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Let algebras be Artin algebras. It is well known that a an algebra has global dimension at most one if and only if the Jacobson radical is projective. As reflexive is a natural generalisation of projective, one might ask when the Jacobson radical $J$ is reflexive. I noted that for Nakayama algebras one has that $J$ is reflexive if and only if the finitistic dimension is at most one. But in general such a statement would probably be too good to be true.

Question: Is there an easy counterexample to $J$ being reflexive implies finitistic dimension at most 1?

(the other direction is not true, see answer below)

Reflexive means that the canonical evaluation map $f_M:M \rightarrow M^{**}$ is an isomorphism where for an algebra $A$: $M^{**}=Hom_A(Hom_A(M,A),A)$. Here $f_M(m)=g$ with $g(h)=h(m)$.

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    $\begingroup$ Could your remind us what "reflexive" means in this context? $\endgroup$
    – Yemon Choi
    Nov 27 '19 at 15:40
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    $\begingroup$ @YemonChoi I added the definition. $\endgroup$
    – Mare
    Nov 27 '19 at 15:58
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The algebra $K[x,y]/(x^2,y^2,xy)$ has finitistic dimension 0 but $J$ is semisimple and not reflexive. I have not yet found an example where $J$ is reflexive but the algebra has finitistic dimension larger than 1.

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