If $C$ is an $B$-algebra and $A$ is a projective $C$-module, then does it follow that the projective dimension of $A$ over $B$ is less than or equal to the projective dimension of $C$ over $B$?
Projective dimension of $A$ over $B$ $\le$ projective dimension of $C$ over $B$?
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ring-theory
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1 Answers
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Yes.
This is simply because $A$, being projective, will be a direct factor of a free $C$-module $F$. If we write $F=A\oplus X$, then $\operatorname{proj.dim}_B(F) = \operatorname{max}(\operatorname{proj.dim}_B(A),\operatorname{proj.dim}_B(X))$. Moreover, since $F$ is a direct sum of copies of $C$, then $\operatorname{proj.dim}_B(F) = \operatorname{proj.dim}_B(C)$. So indeed we get $$ \operatorname{proj.dim}_B(A) \leq \operatorname{proj.dim}_B(C). $$