Mixing
Exercise 4.4.1 in Weibel's 'An Introduction to Homological Algebra'.
$begingroup$
I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?
I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.
commutative-algebra homological-algebra local-rings regular-rings
edited Jan 28 at 21:39
user26857
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
$endgroup$
add a comment |
$begingroup$
I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?
I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.
commutative-algebra homological-algebra local-rings regular-rings
edited Jan 28 at 21:39
user26857
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
$endgroup$
add a comment |
$begingroup$
I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?
I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.
commutative-algebra homological-algebra local-rings regular-rings
edited Jan 28 at 21:39
user26857
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
$endgroup$
I can solve this question on the assumption that the $x_i$s are not zero-divisors since $dim(R/(x)) = dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?
I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $mathfrak{m}$, but I'm not sure if this leads to anything.
commutative-algebra homological-algebra local-rings regular-rings
commutative-algebra homological-algebra local-rings regular-rings
edited Jan 28 at 21:39
user26857
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
edited Jan 28 at 21:39
user26857
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
edited Jan 28 at 21:39
user26857
39.4k124183
edited Jan 28 at 21:39
user26857
39.4k124183
edited Jan 28 at 21:39
user26857
39.4k124183
39.4k124183
asked Jan 28 at 13:54
Jehu314Jehu314
1569
asked Jan 28 at 13:54
Jehu314Jehu314
1569
asked Jan 28 at 13:54
Jehu314Jehu314
1569
1569
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.
answered Jan 28 at 14:06
BenBen
4,313617
$endgroup$
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.
answered Jan 28 at 14:06
BenBen
4,313617
$endgroup$
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
add a comment |
$begingroup$
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.
answered Jan 28 at 14:06
BenBen
4,313617
$endgroup$
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
add a comment |
$begingroup$
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.
answered Jan 28 at 14:06
BenBen
4,313617
$endgroup$
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $mathfrak m/(x_1,ldots, x_i)$ has height $d-i$, which I think is all you need.
answered Jan 28 at 14:06
BenBen
4,313617
edited Jan 28 at 16:38
answered Jan 28 at 14:06
BenBen
4,313617
answered Jan 28 at 14:06
BenBen
4,313617
answered Jan 28 at 14:06
BenBen
4,313617
4,313617
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
add a comment |
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
Yes, if the $x_i$s are not zero divisors.
$endgroup$
– Jehu314
Jan 28 at 14:11
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
$begingroup$
OK, I suggest Matsumura's Commutative Ring Theory Theorem 13.6 for a reference.
$endgroup$
– Ben
Jan 28 at 16:39
add a comment |
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