The group ring of a ring.

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











up vote
2
down vote

favorite












Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.



An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?










share|cite|improve this question

















  • 4




    Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
    – Lord Shark the Unknown
    2 hours ago







  • 1




    I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
    – rschwieb
    2 hours ago










  • @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
    – palio
    2 hours ago










  • @palio No, I did not.
    – Lord Shark the Unknown
    50 mins ago














up vote
2
down vote

favorite












Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.



An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?










share|cite|improve this question

















  • 4




    Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
    – Lord Shark the Unknown
    2 hours ago







  • 1




    I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
    – rschwieb
    2 hours ago










  • @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
    – palio
    2 hours ago










  • @palio No, I did not.
    – Lord Shark the Unknown
    50 mins ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.



An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?










share|cite|improve this question













Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$.
I want to understand this group ring $R[R]$.



An element $xin R[R]$ is written as a finite formal sum
$$x=r_1s_1+r_2s_2+cdots+r_ns_n$$
where both $r_i$ and $s_i$ are in $R$
but since the ring $R$ is closed under sum and multiplication, then it is clear that $xin R$. So can't we just say that the group ring $R[R]$ is equal to $R$?







abstract-algebra group-theory ring-theory group-rings






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









palio

4,02832460




4,02832460







  • 4




    Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
    – Lord Shark the Unknown
    2 hours ago







  • 1




    I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
    – rschwieb
    2 hours ago










  • @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
    – palio
    2 hours ago










  • @palio No, I did not.
    – Lord Shark the Unknown
    50 mins ago












  • 4




    Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
    – Lord Shark the Unknown
    2 hours ago







  • 1




    I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
    – rschwieb
    2 hours ago










  • @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
    – palio
    2 hours ago










  • @palio No, I did not.
    – Lord Shark the Unknown
    50 mins ago







4




4




Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
2 hours ago





Because it isn't. It's better to use different notation, e.g., a typical element would be $sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc.
– Lord Shark the Unknown
2 hours ago





1




1




I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago




I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1
– rschwieb
2 hours ago












@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago




@LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ?
– palio
2 hours ago












@palio No, I did not.
– Lord Shark the Unknown
50 mins ago




@palio No, I did not.
– Lord Shark the Unknown
50 mins ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.






share|cite|improve this answer




















  • in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
    – palio
    2 hours ago






  • 1




    $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
    – David Hill
    2 hours ago






  • 1




    here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
    – David Hill
    2 hours ago






  • 1




    I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
    – palio
    2 hours ago











  • yes, that is correct.
    – David Hill
    2 hours ago

















up vote
4
down vote













It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.



Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.



One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)






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%2f2961054%2fthe-group-ring-of-a-ring%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
    3
    down vote



    accepted










    It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.






    share|cite|improve this answer




















    • in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
      – palio
      2 hours ago






    • 1




      $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
      – David Hill
      2 hours ago






    • 1




      here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
      – David Hill
      2 hours ago






    • 1




      I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
      – palio
      2 hours ago











    • yes, that is correct.
      – David Hill
      2 hours ago














    up vote
    3
    down vote



    accepted










    It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.






    share|cite|improve this answer




















    • in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
      – palio
      2 hours ago






    • 1




      $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
      – David Hill
      2 hours ago






    • 1




      here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
      – David Hill
      2 hours ago






    • 1




      I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
      – palio
      2 hours ago











    • yes, that is correct.
      – David Hill
      2 hours ago












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.






    share|cite|improve this answer












    It would be better to write elements of $R[R]$ in the form $$x=sum_sin Rr_se^s.$$ Since $e^se^t=e^s+t$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    David Hill

    8,3821618




    8,3821618











    • in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
      – palio
      2 hours ago






    • 1




      $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
      – David Hill
      2 hours ago






    • 1




      here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
      – David Hill
      2 hours ago






    • 1




      I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
      – palio
      2 hours ago











    • yes, that is correct.
      – David Hill
      2 hours ago
















    • in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
      – palio
      2 hours ago






    • 1




      $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
      – David Hill
      2 hours ago






    • 1




      here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
      – David Hill
      2 hours ago






    • 1




      I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
      – palio
      2 hours ago











    • yes, that is correct.
      – David Hill
      2 hours ago















    in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
    – palio
    2 hours ago




    in $x=sum_sin Rr_se^s.$ what is $r$ and what is $e$ ? Thanks!
    – palio
    2 hours ago




    1




    1




    $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
    – David Hill
    2 hours ago




    $r_sin R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation).
    – David Hill
    2 hours ago




    1




    1




    here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
    – David Hill
    2 hours ago




    here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character
    – David Hill
    2 hours ago




    1




    1




    I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
    – palio
    2 hours ago





    I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ?
    – palio
    2 hours ago













    yes, that is correct.
    – David Hill
    2 hours ago




    yes, that is correct.
    – David Hill
    2 hours ago










    up vote
    4
    down vote













    It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.



    Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.



    One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)






    share|cite|improve this answer
























      up vote
      4
      down vote













      It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.



      Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.



      One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)






      share|cite|improve this answer






















        up vote
        4
        down vote










        up vote
        4
        down vote









        It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.



        Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.



        One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)






        share|cite|improve this answer












        It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.



        Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.



        One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        rschwieb

        102k1299235




        102k1299235



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2961054%2fthe-group-ring-of-a-ring%23new-answer', 'question_page');

            );

            Post as a guest













































































            Popular posts from this blog

            How to check contact read email or not when send email to Individual?

            Displaying single band from multi-band raster using QGIS

            How many registers does an x86_64 CPU actually have?