I'm studying a bit of homological algebra and I'm now studying about the projective dimension of an $R$-module $M$.
This is how it is defined:
Since the category $R-\operatorname{Mod}$ has enough projectives, for any $R$-module $M$ we can write a chain complex like this:
$$\cdots \longrightarrow P_{n} \overset{d_{n}}{\longrightarrow} P_{n-1} \overset{d_{n-1}}{\longrightarrow} \cdots \overset{d_{2}}{\longrightarrow}P_1 \overset{d_{1}}{\longrightarrow}P_0 \overset{\epsilon}{\longrightarrow}M{\longrightarrow} \space 0$$
This induces the following chain complex:
$$0 \longrightarrow \operatorname{Hom}(P_{0},C) \overset{d_{1}'}{\longrightarrow} \operatorname{Hom}(P_{1},C) \overset{d_{2}'}{\longrightarrow} \cdots \overset{d_{2}}{\longrightarrow} \operatorname{Hom}(P_{n},C) \overset{d_{n}'}{\longrightarrow} \operatorname{Hom}(P_{n+1},C) {\longrightarrow} \cdots $$
where $d_{\space n}^{\space '}(f) = f\circ d_{\space n}$.
The n-th homology of this chain complex is defined as $\operatorname{Ext}^n(A,C)$.
The smallest number that $\operatorname{Ext}^n(A,C)=0$ for all $C$ is called the projective dimension of $M$ and is denoted as $\operatorname{pd}(M)$.
Now my question is whether $\operatorname{Ext}^n(A,C) = 0 \implies \operatorname{Ext}^{n+j}(A,C) = 0$ holds for $j \in \mathbb{N}$ or not?
The correct definition of the projective dimension of $M$ is the smallest $d$ such that $M$ has a projective resolution of length $d$, or equivalently as the smallest $d$ such that $\mathrm{Ext}^{d+1}(M,-)=0$. (And hence $\mathrm{Ext}^d(M,-) \neq 0$.)
The implication $\mathrm{Ext}^n(M,-)= 0 \Rightarrow \mathrm{Ext}^{n+1}(M,-)=0$ is true. See Weibel's book on homological algebra, Lemma 4.1.6.
However, if $C$ is a fixed module, then $\mathrm{Ext}^n(M,C)=0 \Rightarrow \mathrm{Ext}^{n+1}(M,C)=0$ might fail. For example, we have (for some fixed integer $p>1$) $\mathrm{Ext}^0(\mathbb{Z}/p,\mathbb{Z})=\hom(\mathbb{Z}/p,\mathbb{Z})=0$, but $\mathrm{Ext}^1(\mathbb{Z}/p,\mathbb{Z})=\mathbb{Z}/p$.