Let $A$ be an Artinian $R$-algebra .(i.e.there is a ring homomorphism $\alpha:R\rightarrow A$,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!
Let $A$ be the algebra of upper triangular $2\times 2$ matrices over a field $k$, or (equivalently) the path algebra of the quiver $\bullet\to\bullet$.
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.