About Loewy length of syzygy
Suppose $Lambda$ is an Artin algebra with $DeclareMathOperator{rad}{rad}rad^3Lambda=0 $, and $ M $ any finite $Lambda$-module with projective dimension finite. Proof that $Omega M $ has Loewy length at most 2.
This question is from a paper: On the finitistic global dimension conjecture for Artin algebras - Igusa and Todorov. It is a corollary.
My idea is: we can guarantee that $rad^3M=0$ because $rad M=MradLambda$, and
take the projective cover of $M$:
$$Omega M rightarrow P rightarrow M rightarrow 0$$
$ΩM=ker(f)$ is a submodule of $rad(P)$ (where $P$ is the first term of projective cover). And thinking use exactness of this for make something.
abstract-algebra modules representation-theory
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
add a comment |
Suppose $Lambda$ is an Artin algebra with $DeclareMathOperator{rad}{rad}rad^3Lambda=0 $, and $ M $ any finite $Lambda$-module with projective dimension finite. Proof that $Omega M $ has Loewy length at most 2.
This question is from a paper: On the finitistic global dimension conjecture for Artin algebras - Igusa and Todorov. It is a corollary.
My idea is: we can guarantee that $rad^3M=0$ because $rad M=MradLambda$, and
take the projective cover of $M$:
$$Omega M rightarrow P rightarrow M rightarrow 0$$
$ΩM=ker(f)$ is a submodule of $rad(P)$ (where $P$ is the first term of projective cover). And thinking use exactness of this for make something.
abstract-algebra modules representation-theory
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
1
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29
add a comment |
Suppose $Lambda$ is an Artin algebra with $DeclareMathOperator{rad}{rad}rad^3Lambda=0 $, and $ M $ any finite $Lambda$-module with projective dimension finite. Proof that $Omega M $ has Loewy length at most 2.
This question is from a paper: On the finitistic global dimension conjecture for Artin algebras - Igusa and Todorov. It is a corollary.
My idea is: we can guarantee that $rad^3M=0$ because $rad M=MradLambda$, and
take the projective cover of $M$:
$$Omega M rightarrow P rightarrow M rightarrow 0$$
$ΩM=ker(f)$ is a submodule of $rad(P)$ (where $P$ is the first term of projective cover). And thinking use exactness of this for make something.
abstract-algebra modules representation-theory
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
Suppose $Lambda$ is an Artin algebra with $DeclareMathOperator{rad}{rad}rad^3Lambda=0 $, and $ M $ any finite $Lambda$-module with projective dimension finite. Proof that $Omega M $ has Loewy length at most 2.
This question is from a paper: On the finitistic global dimension conjecture for Artin algebras - Igusa and Todorov. It is a corollary.
My idea is: we can guarantee that $rad^3M=0$ because $rad M=MradLambda$, and
take the projective cover of $M$:
$$Omega M rightarrow P rightarrow M rightarrow 0$$
$ΩM=ker(f)$ is a submodule of $rad(P)$ (where $P$ is the first term of projective cover). And thinking use exactness of this for make something.
abstract-algebra modules representation-theory
abstract-algebra modules representation-theory
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
edited Nov 28 '18 at 18:52
Xander Henderson
14.1k103554
14.1k103554
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
asked Nov 27 '18 at 18:15
Júlio César M. Marques
565
565
1
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29
add a comment |
1
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29
1
1
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
I discuss this question to my professor today, the answer is simple.
Take $M$ a f.g. $Lambda$-module with $pdim M < infty$.
Since $MradLambda = radM$, and we have $Omega M subseteq rad P$ (and is a submodule), then $$rad^2Omega M = rad(radOmega M)=radOmega M radLambda=Omega M rad^2Lambda leq rad P rad^2Lambda=Prad^3Lambda=0$$
Then, the Loewy length of $Omega M$ is less or equal 2.
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
add a comment |
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',
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%2f3016122%2fabout-loewy-length-of-syzygy%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
I discuss this question to my professor today, the answer is simple.
Take $M$ a f.g. $Lambda$-module with $pdim M < infty$.
Since $MradLambda = radM$, and we have $Omega M subseteq rad P$ (and is a submodule), then $$rad^2Omega M = rad(radOmega M)=radOmega M radLambda=Omega M rad^2Lambda leq rad P rad^2Lambda=Prad^3Lambda=0$$
Then, the Loewy length of $Omega M$ is less or equal 2.
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
add a comment |
I discuss this question to my professor today, the answer is simple.
Take $M$ a f.g. $Lambda$-module with $pdim M < infty$.
Since $MradLambda = radM$, and we have $Omega M subseteq rad P$ (and is a submodule), then $$rad^2Omega M = rad(radOmega M)=radOmega M radLambda=Omega M rad^2Lambda leq rad P rad^2Lambda=Prad^3Lambda=0$$
Then, the Loewy length of $Omega M$ is less or equal 2.
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
add a comment |
I discuss this question to my professor today, the answer is simple.
Take $M$ a f.g. $Lambda$-module with $pdim M < infty$.
Since $MradLambda = radM$, and we have $Omega M subseteq rad P$ (and is a submodule), then $$rad^2Omega M = rad(radOmega M)=radOmega M radLambda=Omega M rad^2Lambda leq rad P rad^2Lambda=Prad^3Lambda=0$$
Then, the Loewy length of $Omega M$ is less or equal 2.
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
I discuss this question to my professor today, the answer is simple.
Take $M$ a f.g. $Lambda$-module with $pdim M < infty$.
Since $MradLambda = radM$, and we have $Omega M subseteq rad P$ (and is a submodule), then $$rad^2Omega M = rad(radOmega M)=radOmega M radLambda=Omega M rad^2Lambda leq rad P rad^2Lambda=Prad^3Lambda=0$$
Then, the Loewy length of $Omega M$ is less or equal 2.
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
answered Nov 28 '18 at 23:46
Júlio César M. Marques
565
565
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016122%2fabout-loewy-length-of-syzygy%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
1
Any own ideas? Where did you get this question from?
– A. Pongrácz
Nov 27 '18 at 18:20
This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something.
– Júlio César M. Marques
Nov 27 '18 at 18:29