Let $0 \to B \to P \to A \to 0$ be a short exact sequence of $R$-modules with $P$ is projective. By projective dimension of a module, I mean the length of the shortest projective resolution of it. A statement from Passman, A course in Ring Theory, p76 says that if $pd A = n > 0$, then $pd B = n-1$.
In order to see this equality, I assume that $pd A = a$ and $pd B = b$. Then there exist a projective resolution of $B$ such that
$$ 0 \to P_b \to P_{b-1} \to \ldots \to P_1 \to P_0 \to B \to 0 $$
Here by using the first given short exact sequence, I can get a projective resolution of $A$ such that
$$ 0 \to P_b \to P_{b-1} \to \ldots \to P_1 \to P_0 \to P \to A \to 0 $$
and hence I conclude that $a \leq b+1$. I can't see the other inequality: $b+1 \leq a$. I tried to write down a projective resolution of $A$ of length $a$ and used comparison theorem but I still can't see the inequality. Any comment would be so helpful.
In addition, could anyone give an example of projective resolution which is not minimal?
Best Answer
Let's suppose that $B$ has projective resolution $$0\to Q_b\to Q_{b-1}\to\cdots\to Q_1\to Q_0\to B\to0.$$ This breaks up into short exact sequences $$0\to B_{i+1}\to Q_i\to B_i\to0$$ with $B_0=B$ and $B_{b+1}=0$.
Do the same with a projective resolution of $A$: $$0\to P_a\to P_{a-1}\to\cdots\to P_1\to P_0\to A\to0$$ breaks up into short exact sequences $$0\to A_{i+1}\to P_i\to A_i\to0$$ with $A_0=A$ and $A_{a+1}=0$.
Applying Schanuel's lemma to $$0\to B\to P\to A\to 0$$ and $$0\to A_1\to P_0\to A\to 0$$ gives $$P\oplus A_1\cong P_0\oplus B.$$ Now apply Schanuel to $$0\to P_0\oplus B_1\to P_0\oplus Q_0\to P_0\oplus B\to 0$$ and $$0\to P\oplus A_2\to P\oplus P_1\to P\oplus A_1\to 0$$ to get $$P_0\oplus Q_0\oplus P\oplus A_2\cong P\oplus P_1\oplus P_0\oplus B_1.$$ I'll summarise this as $$\text{projective}\oplus A_2\cong\text{projective}\oplus B_1.$$ Iterating this argument gives $$\text{projective}\oplus A_{j+1}\cong\text{projective}\oplus B_j.$$
Now $B_b$ is projective, so that $\text{projective}\oplus A_{b+1} =\text{projective}$. Then $A_{b+1}$ is projective, so the resolution for $A$ ends at $A_{b+1}$ (or before). Then $a\le b+1$.