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

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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 $\operatorname{pd}K+1\leq \operatorname{pd}A$.

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