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Example of a ring where every module of finite projective dimension is free?
A finitely generated, locally free module over a domain which is not projective?A ring such that all projectives are stably free but not all projectives are free? Multiplicative Structures On Free ResolutionsProjective resolution of modules over rings which are regular in codimension nTor dimension in polynomial rings over Artin ringsIs a projective module of constant finite rank finitely generated?When is the category of Gorenstein projective $R$-modules Frobenius?Looking for example of quotient of group algebra by ideal of group ring which fails to be injectiveUsing Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ringWhere is it shown that a countable self-injective ring is semilocal?
$begingroup$
I'm interested in seeing an example of a ring which is not self-injective where every module admitting a finite projective resolution is free, or at least projective.
Note that self-injectivity says that every monomorphism $R to M$ splits (where $M$ is an arbitrary module), whereas the property I'm looking for looks a bit weaker, requiring this only when $M$ is free.
I'm happy to look at noncommutative rings.
ac.commutative-algebra ra.rings-and-algebras homological-algebra
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
$endgroup$
add a comment |
$begingroup$
I'm interested in seeing an example of a ring which is not self-injective where every module admitting a finite projective resolution is free, or at least projective.
Note that self-injectivity says that every monomorphism $R to M$ splits (where $M$ is an arbitrary module), whereas the property I'm looking for looks a bit weaker, requiring this only when $M$ is free.
I'm happy to look at noncommutative rings.
ac.commutative-algebra ra.rings-and-algebras homological-algebra
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
$endgroup$
add a comment |
$begingroup$
I'm interested in seeing an example of a ring which is not self-injective where every module admitting a finite projective resolution is free, or at least projective.
Note that self-injectivity says that every monomorphism $R to M$ splits (where $M$ is an arbitrary module), whereas the property I'm looking for looks a bit weaker, requiring this only when $M$ is free.
I'm happy to look at noncommutative rings.
ac.commutative-algebra ra.rings-and-algebras homological-algebra
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
$endgroup$
I'm interested in seeing an example of a ring which is not self-injective where every module admitting a finite projective resolution is free, or at least projective.
Note that self-injectivity says that every monomorphism $R to M$ splits (where $M$ is an arbitrary module), whereas the property I'm looking for looks a bit weaker, requiring this only when $M$ is free.
I'm happy to look at noncommutative rings.
ac.commutative-algebra ra.rings-and-algebras homological-algebra
ac.commutative-algebra ra.rings-and-algebras homological-algebra
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
edited May 20 at 13:53
Tim Campion
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
asked May 20 at 13:44
Tim CampionTim Campion
15.4k355132
15.4k355132
add a comment |
add a comment |
1 Answer
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The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras.
Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free.
Concrete example:
Take a quiver $Q$ with one point and $n geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2.
So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$.
For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free.
When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras.
You need two results about finite dimensional algebras:
1.For a local algebra, every projective module is free.
- Local algebras have finitistic dimension zero.
The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).
answered May 20 at 14:00
MareMare
3,96031336
$endgroup$
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
add a comment |
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$begingroup$
The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras.
Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free.
Concrete example:
Take a quiver $Q$ with one point and $n geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2.
So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$.
For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free.
When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras.
You need two results about finite dimensional algebras:
1.For a local algebra, every projective module is free.
- Local algebras have finitistic dimension zero.
The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).
answered May 20 at 14:00
MareMare
3,96031336
$endgroup$
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
add a comment |
$begingroup$
The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras.
Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free.
Concrete example:
Take a quiver $Q$ with one point and $n geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2.
So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$.
For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free.
When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras.
You need two results about finite dimensional algebras:
1.For a local algebra, every projective module is free.
- Local algebras have finitistic dimension zero.
The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).
answered May 20 at 14:00
MareMare
3,96031336
$endgroup$
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
add a comment |
$begingroup$
The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras.
Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free.
Concrete example:
Take a quiver $Q$ with one point and $n geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2.
So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$.
For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free.
When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras.
You need two results about finite dimensional algebras:
1.For a local algebra, every projective module is free.
- Local algebras have finitistic dimension zero.
The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).
answered May 20 at 14:00
MareMare
3,96031336
$endgroup$
The classes of ring you look at are precisly the rings of finitistic dimension zero for the question with projective instead of free. I will give a large class of example for finite dimensional algebras.
Let $A$ be a local finite dimensional (over a field $K$) non-selfinjective algebra. Then every non-projective module has infinite projective dimension and thus the only modules of finite projective dimension are the projective modules, which are all free.
Concrete example:
Take a quiver $Q$ with one point and $n geq 2$ loops and let $KQ/I$ the quiver algebra with the relations $I$ such that $I=J^2$ consists of all paths of length 2. Alternatively, this is $K[x_1,...,x_n]/I$ with $I$ the ideal generated by all monomials of length 2.
So the "smallest" example is probably the 3-dimensional algebra $K[x,y]/(x^2,y^2,xy)$.
For finite dimensional algebras, the local finite dimensional non-selfinjective algebras are exactly those with the property that every module of finite projective dimension is free.
When you look instead at the property "every module of finite projective dimension is projective", then you arrive at the finite dimensional algebras with finitistic dimension zero (the finitistic dimension is defined as the supremum of all projective dimensions of modules having finite projective dimension), which is a very large class that contains for example all selfinjective and all local algebras.
You need two results about finite dimensional algebras:
1.For a local algebra, every projective module is free.
- Local algebras have finitistic dimension zero.
The first result is well known and 2. is easy to prove, but I will search for a reference now. (https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 proposition 2.1. states that the finitistic dimension of local algebras is 0, this is not a good reference but I leave it here until a better one is found).
answered May 20 at 14:00
MareMare
3,96031336
edited May 20 at 14:13
answered May 20 at 14:00
MareMare
3,96031336
answered May 20 at 14:00
MareMare
3,96031336
answered May 20 at 14:00
MareMare
3,96031336
3,96031336
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
add a comment |
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Thanks! Could you give a hint or a reference as to why this is so?
$endgroup$
– Tim Campion
May 20 at 14:01
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
Algebra means algebra over a given field $K$ and finite-dimensional means as vector space over $K$?
$endgroup$
– YCor
May 20 at 14:05
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
$begingroup$
@YCor Yes, I added that.
$endgroup$
– Mare
May 20 at 14:07
add a comment |
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