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

Question

The title basically says it all. If $f:R\to S$ is an $R$-algebra and $P$ is a projective $S$-module, then I need to show that $pd_R(P)\le pd_R(S)$. Here $pd_R(A)$ refers to the projective dimension of the $R$-module $A$, i.e. the minimum possible length of a projective resolution $P\to A.$

Unfortunately I don't even know how to start this. I think I'm just hitting a mental roadblock. Of course I can assume that $pd_R(S)$ is finite, say $pd_R(S)=n$, but then what? I can take a projective resolution of $S,$ but I don't really see what that does. I think I want to use what Weibel calls the "pd lemma," which basically says

The following are equivalent for an $R$-module $A$:

1. $pd(A)\le d$
2. $\mathrm{Ext}^n_R(A,B)=0$ for all $n>d$ and all $R$-modules $B$
3. $\mathrm{Ext}^{d+1}_R(A,B)=0$ for all $R$-modules $B$
4. If $\;0\to M_d\to P_{d-1}\to\cdots\to P_0\to A\;$ is any resolution of $A$ with the $P_i$'s projective, then $M_d$ is projective.

There's something easy that I'm missing and I know it because I did the surrounding problems without too much trouble. Any hint would be great.

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This is (part of) Exercise 4.1.3 of Weibel's An Introduction to Homological Algebra.

Answer 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 $$pd_R(P)\le pd_R(S^{(A)})=pd_R(S).$$