What is the relationship between primary decomposition and irreducible decomposition?
$begingroup$
Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module or Noetherian $A$-module. In my commutative algebra class, I was given the following theorems.
The first was about the existence of irreducible decomposition of submodules of $M$ :
Every submodule $N$ of $M$ has an irreducible decomposition $N = bigcap_{i=1}^r N_i$ such that $N_i$ is irreducible.
The irreducible decomposition was called a naive decomposition by the teacher since it did not work well on vector spaces.
Then, I was given the primary decomposition by the following 3 theorems :
1) An irreducible submodule of $M$ is primary.
2) If $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary is an irredundant decomposition, then, $mathbf{Ass}_A(M/N) = {mathfrak{p}_1,dots,mathfrak{p}_r}$.
3) For any shortest primary, irredundant decomposition $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary, if $mathfrak{p}_i$ is minimal in $mathbf{Ass}_A(M/N)$, then, $N_i = phi^{-1}(N_{mathfrak{p}_i})$ where $phi : M rightarrow M_{mathfrak{p}_i}$ is the canonical morphism.
In addtion, 1) implies the existence of irredundant, shortest primary decomposition.
So, my question is what exactly is the relationship between primary decomposition and irreducible decomposition ?
It seemed from 1) that such two decompositions are the same.
Can anyone help me ? Thank you.
For your information, the following are some related definitions.
A proper submodule $N$ of $M$ is irreducible if $N = N_1 cap N_2 implies N = N_1 lor N = N_2$ for any submodules $N_1,N_2$ of $M$.
$N = bigcap_{i=1}^r N_i$ is an irreducible decomposition of $N$ if every $N_i$ is irreducible.
A submodule $N$ of $M$ is $mathfrak{p}$-primary if $mathfrak{p}$ is the only associated prime ideal of $M/N$ over $A$, that is, $mathbf{Ass}_A(M/N) = {mathfrak{p}}$.
The decomposition $N = bigcap_{i=1}^r N_i$ is a primary decomposition if every $N_i$ is primary.
The decomposition $N = bigcap_{i=1}^r N_i$ is shortest primary if $N_i$ are $mathfrak{p}_i$-primary and $forall i neq j, mathfrak{p}_i neq mathfrak{p}_j$.
The decomposition $N = bigcap_{i=1}^r N_i$ is irredundant if the intersection with any $N_i$ omitted does not equal $N$.
Again, thank you for your help. This question may be a little too long, sorry to take your time.
commutative-algebra
$endgroup$
add a comment |
$begingroup$
Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module or Noetherian $A$-module. In my commutative algebra class, I was given the following theorems.
The first was about the existence of irreducible decomposition of submodules of $M$ :
Every submodule $N$ of $M$ has an irreducible decomposition $N = bigcap_{i=1}^r N_i$ such that $N_i$ is irreducible.
The irreducible decomposition was called a naive decomposition by the teacher since it did not work well on vector spaces.
Then, I was given the primary decomposition by the following 3 theorems :
1) An irreducible submodule of $M$ is primary.
2) If $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary is an irredundant decomposition, then, $mathbf{Ass}_A(M/N) = {mathfrak{p}_1,dots,mathfrak{p}_r}$.
3) For any shortest primary, irredundant decomposition $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary, if $mathfrak{p}_i$ is minimal in $mathbf{Ass}_A(M/N)$, then, $N_i = phi^{-1}(N_{mathfrak{p}_i})$ where $phi : M rightarrow M_{mathfrak{p}_i}$ is the canonical morphism.
In addtion, 1) implies the existence of irredundant, shortest primary decomposition.
So, my question is what exactly is the relationship between primary decomposition and irreducible decomposition ?
It seemed from 1) that such two decompositions are the same.
Can anyone help me ? Thank you.
For your information, the following are some related definitions.
A proper submodule $N$ of $M$ is irreducible if $N = N_1 cap N_2 implies N = N_1 lor N = N_2$ for any submodules $N_1,N_2$ of $M$.
$N = bigcap_{i=1}^r N_i$ is an irreducible decomposition of $N$ if every $N_i$ is irreducible.
A submodule $N$ of $M$ is $mathfrak{p}$-primary if $mathfrak{p}$ is the only associated prime ideal of $M/N$ over $A$, that is, $mathbf{Ass}_A(M/N) = {mathfrak{p}}$.
The decomposition $N = bigcap_{i=1}^r N_i$ is a primary decomposition if every $N_i$ is primary.
The decomposition $N = bigcap_{i=1}^r N_i$ is shortest primary if $N_i$ are $mathfrak{p}_i$-primary and $forall i neq j, mathfrak{p}_i neq mathfrak{p}_j$.
The decomposition $N = bigcap_{i=1}^r N_i$ is irredundant if the intersection with any $N_i$ omitted does not equal $N$.
Again, thank you for your help. This question may be a little too long, sorry to take your time.
commutative-algebra
$endgroup$
add a comment |
$begingroup$
Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module or Noetherian $A$-module. In my commutative algebra class, I was given the following theorems.
The first was about the existence of irreducible decomposition of submodules of $M$ :
Every submodule $N$ of $M$ has an irreducible decomposition $N = bigcap_{i=1}^r N_i$ such that $N_i$ is irreducible.
The irreducible decomposition was called a naive decomposition by the teacher since it did not work well on vector spaces.
Then, I was given the primary decomposition by the following 3 theorems :
1) An irreducible submodule of $M$ is primary.
2) If $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary is an irredundant decomposition, then, $mathbf{Ass}_A(M/N) = {mathfrak{p}_1,dots,mathfrak{p}_r}$.
3) For any shortest primary, irredundant decomposition $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary, if $mathfrak{p}_i$ is minimal in $mathbf{Ass}_A(M/N)$, then, $N_i = phi^{-1}(N_{mathfrak{p}_i})$ where $phi : M rightarrow M_{mathfrak{p}_i}$ is the canonical morphism.
In addtion, 1) implies the existence of irredundant, shortest primary decomposition.
So, my question is what exactly is the relationship between primary decomposition and irreducible decomposition ?
It seemed from 1) that such two decompositions are the same.
Can anyone help me ? Thank you.
For your information, the following are some related definitions.
A proper submodule $N$ of $M$ is irreducible if $N = N_1 cap N_2 implies N = N_1 lor N = N_2$ for any submodules $N_1,N_2$ of $M$.
$N = bigcap_{i=1}^r N_i$ is an irreducible decomposition of $N$ if every $N_i$ is irreducible.
A submodule $N$ of $M$ is $mathfrak{p}$-primary if $mathfrak{p}$ is the only associated prime ideal of $M/N$ over $A$, that is, $mathbf{Ass}_A(M/N) = {mathfrak{p}}$.
The decomposition $N = bigcap_{i=1}^r N_i$ is a primary decomposition if every $N_i$ is primary.
The decomposition $N = bigcap_{i=1}^r N_i$ is shortest primary if $N_i$ are $mathfrak{p}_i$-primary and $forall i neq j, mathfrak{p}_i neq mathfrak{p}_j$.
The decomposition $N = bigcap_{i=1}^r N_i$ is irredundant if the intersection with any $N_i$ omitted does not equal $N$.
Again, thank you for your help. This question may be a little too long, sorry to take your time.
commutative-algebra
$endgroup$
Let $A$ be a Noetherian ring and $M$ a finitely generated $A$-module or Noetherian $A$-module. In my commutative algebra class, I was given the following theorems.
The first was about the existence of irreducible decomposition of submodules of $M$ :
Every submodule $N$ of $M$ has an irreducible decomposition $N = bigcap_{i=1}^r N_i$ such that $N_i$ is irreducible.
The irreducible decomposition was called a naive decomposition by the teacher since it did not work well on vector spaces.
Then, I was given the primary decomposition by the following 3 theorems :
1) An irreducible submodule of $M$ is primary.
2) If $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary is an irredundant decomposition, then, $mathbf{Ass}_A(M/N) = {mathfrak{p}_1,dots,mathfrak{p}_r}$.
3) For any shortest primary, irredundant decomposition $N = bigcap_{i=1}^r N_i$ with $N_i$ being $mathfrak{p}_i$-primary, if $mathfrak{p}_i$ is minimal in $mathbf{Ass}_A(M/N)$, then, $N_i = phi^{-1}(N_{mathfrak{p}_i})$ where $phi : M rightarrow M_{mathfrak{p}_i}$ is the canonical morphism.
In addtion, 1) implies the existence of irredundant, shortest primary decomposition.
So, my question is what exactly is the relationship between primary decomposition and irreducible decomposition ?
It seemed from 1) that such two decompositions are the same.
Can anyone help me ? Thank you.
For your information, the following are some related definitions.
A proper submodule $N$ of $M$ is irreducible if $N = N_1 cap N_2 implies N = N_1 lor N = N_2$ for any submodules $N_1,N_2$ of $M$.
$N = bigcap_{i=1}^r N_i$ is an irreducible decomposition of $N$ if every $N_i$ is irreducible.
A submodule $N$ of $M$ is $mathfrak{p}$-primary if $mathfrak{p}$ is the only associated prime ideal of $M/N$ over $A$, that is, $mathbf{Ass}_A(M/N) = {mathfrak{p}}$.
The decomposition $N = bigcap_{i=1}^r N_i$ is a primary decomposition if every $N_i$ is primary.
The decomposition $N = bigcap_{i=1}^r N_i$ is shortest primary if $N_i$ are $mathfrak{p}_i$-primary and $forall i neq j, mathfrak{p}_i neq mathfrak{p}_j$.
The decomposition $N = bigcap_{i=1}^r N_i$ is irredundant if the intersection with any $N_i$ omitted does not equal $N$.
Again, thank you for your help. This question may be a little too long, sorry to take your time.
commutative-algebra
commutative-algebra
asked Dec 14 '18 at 13:34
Yu NingYu Ning
665
665
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the context of algebraic geometry, a decomposition $I=bigcap I_i$ corresponds to a decomposition of $V=bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.
For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)cap(x,y^2).$$
Note both are $(x,y)$-primary, so it is not a shortest decomposition.
$endgroup$
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%2f3039361%2fwhat-is-the-relationship-between-primary-decomposition-and-irreducible-decomposi%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
$begingroup$
In the context of algebraic geometry, a decomposition $I=bigcap I_i$ corresponds to a decomposition of $V=bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.
For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)cap(x,y^2).$$
Note both are $(x,y)$-primary, so it is not a shortest decomposition.
$endgroup$
add a comment |
$begingroup$
In the context of algebraic geometry, a decomposition $I=bigcap I_i$ corresponds to a decomposition of $V=bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.
For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)cap(x,y^2).$$
Note both are $(x,y)$-primary, so it is not a shortest decomposition.
$endgroup$
add a comment |
$begingroup$
In the context of algebraic geometry, a decomposition $I=bigcap I_i$ corresponds to a decomposition of $V=bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.
For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)cap(x,y^2).$$
Note both are $(x,y)$-primary, so it is not a shortest decomposition.
$endgroup$
In the context of algebraic geometry, a decomposition $I=bigcap I_i$ corresponds to a decomposition of $V=bigcup V_i$. So irreducible decomposition simply decomposes a variety into unions of irreducible components. This should be unique in the sense there are only finite number of such components, and this only catches the information of the minimal primes for non-radicals. So primary decomposition should carry slightly more information.
For your question, if $R$ is noetherian, $M$ is f.g. $R$ module, then every irreducible decomposition must also be a primary decomposition, but the converse is not true. For example, $I=(x,y)^2subset R=k[x,y]$ itself is a primary decomposition, with $I$ being $(x,y)$-primary. This is, however, not an irreducible decomposition. We can write $$I=(x^2,y)cap(x,y^2).$$
Note both are $(x,y)$-primary, so it is not a shortest decomposition.
answered Dec 15 '18 at 1:51
Matt KellerMatt Keller
213
213
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%2f3039361%2fwhat-is-the-relationship-between-primary-decomposition-and-irreducible-decomposi%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