$\DeclareMathOperator\pd{pd}$Suppose that $R$ is a commutative ring and $R'$ is a subring of $R$ such that $R$ is a free $R'$-module of finite rank. Assume that both $R$ and $R'$ are regular local rings of the same Krull dimension. Now let $M$ be a finitely generated $R$-module. If $\pd_R(M) \leq 1$, can one also deduce that $\pd_{R'}(M) \leq 1$?


1 Answer 1


This follows from the General Change of Rings Theorem 4.3.1 in Weibel's book: Let $f\colon R'\rightarrow R$ be a ring map, and let $M$ be an $R$-module. Then as an $R'$ module $$\operatorname{pd}_{R'}(M)=\operatorname{pd}_{R}(M)+\operatorname{pd}_{R'}(R).$$ Note $\operatorname{pd}_{R'}(R)=0$ by your assumption that $R$ is a free $R'$ module.

  • $\begingroup$ Thank you very much! $\endgroup$ Commented Jan 24 at 6:46

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