If $f:R\to S$ is an $R$-algebra and $P$ is a projective $S$-module, then $pd_R(P)\le pd_R(S)$.

Solution 1:

$P$ is a projective $S$-module, so it is a direct summand of a free module, say $P\oplus M=S^{(A)}$. The dimension of a sum is the sup of the dimensions of the summands, so $$pd_R(P)\le pd_R(S^{(A)})=pd_R(S).$$

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If $f:R\to S$ is an $R$-algebra and $P$ is a projective $S$-module, then $pd_R(P)\le pd_R(S)$.

Solution 1:

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