Why there exists a non-split sequence with the condition that $pd M=\infty$?

I am reading a paper with the title "On the Finiteness of the Global Dimension for Artinian Rings". Journal of Algebra 251, 475–478 (2002) doi:10.1006/jabr.2001.9130.

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 $Ext_{\Lambda}^{n}(S,T)\neq0$ 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 $Ext_{\Lambda}^{n+1}(S,T)\neq0$ 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 $pd M=\infty$. Where $pd M$ 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 can not 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 $pd M=\infty$.

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

The paper can be seen: https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af

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 $\mathrm{pd} M=\infty?$

Why there exists a non-split sequence with the condition that $pd M=\infty$?

I am reading a paper with the title "On the Finiteness of the Global Dimension for Artinian Rings". Journal of Algebra 251, 475–478 (2002) doi:10.1006/jabr.2001.9130.

I can not 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 $pd M=\infty$. Where $pd M$ denotes the projective dimension of $M$.

The paper can be seen: https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af

Why there exists a non-split sequence with the condition that $pd M=\infty$?

I am reading a paper with the title "On the Finiteness of the Global Dimension for Artinian Rings". Journal of Algebra 251, 475–478 (2002) doi:10.1006/jabr.2001.9130.

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 $Ext_{\Lambda}^{n}(S,T)\neq0$ 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 $Ext_{\Lambda}^{n+1}(S,T)\neq0$ 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 $pd M=\infty$. Where $pd M$ 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 can not 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 $pd M=\infty$.

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

The paper can be seen: https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af https://ac.els-cdn.com/S0021869301991306/1-s2.0-S0021869301991306-main.pdf?_tid=79acfbb0-a69d-11e7-b956-00000aacb362&acdnat=1506858274_82e31ced26098f3aaa365cc477b8b8af