stationary set,club,module theory,Auslander lemma,

Question

Here http://www.ams.org/journals/tran/1990-322-02/S0002-9947-1990-0974514-8/S0002-9947-1990-0974514-8.pdf I do not understand the first two lines of the proof of lemma 9,on page 550:What and why is there "dropping to a club"? Namely,how it follows that it suffices to prove our claim that $E$ below is stationary in $\kappa$. And how it follows from Auslander lemma that $(ii)$ continues to hold? A detailed explanation of these two lines will be welcome.

Answer 1

I can’t help with the Auslander lemma, but I can explain the part about dropping to a club. The assumption is that the set

$$E=\left\{\alpha<\kappa:\exists\beta>\alpha\text{ with }\operatorname{Ext}_R^1(M_\beta/M_\alpha,H_\beta/H_\alpha)\ne 0\right\}$$

is stationary in $\kappa$. For each $\alpha<\kappa$ let $\varphi(\alpha)>\alpha$ be such that $\operatorname{Ext}_R^1\big(M_\varphi(\alpha)/M_\alpha,H_\varphi(\alpha)/H_\alpha\big)\ne 0$. Recursively define $\alpha_\xi$ for $\xi<\kappa$ by $\alpha_0=0$, $\alpha_{\xi+1}=\varphi(\alpha_\xi)$, and $\alpha_\eta=\sup_{\xi<\eta}\alpha_\xi$ if $\eta<\kappa$ is a limit ordinal. Then $\{\alpha_\xi:\xi<\kappa\}$ is a club (closed unbounded subset) in $\kappa$, so $E\cap\{\alpha_\xi:\xi<\kappa\}$ is still stationary in $\kappa$.

We now replace the sequence $\langle M_\alpha:\alpha<\kappa\rangle$ by the sequence $\langle M_{\alpha_\xi}:\xi<\alpha\rangle$. This sequence clearly satisfies all of the conditions of the lemma except possibly (ii), and apparently there is some lemma by Auslander that implies that it satisfies (ii) as well.

For each $\xi<\kappa$ we have $\alpha_{\xi+1}=\varphi(\alpha_\xi)$, so $\operatorname{Ext}_R^1(M_{\alpha_{\xi+1}}/M_{\alpha_\xi},H_{\alpha_{\xi+1}}/H_{\alpha_\xi})\ne 0$. Thus, if simply re-index $\langle M_{\alpha_\xi}:\xi<\kappa\rangle$ as $\langle M_\alpha:\alpha<\kappa\rangle$, forgetting about the original $\kappa$-sequence of submodules altogether, we have $\operatorname{Ext}_R^1(M_{\alpha+1}/M_\alpha,H_{\alpha+1}/H_\alpha)\ne 0$ for each $\alpha<\kappa$. Finally, the definition of the $G_\alpha$ and $H_\alpha$ ensures that $H_{\alpha+1}/H_\alpha=G_\alpha$, so we may as well assume that

$$E=\left\{\alpha<\kappa:\operatorname{Ext}_R^1(M_{\alpha+1}/M_\alpha,G_\alpha)\ne 0\right\}\;.$$