$\require{AMScd}$I have a question regarding a claim in A first course of homological algebra by Northcott. I think it's very easy, since the author didn't provide a proof, and just kind of claimed it. It's on page 76. And it says:
Let $0 \to A_1 \to P \to A \to 0$ is exact, and $P$ is projective, then $\mbox{Pd}(A) = \mbox{Pd}(A_1) + 1$, provided that $\mbox{Pd}(A) > 0$.
Where $\mbox{Pd}(A)$ denotes the projective dimension of $A$, i.e if $A$ has a projective resolution $0 \to P_n \to P_{n-1} \to... \to P_1 \to P_0 \to A \to 0$, where all $P_i$'s are projective, and there's no shorter resolution. Then we'll write $\mbox{Pd}(A) = n$.
Question
Ok, so how can I prove it?
I can get to the point that $\mbox{Pd}(A) \le \mbox{Pd}(A_1) + 1$, by assuming that $\mbox{Pd}(A_1) = n$, i.e there exists a projective resolution of $A_1$, which looks like this: $0 \to P_n \to P_{n-1} \to... \to P_1 \to P_0 \to A_1 \to 0$, since $P_0 \to A_1$ is epic, and $A_1 \to P$ is monic, I can combine the 2 mappings, and arrive the following resolution: $0 \to P_n \to P_{n-1} \to... \to P_1 \to P_0 \to P \to A \to 0$, which means that $\mbox{Pd}(A) \le n+1$.
But I cannot prove that $\mbox{Pd}(A) = n+1$. Here's how I start, assume that $A$ has a resolution: $0 \to P_n \to P_{n-1} \to... \to P_1 \to P_0 \to A \to 0$, and I'll try to construct a resolution for $A'$ that lacks $P_0$, i.e, what I'm trying to prove is, if $\mbox{Pd}(A) < n+1$, then $\mbox{Pd}(A_1) < n$ (which is, of course not right, since we assume $\mbox{Pd}(A_1) = n$). So here's how it goes:
$\begin{CD}@. @. @. 0 @>>> A_1 @>\chi>> P @>\sigma>> A @>>> 0\\ @. @. @. @. @. @. @| \\ 0 @>>> P_n @>\chi_n>> P_{n-1} @>\chi_{n-1}>> ... @>\chi_2>> P_1 @>\chi_1>> P_0 @>\chi_0>> A @>0>> 0 \\ \end{CD}$
Since $\sigma$ is epic, and $P_0$ is projective, hence there exists: $\gamma_0: P_0 \to P$, such that the square commutes:
$\begin{CD}@. @. @. 0 @>>> A_1 @>\chi>> P @>\sigma>> A @>>> 0\\ @. @. @. @. @. @A\gamma_0AA @| \\ 0 @>>> P_n @>\chi_n>> P_{n-1} @>\chi_{n-1}>> ... @>\chi_2>> P_1 @>\chi_1>> P_0 @>\chi_0>> A @>0>> 0 \\ \end{CD}$
Since $\sigma (\gamma_0 \chi_1) = 0$, and the former row is exact, hence there exists $\gamma_1: P_1 \to A$, such that the square commutes.
$\begin{CD}@. @. @. 0 @>>> A_1 @>\chi>> P @>\sigma>> A @>>> 0\\ @. @. @. @. @A\gamma_1AA @A\gamma_0AA @| \\ 0 @>>> P_n @>\chi_n>> P_{n-1} @>\chi_{n-1}>> ... @>\chi_2>> P_1 @>\chi_1>> P_0 @>\chi_0>> A @>0>> 0 \\ \end{CD}$
One problem remains is that $\gamma_1$ is not epic, so, I cannot have the resolution: $0 \to P_n \to ... \to P_1 \to A_1 \to 0$.
Can my proof be improved, or should I tackle it in a complete different way?
Thank you guys very much in advance,
And have a good day,
Edit
I still don't see how we can use Schanuel's lemma to tackle this problem, ok so I assume $\mbox{Pd}(A) < n + 1$, i.e, I'll have the resolution $\begin{CD} @. @. @. @. @. 0 \\ @. @. @. @. @. @VVV \\ 0 @>>> P_n @>>> P_{n - 1} @>>> ... @>>> P_0 @>>> A_1 @>>> 0 \\ @. @. @. @. @. @VVV \\ @. @. @. @. @. P \\ @. @. @. @. @. @VVV \\ 0 @>>> Q_n @>>> Q_{n - 1} @>>> ... @>>> Q_0 @>>> A @>>> 0 \\ @. @. @. @. @. @VVV \\ @. @. @. @. @. 0 \end{CD}$
How can I apply Schanuel's lemma to find a resolution for $A_1$ that has length no larger than $n - 1$?
Given a minimal resolution $Q^\bullet$ of $A_1$, tagging it onto $P \rightarrow A$ gives a resolution of $A$, so $\dim A_1 + 1 \geqslant \dim A$.
To see that $\dim A_1 + 1 \leqslant \dim A$, compare $Q^\bullet$ with a minimal resolution: $$0 \rightarrow P_{\dim A} \rightarrow \ldots P_1 \rightarrow P_0 \rightarrow A \rightarrow 0$$$$0 \rightarrow \ker Q_{\dim A - 2} \rightarrow \ldots Q_0 \rightarrow P \rightarrow A \rightarrow 0$$By the Generalized Schanuel's lemma, we know that $\ker Q_{\dim A - 2}$ is projective, so $0 \rightarrow \ker Q_{\dim A - 2} \rightarrow \ldots Q_0 \rightarrow A_1 \rightarrow 0$ is a resolution of $A_1$ of length $\dim A - 1$, as desired.