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 at 14:37
Rtk427
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 at 14:37
Rtk427
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 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 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 at 14:37
Rtk427
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 at 14:37
Rtk427
245
asked Nov 29 at 14:37
Rtk427
245
asked Nov 29 at 14:37
Rtk427
245
asked Nov 29 at 14:37
Rtk427
245
asked Nov 29 at 14:37
Rtk427
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 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 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 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 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 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 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 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 at 15:11
add a comment |
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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 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 at 15:11