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Rings over which torsion free module is projective
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If R is an integral domain with the property that any finitely generated torsion free R-module becomes projective. Certainly R could be Dedekind domain. But is it necessary that R must be Dedekind domain?
abstract-algebra commutative-algebra algebraic-number-theory
asked 2 days ago
Sunny Rathore
49327
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up vote
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If R is an integral domain with the property that any finitely generated torsion free R-module becomes projective. Certainly R could be Dedekind domain. But is it necessary that R must be Dedekind domain?
abstract-algebra commutative-algebra algebraic-number-theory
asked 2 days ago
Sunny Rathore
49327
$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
What are Prüfer domains?
– Sunny Rathore
yesterday
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If R is an integral domain with the property that any finitely generated torsion free R-module becomes projective. Certainly R could be Dedekind domain. But is it necessary that R must be Dedekind domain?
abstract-algebra commutative-algebra algebraic-number-theory
asked 2 days ago
Sunny Rathore
49327
If R is an integral domain with the property that any finitely generated torsion free R-module becomes projective. Certainly R could be Dedekind domain. But is it necessary that R must be Dedekind domain?
abstract-algebra commutative-algebra algebraic-number-theory
asked 2 days ago
Sunny Rathore
49327
asked 2 days ago
Sunny Rathore
49327
asked 2 days ago
Sunny Rathore
49327
asked 2 days ago
Sunny Rathore
49327
49327
$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
What are Prüfer domains?
– Sunny Rathore
yesterday
add a comment |Â
$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
What are Prüfer domains?
– Sunny Rathore
yesterday
$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
What are Prüfer domains?
– Sunny Rathore
yesterday
What are Prüfer domains?
– Sunny Rathore
yesterday
add a comment |Â
1 Answer
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Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.
Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:
(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)
An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.
Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff
(3) f.g. submodules of projective modules are projective.
You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).
As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every
f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.
answered 7 hours ago
Badam Baplan
3,246721
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.
Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:
(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)
An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.
Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff
(3) f.g. submodules of projective modules are projective.
You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).
As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every
f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.
answered 7 hours ago
Badam Baplan
3,246721
add a comment |Â
up vote
0
down vote
Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.
Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:
(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)
An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.
Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff
(3) f.g. submodules of projective modules are projective.
You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).
As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every
f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.
answered 7 hours ago
Badam Baplan
3,246721
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.
Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:
(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)
An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.
Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff
(3) f.g. submodules of projective modules are projective.
You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).
As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every
f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.
answered 7 hours ago
Badam Baplan
3,246721
Prüfer domains constitute a very important class of rings which are central to a few branches of commutative ring theory, in particular multiplicative ideal theory. They are the non-noetherian analogue of Dedekind domains.
Prüfer domains are most commonly characterized as domains $D$ satisfying either of the following equivalent properties:
(1) every f.g. ideal $I$ of $D$ is invertible (i.e. there exists a fractional ideal $J$ such that $IJ = D$
(2) $D_P$ is locally a valuation domain (i.e. a domain in which the ideals are totally ordered by inclusion)
An easy to find reference for this equivalence is theorems 58-64 of Kaplansky's Commutative Rings.
Note that in a domain, f.g. ideals are invertible iff they are projective, which hints at a module-theoretic characterization. One such characterization is that $D$ is Prüfer iff
(3) f.g. submodules of projective modules are projective.
You can read more about that in T.Y. Lam's Lectures on Modules and Rings (section 2E is all about (semi)-hereditary rings, which generalize Prüfer and Dedekind domains to rings with zero-divisors).
As Lam notes in the end of that section, this easily leads to the characterization of Prüfer domains in terms f.g. torsionfree modules being projective, because every
f.g. torsionfree module in a domain can be embedded in a (finite-rank) free module.
answered 7 hours ago
Badam Baplan
3,246721
answered 7 hours ago
Badam Baplan
3,246721
answered 7 hours ago
Badam Baplan
3,246721
answered 7 hours ago
Badam Baplan
3,246721
3,246721
add a comment |Â
add a comment |Â
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$R$ could be a field. If $R$ is not a field, then $R$ is a Dedekind domain, if it is Noetherian.
– Mohan
2 days ago
what if R is not Noetherian?
– Sunny Rathore
2 days ago
Not necessarily! This property characterizes the Prüfer domains.
– user26857
2 days ago
What are Prüfer domains?
– Sunny Rathore
yesterday