Augmentation Ideal of Group Ring of Free Group.
$begingroup$
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
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
$endgroup$
add a comment |
$begingroup$
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
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
$endgroup$
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
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
edited Dec 30 '18 at 16:14
Shaun
11.1k113688
11.1k113688
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,202104074
8,202104074
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $I$.
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
$endgroup$
add a comment |
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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 $I$.
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
$endgroup$
add a comment |
$begingroup$
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 $I$.
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
$endgroup$
add a comment |
$begingroup$
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 $I$.
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
$endgroup$
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 $I$.
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
answered Dec 31 '18 at 14:07
Pierre-Guy PlamondonPierre-Guy Plamondon
8,91511739
8,91511739
add a comment |
add a comment |
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$begingroup$
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