Submodules of modules of finite projective dimension over regular ring The Next CEO of Stack OverflowProjective module over $R[X]$Depth of finite projective modules over a nonlocal ringNoetherian module over noetherian ringModules with maximal submodules and projective dimensionShort exact sequence of modules over a Noetherian local ring of depth $1$.projective dimension over local ringprojective resolution of finitely generated modulesOne of characterizations of projective modules over noetherian ring of finite global dimensionAny module over a regular local ring has finite free resolutionProjective dimension of module over regular ring is always finite?
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Submodules of modules of finite projective dimension over regular ring
The Next CEO of Stack OverflowProjective module over $R[X]$Depth of finite projective modules over a nonlocal ringNoetherian module over noetherian ringModules with maximal submodules and projective dimensionShort exact sequence of modules over a Noetherian local ring of depth $1$.projective dimension over local ringprojective resolution of finitely generated modulesOne of characterizations of projective modules over noetherian ring of finite global dimensionAny module over a regular local ring has finite free resolutionProjective dimension of module over regular ring is always finite?
$begingroup$
Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, however not every $R$-module may have finite projective dimension.
My question is: Let $M$ be an $R$-module of finite projective dimension. Then, does every submodule also have finite projective dimension ? If this is not true in general, what if we also assume $R$ is an integral domain ?
commutative-algebra homological-algebra projective-module regular-rings
asked Mar 26 at 12:54
user521337user521337
1,2081417
$endgroup$
add a comment |
$begingroup$
Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, however not every $R$-module may have finite projective dimension.
My question is: Let $M$ be an $R$-module of finite projective dimension. Then, does every submodule also have finite projective dimension ? If this is not true in general, what if we also assume $R$ is an integral domain ?
commutative-algebra homological-algebra projective-module regular-rings
asked Mar 26 at 12:54
user521337user521337
1,2081417
$endgroup$
$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49
add a comment |
$begingroup$
Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, however not every $R$-module may have finite projective dimension.
My question is: Let $M$ be an $R$-module of finite projective dimension. Then, does every submodule also have finite projective dimension ? If this is not true in general, what if we also assume $R$ is an integral domain ?
commutative-algebra homological-algebra projective-module regular-rings
asked Mar 26 at 12:54
user521337user521337
1,2081417
$endgroup$
Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, however not every $R$-module may have finite projective dimension.
My question is: Let $M$ be an $R$-module of finite projective dimension. Then, does every submodule also have finite projective dimension ? If this is not true in general, what if we also assume $R$ is an integral domain ?
commutative-algebra homological-algebra projective-module regular-rings
commutative-algebra homological-algebra projective-module regular-rings
asked Mar 26 at 12:54
user521337user521337
1,2081417
asked Mar 26 at 12:54
user521337user521337
1,2081417
edited Mar 26 at 13:01
user521337
asked Mar 26 at 12:54
user521337user521337
1,2081417
asked Mar 26 at 12:54
user521337user521337
1,2081417
asked Mar 26 at 12:54
user521337user521337
1,2081417
1,2081417
$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49
add a comment |
$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49
$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.
Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence
$$0longrightarrow Klongrightarrow Plongrightarrow Nlongrightarrow0.$$
Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.
Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence
$$0longrightarrow Klongrightarrow Plongrightarrow Nlongrightarrow0.$$
Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
add a comment |
$begingroup$
For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.
Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence
$$0longrightarrow Klongrightarrow Plongrightarrow Nlongrightarrow0.$$
Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
add a comment |
$begingroup$
For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.
Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence
$$0longrightarrow Klongrightarrow Plongrightarrow Nlongrightarrow0.$$
Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
For any ring $R$, if there is an $R$-module $N$ of infinite projective dimension then there is a projective $R$-module (which certainly has finite projective dimension!) with a submodule of infinite projective dimension.
Indeed, choose an epimorphism from a projective module $P$ to $N$, and let $K$ be the kernel, so we have a short exact sequence
$$0longrightarrow Klongrightarrow Plongrightarrow Nlongrightarrow0.$$
Since $N$ has infinite projective dimension and $P$ is projective, $K$ also has infinite projective dimension.
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
answered 2 days ago
Jeremy RickardJeremy Rickard
16.9k11746
16.9k11746
add a comment |
add a comment |
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$begingroup$
First I'd like to ask why deleted the question about the finiteness of projective dimension of ideals in the same class of rings.
$endgroup$
– user26857
Mar 26 at 15:36
$begingroup$
@user26857: because going by Mohan's suggestion, I should read it in a book and I found it in Lam's Lectures on Rings and Modules
$endgroup$
– user521337
Mar 26 at 16:32
$begingroup$
Found what? An example of ideal of infinite projective dimension?
$endgroup$
– user26857
Mar 26 at 20:52
$begingroup$
@user26857: No no, found that every ideal does indeed have finite projective dimension, in fact as is proved in the book by Lam, being regular is equivalent to the property that every finitely generated module over the ring has finite projective dimension
$endgroup$
– user521337
Mar 26 at 21:49