroup$
    @ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
    $endgroup$
    – Keshav Srinivasan
    5 hours ago














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Composition_ring
    $endgroup$
    – Chris Culter
    6 hours ago






  • 1




    $begingroup$
    @ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
    $endgroup$
    – Keshav Srinivasan
    5 hours ago








1




1




$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
6 hours ago




$begingroup$
en.wikipedia.org/wiki/Composition_ring
$endgroup$
– Chris Culter
6 hours ago




1




1




$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
5 hours ago




$begingroup$
@ChrisCulter Thanks, I guess that's the notion I was groping towards. I just posted a follow-up question: math.stackexchange.com/q/3075029/71829
$endgroup$
– Keshav Srinivasan
5 hours ago










1 Answer
1






active

oldest

votes


















11












$begingroup$

Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).



The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.






share|cite|improve this answer









$endgroup$













    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    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%2f3074950%2fwhat-algebraic-structure-does-the-set-of-endomorphisms-of-a-ring-have%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11












    $begingroup$

    Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).



    The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.






    share|cite|improve this answer









    $endgroup$


















      11












      $begingroup$

      Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).



      The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.






      share|cite|improve this answer









      $endgroup$
















        11












        11








        11





        $begingroup$

        Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).



        The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.






        share|cite|improve this answer









        $endgroup$



        Of your three operations, only $cdot$ is actually a valid operation. In general, if $f$ and $g$ are ring endomorphisms, then your $f+g$ and $f*g$ are not ring endomorphisms: $f+g$ will typically not preserve multiplication and $f*g$ typically will not preserve addition (or multiplication, if $R$ is not commutative).



        The natural structure that $operatorname{End}(R)$ has under the composition operation $cdot$ is a monoid: composition is associative and has an identity element (the identity map). This is not special to rings but is true of the set of endomorphisms of pretty much any kind of mathematical object.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        Eric WofseyEric Wofsey

        181k12209337




        181k12209337






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074950%2fwhat-algebraic-structure-does-the-set-of-endomorphisms-of-a-ring-have%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to make a Squid Proxy server?

            Image
            1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari
            Read more

            Is this a new Fibonacci Identity?

            Image
            6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w
            Read more

            Touch on Surface Book

            Image
            0 How can I make touch work on my Surface Book? About my Surface Book: Model: Intel Core i7-6600 U @ 2.60 GHz with 8 GB Ram and Nvidia GPU OS: Ubuntu 17.04 Latest, kernel 4.10.0-13-generic System outputs on Ubuntu: ~$ xinput list ⎡ Virtual core pointer id=2 [master pointer (3)] ⎜ ↳ Virtual core XTEST pointer id=4 [slave pointer (2)] ⎜ ↳ USB OPTICAL MOUSE id=8 [slave pointer (2)] ⎜ ↳ Microsoft Surface Keyboard Consumer Control id=10 [slave pointer (2)] ⎜ ↳ Microsoft Surface Keyboard Touchpad id=11 [slave pointer (2)] ⎣ Virtual core keyboard id=3 [master keyboard (2)] ↳ Virtual core XTEST keyboard id=5 [slave keyboard (3)]
            Read more