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
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
$endgroup$
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
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
$endgroup$
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
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
$endgroup$
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
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
edited Dec 30 '18 at 16:14
Shaun
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
edited Dec 30 '18 at 16:14
Shaun
10.7k113687
edited Dec 30 '18 at 16:14
Shaun
10.7k113687
edited Dec 30 '18 at 16:14
Shaun
10.7k113687
10.7k113687
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
asked Dec 30 '18 at 16:07
yoyosteinyoyostein
8,175104074
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
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 |
Your Answer
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056964%2faugmentation-ideal-of-group-ring-of-free-group%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
$begingroup$
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 |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056964%2faugmentation-ideal-of-group-ring-of-free-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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