Augmentation Ideal of Group Ring of Free Group.












2












$begingroup$


Let $RG$ be a group ring, where $R$ is a commutative ring. Let $I=kerepsilon$ be the augmentation ideal.



It is known that any element $alphain I$ can be written in the form $$alpha=sum_{gin G}a_g(g-1),$$ where $a_gin R$.



Suppose further that $G$ is a free group with generators $g_1,g_2,dots,g_n$. With this extra information, how can we improve on the previous result?



For example is it true that $alpha$ can be written in the form $$alpha=sum_{i=1}^nleft( a_g(g_i-1)+b_g(g_i^{-1}-1)right)?$$



Thanks.






abstract-algebra group-theory ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 30 '18 at 17:01
















2












$begingroup$


Let $RG$ be a group ring, where $R$ is a commutative ring. Let $I=kerepsilon$ be the augmentation ideal.



It is known that any element $alphain I$ can be written in the form $$alpha=sum_{gin G}a_g(g-1),$$ where $a_gin R$.



Suppose further that $G$ is a free group with generators $g_1,g_2,dots,g_n$. With this extra information, how can we improve on the previous result?



For example is it true that $alpha$ can be written in the form $$alpha=sum_{i=1}^nleft( a_g(g_i-1)+b_g(g_i^{-1}-1)right)?$$



Thanks.






abstract-algebra group-theory ring-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 30 '18 at 17:01














2












2








2


1



$begingroup$


Let $RG$ be a group ring, where $R$ is a commutative ring. Let $I=kerepsilon$ be the augmentation ideal.



It is known that any element $alphain I$ can be written in the form $$alpha=sum_{gin G}a_g(g-1),$$ where $a_gin R$.



Suppose further that $G$ is a free group with generators $g_1,g_2,dots,g_n$. With this extra information, how can we improve on the previous result?



For example is it true that $alpha$ can be written in the form $$alpha=sum_{i=1}^nleft( a_g(g_i-1)+b_g(g_i^{-1}-1)right)?$$



Thanks.






abstract-algebra group-theory ring-theory







share|cite|improve this question











$endgroup$




Let $RG$ be a group ring, where $R$ is a commutative ring. Let $I=kerepsilon$ be the augmentation ideal.



It is known that any element $alphain I$ can be written in the form $$alpha=sum_{gin G}a_g(g-1),$$ where $a_gin R$.



Suppose further that $G$ is a free group with generators $g_1,g_2,dots,g_n$. With this extra information, how can we improve on the previous result?



For example is it true that $alpha$ can be written in the form $$alpha=sum_{i=1}^nleft( a_g(g_i-1)+b_g(g_i^{-1}-1)right)?$$



Thanks.






abstract-algebra group-theory ring-theory




abstract-algebra group-theory ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 16:14









Shaun

10.7k113687




10.7k113687










asked Dec 30 '18 at 16:07









yoyosteinyoyostein

8,175104074




8,175104074








  • 3




    $begingroup$
    It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 30 '18 at 17:01














  • 3




    $begingroup$
    It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 30 '18 at 17:01








3




3




$begingroup$
It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
$endgroup$
– Pierre-Guy Plamondon
Dec 30 '18 at 17:01




$begingroup$
It seems the element $g_1^2 - g_1 in I$ is not of the required form. Or am I missing something?
$endgroup$
– Pierre-Guy Plamondon
Dec 30 '18 at 17:01










1 Answer
1






active

oldest

votes


















2












$begingroup$

Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $mathcal{B} = { g-1 | gin G}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.




Proof that $mathcal{B} = { g-1 | gin G}$ is a basis of $I$ as an $R$-module.



To see that $mathcal{B}$ generates $I$, let $x=sum_g b_g g in I$. Since $xin I$, we have that $b_1 = -sum_{gneq 1} b_g$. Therefore $$ x = sum_{gneq 1} b_g(g-1).$$
Thus $mathcal{B}$ generates $I$.



To see that the elements of $mathcal{B}$ are linearly independent, assume that $$ sum_{gneq 1} a_g(g-1) =0. $$ Written in the basis ${g | gin G}$ of $RG$, this element is $$(-sum_{gneq 1}a_g) 1 + sum_{gneq 1}a_g g.$$ Thus, $a_g=0$ for all $gin G setminus {1}$.



Therefore, $mathcal{B}$ is a basis of the $R$-module $I$.






share|cite|improve this answer









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    $begingroup$

    Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $mathcal{B} = { g-1 | gin G}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.




    Proof that $mathcal{B} = { g-1 | gin G}$ is a basis of $I$ as an $R$-module.



    To see that $mathcal{B}$ generates $I$, let $x=sum_g b_g g in I$. Since $xin I$, we have that $b_1 = -sum_{gneq 1} b_g$. Therefore $$ x = sum_{gneq 1} b_g(g-1).$$
    Thus $mathcal{B}$ generates $I$.



    To see that the elements of $mathcal{B}$ are linearly independent, assume that $$ sum_{gneq 1} a_g(g-1) =0. $$ Written in the basis ${g | gin G}$ of $RG$, this element is $$(-sum_{gneq 1}a_g) 1 + sum_{gneq 1}a_g g.$$ Thus, $a_g=0$ for all $gin G setminus {1}$.



    Therefore, $mathcal{B}$ is a basis of the $R$-module $I$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $mathcal{B} = { g-1 | gin G}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.




      Proof that $mathcal{B} = { g-1 | gin G}$ is a basis of $I$ as an $R$-module.



      To see that $mathcal{B}$ generates $I$, let $x=sum_g b_g g in I$. Since $xin I$, we have that $b_1 = -sum_{gneq 1} b_g$. Therefore $$ x = sum_{gneq 1} b_g(g-1).$$
      Thus $mathcal{B}$ generates $I$.



      To see that the elements of $mathcal{B}$ are linearly independent, assume that $$ sum_{gneq 1} a_g(g-1) =0. $$ Written in the basis ${g | gin G}$ of $RG$, this element is $$(-sum_{gneq 1}a_g) 1 + sum_{gneq 1}a_g g.$$ Thus, $a_g=0$ for all $gin G setminus {1}$.



      Therefore, $mathcal{B}$ is a basis of the $R$-module $I$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $mathcal{B} = { g-1 | gin G}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.




        Proof that $mathcal{B} = { g-1 | gin G}$ is a basis of $I$ as an $R$-module.



        To see that $mathcal{B}$ generates $I$, let $x=sum_g b_g g in I$. Since $xin I$, we have that $b_1 = -sum_{gneq 1} b_g$. Therefore $$ x = sum_{gneq 1} b_g(g-1).$$
        Thus $mathcal{B}$ generates $I$.



        To see that the elements of $mathcal{B}$ are linearly independent, assume that $$ sum_{gneq 1} a_g(g-1) =0. $$ Written in the basis ${g | gin G}$ of $RG$, this element is $$(-sum_{gneq 1}a_g) 1 + sum_{gneq 1}a_g g.$$ Thus, $a_g=0$ for all $gin G setminus {1}$.



        Therefore, $mathcal{B}$ is a basis of the $R$-module $I$.






        share|cite|improve this answer









        $endgroup$



        Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $mathcal{B} = { g-1 | gin G}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.




        Proof that $mathcal{B} = { g-1 | gin G}$ is a basis of $I$ as an $R$-module.



        To see that $mathcal{B}$ generates $I$, let $x=sum_g b_g g in I$. Since $xin I$, we have that $b_1 = -sum_{gneq 1} b_g$. Therefore $$ x = sum_{gneq 1} b_g(g-1).$$
        Thus $mathcal{B}$ generates $I$.



        To see that the elements of $mathcal{B}$ are linearly independent, assume that $$ sum_{gneq 1} a_g(g-1) =0. $$ Written in the basis ${g | gin G}$ of $RG$, this element is $$(-sum_{gneq 1}a_g) 1 + sum_{gneq 1}a_g g.$$ Thus, $a_g=0$ for all $gin G setminus {1}$.



        Therefore, $mathcal{B}$ is a basis of the $R$-module $I$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 14:07









        Pierre-Guy PlamondonPierre-Guy Plamondon

        8,91511739




        8,91511739






























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