Let $A$ be a two-sided artinian ring. Recall that a module $M$ is said to have dominant dimension at least $n$ in case the terms $I_i$ in the minimal injective coresolution of $M$ are projective for $i=0,1...,n-1$. $A$ is said to have dominant dimension at least $n$ in case the regular module $A$ has dominant dimension at least $n$ (which is equivalent to the bimodule $A$ having dominant dimension at least $n$). $M$ is said to be $n$-torsionfree in case $Ext_A^i(Tr(M),A)=0$ for $i=1,...,n$.
Note that being 1-torsionfree is the same as being torsionfree and being 2-torsionfree is the same as being reflexive.
Here $Tr(M)$ is the Auslander-Bridger transpose of $M$.
Consider the 3 conditions:
a) $A$ has dominant dimension at least $n$.
b) $A$ as a bimodule is $n$-torsionfree.
c) $A$ is an $n$-syzygy as a bimodule, that is $A \cong \Omega^n(U)$ for some other $A$-bimodule $U$.
Note that an equivalent condition to condition a) for $n=1,2$ was provided by Colby and Fuller in https://www.jstor.org/stable/2043764?seq=1 in terms of conditions on the double dual functor, which also gives a strong link to condition b).
Question: Is it true that the conditions a), b) and c) are equivalent? I can show that they are equivalent for $A$ being a finite dimensional algebra. But I had to use that $A$ has a duality and one deep result of Auslander-Reiten (and some other recent results on the dominant dimension). I wonder whether a more direct proof is possible for general artinian rings.
We always have that a) implies b) and that b) implies c). So the question is only about whether c) implies a).
