# The projective dimension of modules in a short exact sequence

Let $$0\to K\to P\to A\to 0$$ be a short exact sequence of right modules with $$P$$ projective and $$A$$ not projective. Suppose $$\text{pd}A<\infty$$ and $$\text{pd}K<\infty$$, where $$\text{pd}$$ is the projective dimension. Then $$1+\operatorname{pd}K=\operatorname{pd}A$$.

This is supposed to follow easily from the definition of projective dimension and the generalized Schanuel's lemma, but I don't even know where to begin. I can extend a projective resolution of $$K$$ to one of $$A$$, showing $$1+\operatorname{pd}K\geq\operatorname{pd}A$$, but not the other. For reference, the lemma is:

Let $$A$$ be a right module and suppose we have two exact sequences

$$0\to K_n\to P_n\to \cdots\to P_1\to P_0\to A\to 0$$

$$0\to K_n'\to P_n'\to \cdots\to P_1'\to P_0'\to A\to 0$$

with each $$P_i,P_i'$$ projective right modules. Then we have an isomorphism if $$n$$ is even

$$K_n\oplus P_n'\oplus P_{n-1}\oplus\cdots\oplus P_0'\cong K_n'\oplus P_n\oplus P_{n-1}'\oplus\cdots\oplus P_0$$

And if $$n$$ is odd $$K_n\oplus P_n'\oplus P_{n-1}\oplus\cdots\oplus P_0\cong K_n'\oplus P_n\oplus P_{n-1}'\oplus\cdots\oplus P_0'$$

After some more thought I found the answer, so I'll post the proof of this trivial lemma.

Let $$0\to Q_n\to Q_{n-1}\to\cdots\to Q_1\to Q_0\to A\to 0$$ be a minimal projective resolution of $$A$$. We can extend the short exact sequence $$0\to K\to P\to A\to 0$$ to an almost-projective resolution of $$A$$, i.e.:

$$0\to L_n\to P_{n-1}\to\cdots\to P_1\to P\to A\to 0$$

where each $$P_i$$ is projective and $$L_n$$ not necessarily so. But by the generalized Schanuel lemma, this also is a projective resolution with $$L_n$$ projective. Since the image of $$P_1\to P$$ is precisely $$K$$, then the above is also a projective resolution of $$K$$, and so $$\operatorname{pd}K+1\leq \operatorname{pd}A$$.