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Why does there exist a non-split sequence with the condition that $\mathrm{pd} M = \infty$?

Remarks.

  • I am reading

Andrzej Skowroński, Sverre O.Smalø, Dan Zacharia: On the Finiteness of the Global Dimension for Artinian Rings. Journal of Algebra 251, 475–478 (2002).

wherein on p. 476 there is the

Theorem. Let $\Lambda$ be a left artinian ring and assume that each indecomposable finitely generated left $\Lambda$-module has finite projective dimension or finite injective dimension. Then $\Lambda$ has finite left global dimension.

Proof of the Theorem. Assume that $\Lambda$ is a left artinian ring of infinite global dimension where each indecomposable module has finite projective dimension or finite injective dimension. We shall show that this leads to a contradiction. Clearly we may assume that no simple module has both infinite projective dimension and infinite injective dimension. Let $S$ be a simple module of infinite projective dimension. Recall that if $$\cdots \longrightarrow P_{n} \longrightarrow P_{n-1} \longrightarrow \cdots \longrightarrow P_{1}\longrightarrow P_{0} \longrightarrow S\longrightarrow 0$$ is a minimal projective resolution of S and T is a simple $\Lambda$-module, then $\mathrm{Ext}_{\Lambda}^{n}(S,T)\neq0$ if and only if the projective cover of $T$ is a direct summand of $P_{n}$. Since $S$ has infinite projective dimension and sup $\{{\rm id}\ Y$ $|$ $Y$ is simple and of finite injective dimension$\}$ is finite, say equal to $n$, each simple module $T$ with $\mathrm{Ext}_{\Lambda}^{n+1}(S,T)\neq0$ is of infinite injective dimension. Now since $S$ has infinite projective dimension, there exists a direct summand M of $\Omega^{n}(S)$ and a nonsplit exact sequence $0 \longrightarrow T \longrightarrow E\longrightarrow M \longrightarrow 0$ with $\mathrm{pd} M=\infty$. Where $\mathrm{pd} M$ denotes the projective dimension of $M$. ${\rm id} Y$ denotes the injective dimension of $M$. $ \Omega^{n}(S)$ is the $n$th syzygy of $S$.

I cannot understand that why there exists a direct summand $M$ of $ \Omega^{n}(S)$ and a nonsplit exact sequence $0 \longrightarrow T \longrightarrow E\longrightarrow M \longrightarrow 0$ with $\mathrm{pd} M=\infty$.

In fact, since $\mathrm{Ext}_{\Lambda}^{1}(\Omega^{n}(S),T)=\mathrm{Ext}_{\Lambda}^{n+1}(S,T)\neq0$, we know that there exists a direct summand $M$ of $ \Omega^{n}(S)$ such that $\mathrm{Ext}_{\Lambda}^{1}(M,T)\neq0$.But why $\mathrm{pd} M=\infty?$

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Since $S$ has infinite projective dimension, there is some indecomposable summand $M$ of $\Omega^n(S)$ that has infinite projective dimension.

For simple modules $T$ of finite injective dimension, $\text{Ext}^1(\Omega^n(S),T)=0$ and so $\text{Ext}^1(M,T)=0$.

But there is some simple module $T$ for which $\text{Ext}^1(M,T)\neq0$, and so this $T$ must have infinite injective dimension.

In other words, don't choose $M$ with the $\text{Ext}$ condition and try to prove that it has infinite projective dimension. Choose $M$ with infinite projective dimension and prove that it has the $\text{Ext}$ condition.

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    $\begingroup$ Oh, so you choose $M$ before you choose $T$. It was unclear from the text that you were allowed to do that. $\endgroup$ – Dag Oskar Madsen Oct 2 '17 at 12:33
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    $\begingroup$ @DagOskarMadsen I agree that the text isn't very clear. But I think my interpretation is sufficient for everything that follows in the paper. $\endgroup$ – Jeremy Rickard Oct 2 '17 at 12:40

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