I want to prove that if $M$ is an $A$-module with finite projective dimension and $N$ is an $A$-module that is a direct summand of $M$ then the projective dimension of $N$ is less or equal to that of $M$.
I tried using the exact sequence $$0\rightarrow N \rightarrow M \rightarrow N'\rightarrow 0$$ with no results.
Write $M=N\oplus N'$ and set $n=\mathrm{pd}(M)$. Let $$0\to X_n\to P_{n-1}\to\cdots\to P_0\to N\to 0$$ be an exact sequence with $P_i$ projective modules, and similarly $$0\to X'_n\to P'_{n-1}\to\cdots\to P'_0\to N'\to 0$$ be an exact sequence with $P'_i$ projective modules. Then take the direct sum of these two sequences and since $\mathrm{pd}(M)=n$ conclude that $X_n\oplus X'_n$ is projective, so both are projective. Now it follows $\mathrm{pd}(N)\le n$.