For a Cohen-Macaulay local ring grade and height are same
In Matsumura's book ' Commutative ring theory' Theorem 17.4, page 135 its been proved that in a Noetheriam local ring $(A,mathfrak m)$ for any proper ideal $I$, grade $I=$ ht$I$, where grade of an ideal is defined to be the maximal length of $A$ regular sequence contained inside $I$ and height of an ideal is defined to be the infimum of of ht$p$ where $p$ is a minimal prime ideal containing $I$. In general grade $leq$ ht. For the other inequality let ht$I=r$. Then the book says that we can find $a_1,ldots,a_r$ such that ht$(a_1,ldots,a_i)=i$ for $1leq ileq r$. I can't see how can we get this.
I have an argument which directly says that we can find $a_1,ldots ,a_r$ inside $I$ which is a regular sequence, where ht$I=r$. But my argument does not use that $A$ is Cohen-Macaulay, hence must be flawed. But I cannot find where it is wrong.
Let $rgeq 1$
Let $p_1,ldots ,p_k$ are minimal prime ideals of $A$. Then $Insubseteq p_i$ for each i and hence by prime avoidance $Insubseteqcup p_i$. Therefore $exists a_1in I $ such that $a_1$ is not a zero-divisor and hence regular.
Now let $rgeq 2$
We consider $A'=A/(a_1)$. Let $overline{q_1},ldots overline{q_n}$ are minimal prime ideals of $A'$. Then each $q_i$ is a minimal prime ideal containing $(a_1)$ and hence ht$q_i=1$. Therefore $overline Insubseteqcupoverline{q_i}$ as ht$Igeq 2$. Then we can choose $overline{a_2}inoverline{I}$ such that it is a zero-divisor in $A'$. Hence $a_1,a_2$ is an $A$ sequence such that ht$(a_1,a_2)=2$. We continue this way...
Thank you in advance.
commutative-algebra cohen-macaulay
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
add a comment |
In Matsumura's book ' Commutative ring theory' Theorem 17.4, page 135 its been proved that in a Noetheriam local ring $(A,mathfrak m)$ for any proper ideal $I$, grade $I=$ ht$I$, where grade of an ideal is defined to be the maximal length of $A$ regular sequence contained inside $I$ and height of an ideal is defined to be the infimum of of ht$p$ where $p$ is a minimal prime ideal containing $I$. In general grade $leq$ ht. For the other inequality let ht$I=r$. Then the book says that we can find $a_1,ldots,a_r$ such that ht$(a_1,ldots,a_i)=i$ for $1leq ileq r$. I can't see how can we get this.
I have an argument which directly says that we can find $a_1,ldots ,a_r$ inside $I$ which is a regular sequence, where ht$I=r$. But my argument does not use that $A$ is Cohen-Macaulay, hence must be flawed. But I cannot find where it is wrong.
Let $rgeq 1$
Let $p_1,ldots ,p_k$ are minimal prime ideals of $A$. Then $Insubseteq p_i$ for each i and hence by prime avoidance $Insubseteqcup p_i$. Therefore $exists a_1in I $ such that $a_1$ is not a zero-divisor and hence regular.
Now let $rgeq 2$
We consider $A'=A/(a_1)$. Let $overline{q_1},ldots overline{q_n}$ are minimal prime ideals of $A'$. Then each $q_i$ is a minimal prime ideal containing $(a_1)$ and hence ht$q_i=1$. Therefore $overline Insubseteqcupoverline{q_i}$ as ht$Igeq 2$. Then we can choose $overline{a_2}inoverline{I}$ such that it is a zero-divisor in $A'$. Hence $a_1,a_2$ is an $A$ sequence such that ht$(a_1,a_2)=2$. We continue this way...
Thank you in advance.
commutative-algebra cohen-macaulay
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11
add a comment |
In Matsumura's book ' Commutative ring theory' Theorem 17.4, page 135 its been proved that in a Noetheriam local ring $(A,mathfrak m)$ for any proper ideal $I$, grade $I=$ ht$I$, where grade of an ideal is defined to be the maximal length of $A$ regular sequence contained inside $I$ and height of an ideal is defined to be the infimum of of ht$p$ where $p$ is a minimal prime ideal containing $I$. In general grade $leq$ ht. For the other inequality let ht$I=r$. Then the book says that we can find $a_1,ldots,a_r$ such that ht$(a_1,ldots,a_i)=i$ for $1leq ileq r$. I can't see how can we get this.
I have an argument which directly says that we can find $a_1,ldots ,a_r$ inside $I$ which is a regular sequence, where ht$I=r$. But my argument does not use that $A$ is Cohen-Macaulay, hence must be flawed. But I cannot find where it is wrong.
Let $rgeq 1$
Let $p_1,ldots ,p_k$ are minimal prime ideals of $A$. Then $Insubseteq p_i$ for each i and hence by prime avoidance $Insubseteqcup p_i$. Therefore $exists a_1in I $ such that $a_1$ is not a zero-divisor and hence regular.
Now let $rgeq 2$
We consider $A'=A/(a_1)$. Let $overline{q_1},ldots overline{q_n}$ are minimal prime ideals of $A'$. Then each $q_i$ is a minimal prime ideal containing $(a_1)$ and hence ht$q_i=1$. Therefore $overline Insubseteqcupoverline{q_i}$ as ht$Igeq 2$. Then we can choose $overline{a_2}inoverline{I}$ such that it is a zero-divisor in $A'$. Hence $a_1,a_2$ is an $A$ sequence such that ht$(a_1,a_2)=2$. We continue this way...
Thank you in advance.
commutative-algebra cohen-macaulay
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
In Matsumura's book ' Commutative ring theory' Theorem 17.4, page 135 its been proved that in a Noetheriam local ring $(A,mathfrak m)$ for any proper ideal $I$, grade $I=$ ht$I$, where grade of an ideal is defined to be the maximal length of $A$ regular sequence contained inside $I$ and height of an ideal is defined to be the infimum of of ht$p$ where $p$ is a minimal prime ideal containing $I$. In general grade $leq$ ht. For the other inequality let ht$I=r$. Then the book says that we can find $a_1,ldots,a_r$ such that ht$(a_1,ldots,a_i)=i$ for $1leq ileq r$. I can't see how can we get this.
I have an argument which directly says that we can find $a_1,ldots ,a_r$ inside $I$ which is a regular sequence, where ht$I=r$. But my argument does not use that $A$ is Cohen-Macaulay, hence must be flawed. But I cannot find where it is wrong.
Let $rgeq 1$
Let $p_1,ldots ,p_k$ are minimal prime ideals of $A$. Then $Insubseteq p_i$ for each i and hence by prime avoidance $Insubseteqcup p_i$. Therefore $exists a_1in I $ such that $a_1$ is not a zero-divisor and hence regular.
Now let $rgeq 2$
We consider $A'=A/(a_1)$. Let $overline{q_1},ldots overline{q_n}$ are minimal prime ideals of $A'$. Then each $q_i$ is a minimal prime ideal containing $(a_1)$ and hence ht$q_i=1$. Therefore $overline Insubseteqcupoverline{q_i}$ as ht$Igeq 2$. Then we can choose $overline{a_2}inoverline{I}$ such that it is a zero-divisor in $A'$. Hence $a_1,a_2$ is an $A$ sequence such that ht$(a_1,a_2)=2$. We continue this way...
Thank you in advance.
commutative-algebra cohen-macaulay
commutative-algebra cohen-macaulay
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
asked Nov 29 '18 at 14:37
Rtk427Rtk427
245
245
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11
add a comment |
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11
add a comment |
0
active
oldest
votes
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%2f3018704%2ffor-a-cohen-macaulay-local-ring-grade-and-height-are-same%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3018704%2ffor-a-cohen-macaulay-local-ring-grade-and-height-are-same%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
In order to find such a sequence, you need to avoid the set of minimal primes. To have a regular sequence, one needs to avoid the set of associated primes. In general, these two sets are different. Cohen’s characterization was that if $I$ is of height $h$ and generated by $h$ elements, then all the assoaiated primes of I is minimal and of height $h$.
– Youngsu
Nov 30 '18 at 4:18
@Youngsu Thank you. I now understand that I was only showing that each element was not in the minimal primes but zero divisors are the union of all associated primes and there might be embedded associated primes.
– Rtk427
Dec 2 '18 at 15:11