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Is Artinian algebra with finite injective dimension of itself is self-injective？
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Let $A$ be an Artinian $R$-algebra .(i.e.there is a ring homomorphism $alpha:Rrightarrow A$,where $R$ is a commutative Artinian ring,the image of $alpha$ lies in the center of $A$ and $A$ is finite generated $R$ module.)
I have a question:if the injective dimension of $A$ as left $A$ module is finite,is $A$ self-injective?
If $A$ is commutative,then the the Krull dimension of $A$ is zero.so it is well-known that $A$ is self-injective (since $id_AA$ is finite and the Krull dimension is finite,we have $dimA=id_AA$.
so is this true if $A$ noncommutative?if not,if we add that the injective dimension of $A$ as right $A$ module is also finite,is $A$ self-injective？
Thank you in advance!
ring-theory representation-theory homological-algebra noncommutative-algebra
asked Jul 15 at 4:33
Sky
939210
add a comment |Â
up vote
1
down vote
favorite
Let $A$ be an Artinian $R$-algebra .(i.e.there is a ring homomorphism $alpha:Rrightarrow A$,where $R$ is a commutative Artinian ring,the image of $alpha$ lies in the center of $A$ and $A$ is finite generated $R$ module.)
I have a question:if the injective dimension of $A$ as left $A$ module is finite,is $A$ self-injective?
If $A$ is commutative,then the the Krull dimension of $A$ is zero.so it is well-known that $A$ is self-injective (since $id_AA$ is finite and the Krull dimension is finite,we have $dimA=id_AA$.
so is this true if $A$ noncommutative?if not,if we add that the injective dimension of $A$ as right $A$ module is also finite,is $A$ self-injective？
Thank you in advance!
ring-theory representation-theory homological-algebra noncommutative-algebra
asked Jul 15 at 4:33
Sky
939210
In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be an Artinian $R$-algebra .(i.e.there is a ring homomorphism $alpha:Rrightarrow A$,where $R$ is a commutative Artinian ring,the image of $alpha$ lies in the center of $A$ and $A$ is finite generated $R$ module.)
I have a question:if the injective dimension of $A$ as left $A$ module is finite,is $A$ self-injective?
If $A$ is commutative,then the the Krull dimension of $A$ is zero.so it is well-known that $A$ is self-injective (since $id_AA$ is finite and the Krull dimension is finite,we have $dimA=id_AA$.
so is this true if $A$ noncommutative?if not,if we add that the injective dimension of $A$ as right $A$ module is also finite,is $A$ self-injective？
Thank you in advance!
ring-theory representation-theory homological-algebra noncommutative-algebra
asked Jul 15 at 4:33
Sky
939210
Let $A$ be an Artinian $R$-algebra .(i.e.there is a ring homomorphism $alpha:Rrightarrow A$,where $R$ is a commutative Artinian ring,the image of $alpha$ lies in the center of $A$ and $A$ is finite generated $R$ module.)
I have a question:if the injective dimension of $A$ as left $A$ module is finite,is $A$ self-injective?
If $A$ is commutative,then the the Krull dimension of $A$ is zero.so it is well-known that $A$ is self-injective (since $id_AA$ is finite and the Krull dimension is finite,we have $dimA=id_AA$.
so is this true if $A$ noncommutative?if not,if we add that the injective dimension of $A$ as right $A$ module is also finite,is $A$ self-injective？
Thank you in advance!
ring-theory representation-theory homological-algebra noncommutative-algebra
asked Jul 15 at 4:33
Sky
939210
asked Jul 15 at 4:33
Sky
939210
asked Jul 15 at 4:33
Sky
939210
asked Jul 15 at 4:33
Sky
939210
939210
In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23
add a comment |Â
In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23
In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23
In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23
add a comment |Â
1 Answer
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Let $A$ be the algebra of upper triangular $2times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $bullettobullet$.
Then the left and right injective dimensions of $A$ as an $A$-module are both one. So $A$ has finite injective dimension on both sides, but is not self-injective.
In fact, any non-semisimple finite dimensional $k$-algebra of finite global dimension will do.
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $A$ be the algebra of upper triangular $2times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $bullettobullet$.
Then the left and right injective dimensions of $A$ as an $A$-module are both one. So $A$ has finite injective dimension on both sides, but is not self-injective.
In fact, any non-semisimple finite dimensional $k$-algebra of finite global dimension will do.
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
add a comment |Â
up vote
3
down vote
accepted
Let $A$ be the algebra of upper triangular $2times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $bullettobullet$.
Then the left and right injective dimensions of $A$ as an $A$-module are both one. So $A$ has finite injective dimension on both sides, but is not self-injective.
In fact, any non-semisimple finite dimensional $k$-algebra of finite global dimension will do.
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $A$ be the algebra of upper triangular $2times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $bullettobullet$.
Then the left and right injective dimensions of $A$ as an $A$-module are both one. So $A$ has finite injective dimension on both sides, but is not self-injective.
In fact, any non-semisimple finite dimensional $k$-algebra of finite global dimension will do.
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
Let $A$ be the algebra of upper triangular $2times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $bullettobullet$.
Then the left and right injective dimensions of $A$ as an $A$-module are both one. So $A$ has finite injective dimension on both sides, but is not self-injective.
In fact, any non-semisimple finite dimensional $k$-algebra of finite global dimension will do.
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
answered Jul 15 at 9:07
Jeremy Rickard
15.5k11539
15.5k11539
add a comment |Â
add a comment |Â
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In fact the algebras you consider (i.e. finite injective dimension over itself from each side) are called Gorenstein algebras or Iwanaga-Gorenstein algebras. There is a lot of theory concerning them.
– Julian Kuelshammer
Jul 15 at 14:23