Mixing
Sufficient conditions on $M$ such that the intersection of infinitely many $mathfrak{m} M$, $mathfrak{m}$...
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Let $R$ be a one-dimensional, noetherian and reduced ring. Let $M$ be a finitely generated and torsion-free $R$-module. I am looking for sufficient conditions on $M$ such that
$$bigcap_{mathfrak{m}} mathfrak{m} M = 0$$
whenever $mathfrak{m}$ runs through an infinite set of maximal ideals of $R$.
I was wondering if there is a well-known type of modules that satisfy this type of condition. This could be interpreted as follows: For every $m in M$ the set of zeros of $m$ on $operatorname{Spec}(R)$ is finite. Here the zeros of $m$ are given by those $mathfrak{m}$ such that $m in mathfrak{m} M$.
commutative-algebra modules
asked Jan 22 at 7:48
windsheafwindsheaf
607312
$endgroup$
add a comment |
$begingroup$
Let $R$ be a one-dimensional, noetherian and reduced ring. Let $M$ be a finitely generated and torsion-free $R$-module. I am looking for sufficient conditions on $M$ such that
$$bigcap_{mathfrak{m}} mathfrak{m} M = 0$$
whenever $mathfrak{m}$ runs through an infinite set of maximal ideals of $R$.
I was wondering if there is a well-known type of modules that satisfy this type of condition. This could be interpreted as follows: For every $m in M$ the set of zeros of $m$ on $operatorname{Spec}(R)$ is finite. Here the zeros of $m$ are given by those $mathfrak{m}$ such that $m in mathfrak{m} M$.
commutative-algebra modules
asked Jan 22 at 7:48
windsheafwindsheaf
607312
$endgroup$
$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44
add a comment |
$begingroup$
Let $R$ be a one-dimensional, noetherian and reduced ring. Let $M$ be a finitely generated and torsion-free $R$-module. I am looking for sufficient conditions on $M$ such that
$$bigcap_{mathfrak{m}} mathfrak{m} M = 0$$
whenever $mathfrak{m}$ runs through an infinite set of maximal ideals of $R$.
I was wondering if there is a well-known type of modules that satisfy this type of condition. This could be interpreted as follows: For every $m in M$ the set of zeros of $m$ on $operatorname{Spec}(R)$ is finite. Here the zeros of $m$ are given by those $mathfrak{m}$ such that $m in mathfrak{m} M$.
commutative-algebra modules
asked Jan 22 at 7:48
windsheafwindsheaf
607312
$endgroup$
Let $R$ be a one-dimensional, noetherian and reduced ring. Let $M$ be a finitely generated and torsion-free $R$-module. I am looking for sufficient conditions on $M$ such that
$$bigcap_{mathfrak{m}} mathfrak{m} M = 0$$
whenever $mathfrak{m}$ runs through an infinite set of maximal ideals of $R$.
I was wondering if there is a well-known type of modules that satisfy this type of condition. This could be interpreted as follows: For every $m in M$ the set of zeros of $m$ on $operatorname{Spec}(R)$ is finite. Here the zeros of $m$ are given by those $mathfrak{m}$ such that $m in mathfrak{m} M$.
commutative-algebra modules
commutative-algebra modules
asked Jan 22 at 7:48
windsheafwindsheaf
607312
asked Jan 22 at 7:48
windsheafwindsheaf
607312
asked Jan 22 at 7:48
windsheafwindsheaf
607312
asked Jan 22 at 7:48
windsheafwindsheaf
607312
asked Jan 22 at 7:48
windsheafwindsheaf
607312
607312
$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44
add a comment |
$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44
$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44
add a comment |
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$begingroup$
Can you do this for a free module? Then can you reduce your question to this case? Hint: the natural map $M$ to its double dual is injective, since $M$ has no torsion.
$endgroup$
– Mohan
Jan 22 at 14:58
$begingroup$
@Mohan By what you are saying, I see that the statement is true for free modules and hence for those who can be embedded in a free module. Thus torsionless modules satisfy the statement. But torsion-free does not imply torsionless whenever the ground ring is not a domain. Am I missing something?
$endgroup$
– windsheaf
Jan 23 at 16:44