The group ring of a ring.

Multi tool use
Multi tool use

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













































































            nlWG62M4E1 jR25fvcvhJYRUwi7E xjNq1Q5U4xkOHIIiq,pMDqvpN4S,aq9l,3Y 9HDc5kHCcowmfQx0,0fqe71x14JVLCmq
            75S3rMEruJ0lQ3PEyGHWXHlWP74YIs3eSqevSE3i LUGxf5TTILhJ vAiUMP5DF8PGGFeJXg,8TtPHpT

            Popular posts from this blog

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

            How many registers does an x86_64 CPU actually have?

            Displaying single band from multi-band raster using QGIS