The situation I'm considering is quite involved. All rings are noetherian commutative with $1$. All modules are finitely generated.
First of all we fix a non reduced local ring $A$ where all zero divisors are nilpotent element. Let $A^\prime$ be a quotient of $A$ by a principal ideal $t$, where $t$ is not a zero divisor in $A$.
Let $M$ be an $A$-module and suppose that one has a presentation of $M$ given by an exact sequence $$ 0\to N'\to N\xrightarrow{\cdot(h)} N \to M\to 0$$ of $A$-modules, where the middle arrow is the multiplication by an element $h\in A$ that is a zero divisor of $N$ (I'm just saying that I don't want to assume $N'=\ker (\cdot h)$ to be $0$).
Assume the following extra conditions: a) $N$ is an $A'$-module of finite projective dimension. b) $N'$ is an $A'$-module.
Questions:
1) is it true that $N'$ has finite projective dimension as $A'$-module?
2) is it true that $M$ is an $A'$-module of finite projective dimension?
3) can we deduce the existence of finite projective resolutions for the above objects as $A$-modules?
In case the questions do have negative answers, in which way one can add assumptions on the objects so that things do magically become true?
EDIT: I realized that I can actually ask a more precise question. I keep the previous notation, but I write specific examples for the modules:
$$M=A/(t - x^b) $$ for $x$ nilpotent (say with $x^m=0$ and $x^n\neq 0$ for $n<m$ and $b<m$).
$$N = A/(t^a-x^{ab}) = A/(t^a)$$ for some $a$ with $ab>m$. Finally, I can take for $h$ an element of $A$ such that $h(t-x^b) = (t^a-x^{ab})$.
So I keep the general questions, but I'm curious to see an answer also in this specific case.