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Let $A$ be an algebra over a field k. A module $_AM$ is called a generator if $\textrm{add}(A) \subseteq \textrm{add}(M)$, a cogenerator if $\textrm{add}\big(D(A)\big) \subseteq \textrm{add}(M)$. $M$ is $n$-rigid if $\textrm{Ext}_A^i(M,M)=0$ for $1 \leq i \leq n$.

Now suppose $_AM$ is a generator-cogenerator which is $n$-rigid and neither projective nor injective. $\Lambda := \textrm{End}_A(M)$. If $\Lambda$ is at most $(n+m+2)$-Gorenstein, that is the injective dimension $\textrm{inj.dim}(_{\Lambda} \Lambda)=\textrm{inj.dim} (\Lambda_{\Lambda}) \leq (n+m+2)$, then for any $\Lambda$-module $N$, how to get that $\Omega^2_{\Lambda}(N) \cong \textrm{Hom}_A(M,Y)$ for some $A$-module $Y$(here, $\Omega$ is the syzygy functor)?

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This is an application of the Yoneda lemma. There is an exact sequence

$$0\to\Omega^2_\Lambda(N)\to P_1\to P_0\to N\to0$$

with $P_i$ projective. By definition of $\Lambda$, it follows that $P_i=\operatorname{Hom}_A(M,M_i)$ for some $M_i\in\operatorname{add}{M}$. (Here I am making the additional assumption that $N$ is finitely presented, which you probably want; if, for example, $A$ and $M$ are finite-dimensional over $k$, then this is automatic.) By Yoneda, the map $P_1\to P_0$ is the image under the left exact functor $\operatorname{Hom}_A(M,-)$ of some $f\colon M_1\to M_0$, and so its kernel is $\operatorname{Hom}_A(M,Y)$, where $Y$ is the kernel of $f$.

Note that this didn't use any of the assumptions on $M$, or the Gorenstein-ness of $\Lambda$.

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  • $\begingroup$ Thank you, sir. I have other questions: (1)if $N$ is finitely presented as $\Lambda$-module, then the projective dimension $proj.dim(_{\Lambda}N) \leq 2$ always hold? (2) If we have an exact sequence of the form $0 \rightarrow Hom_A(M,N') \rightarrow Hom_A(M,N) \rightarrow Hom_A(M,N'') \rightarrow 0$ as $\Lambda$-modules, could we get an exact sequence $0 \rightarrow N' \rightarrow N \rightarrow N'' \rightarrow 0$ as $A$-modules? $\endgroup$
    – Xiaosong Peng
    Commented Mar 20, 2017 at 13:34
  • $\begingroup$ If you have other questions, you should usually ask them as new questions! However, in this case the answers are fairly quick: (1) No (and even almost never: under your assumptions, if $\operatorname{add}(M)\ne\operatorname{mod}(A)$ then $\operatorname{gldim}(\Lambda)>2$). (2) Also no in general; all three Hom-spaces could be $0$ without needing any relationship between $N$, $N'$ and $N''$, but if $M$ is a generator then yes, because the second sequence is obtained from the first by the (exact) restriction functor to $A$-modules. $\endgroup$
    – Matthew Pressland
    Commented Mar 20, 2017 at 14:03

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