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Let $\Lambda$ be an artin algebra. We denote by $mod \Lambda$ the category of all finitely generated $\Lambda$-modules. A module $X \in mod \Lambda$ is called torsionless if it is a submodule of free modules.

There are two conclusions I don't know how to prove:

  1. Let $A$ be a torsionless $\Lambda^{op}$-module, then $Ext_{\Lambda} ^1(TrA, \Lambda)=0$. ($TrA$ means the transpose of $A$)
  2. $Ext_{\Lambda^{op}} ^1(A, \Lambda ^ {op})=0$ for all torsionless $\Lambda^{op}$-modules $A$ is equivalent to $id_{\Lambda^{op}} \Lambda^{op} \leq 1$.($id_{\Lambda^{op}} \Lambda^{op} $ means the injective dimension of $\Lambda^{op}$)

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  1. THis is a standard fact about torsionless modules, which you can find (together with a characterisation of reflexive modules) in the book by auslander reiten smalo (look in the index for torsionless or reflexive modules).
  2. Every module of the form $\Omega^{1}(N)$ is torsionless for some module $N$. Then $Ext^{1}(\Omega^{1}(N),\Lambda)=Ext^{2}(N, \Lambda)=0$ for all $N$, which exactly means that the injective dimension of the algebra is at most 1.