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Let $R$ be an arbitrary ring. Let $D$ be the class of $R$-modules of projective dimension less than or equal to a natural number $n$. If $L$ is the direct union of a continuous chain of submodules ${L_{\alpha},\alpha < \lambda}$ for some ordinal number $\lambda$ (this means that $L=\bigcup_{\alpha}L_{\alpha},\ L_{\alpha}\subseteq L_{\alpha'}$ if $\alpha \leq \alpha' <\lambda$ and $\ L_{\beta}=\bigcup_{\alpha <\beta} L_\alpha$ when $\beta < \lambda $ is a limit ordinal) with $L_{0}\in D$ and $L_{\alpha +1}/L_{\alpha}\in D, \forall \alpha<\lambda,$ can one show that $L \in D$?

PS: We know that when $R$ is a perfect ring, then $D$ is closed under direct limits, then we can prove the above by transfinite induction. But if $R$ is not perfect, how can we show that?

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The projective dimension of $L$ is also $\leq n$. This result is known as Auslander's Lemma. – Fernando Muro Aug 7 '14 at 13:42

When $n=0$, the module $L$ is isomorphic to the direct sum of $L_{\alpha+1}/L_\alpha$, so it is projective. For an arbitrary natural number $n$, argue by induction. Let $P_\alpha$ denote the free $R$-module with generators indexed by the elements of $L_\alpha$ and $P$ denote the free $R$-module with generators indexed by the elements of $L$. Then $P$ is the direct union of $P_\alpha$. There is a natural morphism of $R$-modules $P\to L$ mapping $P_\alpha$ surjectively onto $L_\alpha$. Let $M$ be the kernel of the morphism $P\to L$ and $M_\alpha$ be the kernel of the morphism $P_\alpha\to L_\alpha$. Then $M$ is the direct union of $M_\alpha$ and the quotient modules $M_{\alpha+1}/M_\alpha$, being isomorphic to the kernels of the morphisms $P_{\alpha+1}/P_\alpha\to L_{\alpha+1}/L_\alpha$, have projective dimensions not exceeding $n-1$.

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