Given a finite dimensional algebra $A$ with Jacobson radical $J$. Is the global dimension of $A$ equal to the injective dimension of $J$? I can prove this for algebras with finite global dimension and in general it should be equivalent to the inequality $injdim(J) \geq injdim(J^2)-1$, see my answer.


Here a proof that works for a large class of algebra including all algebras of finite global dimension in case I made no mistake. We assume for simplicity that the algebras are given by quiver and relations (which is no loss of generality in case the field is algebraically closed) and the algebra is not semi-simple or local (in which case it is trivial that the global dimension equals the injective dimension of the Jacobson radical).

I use the well known lemma that in a short exact sequence $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ one has $pd(X) \leq max(pd(Y),pd(Z)-1)$. Dualising that should give that $injdim(Z) \leq max( injdim(Y), injdim(X)-1)$.

Lemma: Let $A$ be a general algebra, then the injective dimension of $J/J^2$ equals the global dimension of $A$ when $J$ denotes the Jacobson radical of $A$.

Proof: Note that $J/J^2$ is a direct sum of simple modules. Let $S$ be a simple module that is not injective corresponding to the point $i$. Then there is an arrow starting at a point $j$ to $i$ or else $S$ would be injective, since $S$ is the socle of $D(Ae_i)$, which has a vector space basis corresponding to all paths that end at $i$ in the algebra. Now $S$ is a direct summand of $J/J^2$ since $e_j J/J^2 e_i \neq 0$ and $J/J^2$ is semisimple. But the global dimension of an algebra equals the supremum of the injective dimension of simple modules. Thus there is a simple non-injective module with injective dimension equal to the global dimension that is a summand of $J/J^2$, showing that $J/J^2$ has injective dimension equal to the global dimension.

proposition: Let $A$ be an algebra with finite global dimension or such that the injective dimension of $J$ is larger than or equal to the injective dimension of $J^2$ minus one. Then the injective dimension of $J$ is equal to the global dimension of $A$.

proof: We have the short exact sequence $0 \rightarrow J^2 \rightarrow J \rightarrow J/J^2 \rightarrow 0$, which gives that $injdim(J/J^2) \leq max(injdim(J),injdim(J^2)-1)$, which shows that the injective dimension of $J$ is at least the injective dimension of $J/J^2$ in case of our second assumption. Assume now that the global dimension is finite. Then $J/J^2$ has injective dimension equal to the global dimension and thus also $J$ has then injective dimension equal to the global dimension $n$ since $injdim(J^2)-1 \leq n-1$ and the above inequality. In case the global dimension is infinite, we use the second assumption to see that the injective dimension of $J$ is also infinite.

Maybe someone sees how to remove in a nice way the assumption that the algebra is a quiver algebra for the lemma or how to prove $injdim(J^2)-1 \leq injdim(J)$ (or the weaker statement that the infinity of $injdim(J^2)$ implies that $injdim(J)$ is also infinite)?

edit: probably one can remove the assumption to have quiver algebras by noting that $e_j J/J^2 e_i \neq 0$ iff $Ext^1(S_i , S_j) \neq 0$. So $S_i$ being a non-injective module is equivalent to $Ext^1(S_i, A/J)$ nonzero which is equivalent to $e_j J/J^2 e_i \neq 0$ for some $j$. So the proof should work in fact for any Artin algebra with the above assumptions. I have to find a reference for $e_j J/J^2 e_i \neq 0$ iff $Ext^1(S_i , S_j) \neq 0$ for Artin algebras, is there any? (or even for general finite dimensional algebras)


Here a proof for Nakayama algebras of 1. I wonder whether one can modify this to get a proof for a larger class of algebras:

First note that 1. is trivially true for selfinjective algebras and thus we can assume the algebra to be nonselfinjective. Let $A$ be a (nonselfinjective) Nakayama algebra with global dimension $g$. Let $P$ be an indecomposable projective module with injective dimension g. Then $P$ is not injective and thus there is an embedding $P \rightarrow I$ where $I$ is the injective envelope of $P$. But $I$ is projective, since Nakayama algebras are QF-3. Let $I=e_i A$. Then we can write $P=e_i J^k$ for some $k$. But with $e_i J^k$, all the modules $e_i J^l $ for $l=0,1,2,...k$ are also projective. Thus we can write P as the radical of the projective module $e_i J^{k-1}$. Therefore, there exists a direct summand of the Jacobson radical with injective dimension $g$ and thus the injective dimension of the Jacobson radical is itself equal to $g$.


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