Modules with projective dimension $n$ have not vanishing $\mathrm{Ext}^n$ 
Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with projective dimension $n$. Then for every finitely generated $R$-module $N$ we have $\mathrm{Ext}^n(M,N)\neq 0$. Why?

By definition, if the projective dimension is $n$ this means that $\mathrm{Ext}^{n+1}(M,-)=0$ and $\mathrm{Ext}^n(M,-)\neq 0$, so there exists an $N$ such that $\mathrm{Ext}^n(M,N)\neq 0$. Why is this true for every $N$?
I found this theorem on this notes, page 6 proposition 9. Did I misunderstand it? Is there a similar statement?
This result is used on page 41, implication 3 implies 4, and is used in the case $N=R$. Is it true in this case?
 A: Take $R=\mathbb Z\times\mathbb Z$, consider the elements $e_1=(1,0)$, $e_2=(0,1)\in R$, and the modules  $M=R/(2e_1)$ and $N=Re_2$. Show that the projective dimension of $M$ is $1$ and compute $\operatorname{Ext}_R^1(M,N)$.
A: Yes, for $N=R$ the result is true.

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with $\mathrm{pd}_R(M)=n$. Then $\mathrm{Ext}_R^n(M,R)\neq 0$. 

Proof. First consider $R$ local with maximal ideal $\mathfrak{m}$. Let $F_{\bullet}$ be a minimal free resolution of $M$ and $\varphi_n:F_n\rightarrow F_{n-1}$ the last homomorphism. One knows that the elements of the matrix associated to $\varphi_n$ are all in $\mathfrak{m}$. $\mathrm{Ext}_R^n(M,R)$ is the cokernel of the dual map $\varphi_n^*:F_{n-1}^*\rightarrow F_n^*$ whose matrix is the transpose of the matrix associated to $\varphi_n$. Therefore the elements of the matrix associated to $\varphi_n^*$ are also in $\mathfrak{m}$, and this shows that $\mathrm{Coker}\ \varphi_n^*\neq 0$.
Now it is easy to deduce the general case from the local case via the well known fact that $\mathrm{pd}_R(M)= \sup_{\mathfrak{m}\in\mathrm{Max}(R)}\mathrm{pd}_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$.
A: Suppose $R$ is local noetherian with maximal ideal $\def\m{\mathfrak m}\m$ and residue field $k$, and let $M$ be a f.g. $R$-module with $\operatorname{pdim}M=n$. Then $\def\Ext{\operatorname{Ext}}\Ext_R^{n}(M,k)\neq0$.
Let $N$ be a non-zero f.g. module. There is a submodule $N'\subseteq N$ such that $N/N'\cong k$ as $R$-modules. From the short exact sequence $0\to N'\to N\to k\to 0$ we get an exact sequence $$\Ext^n_R(M,N)\to\Ext^n_R(M,k)\to\Ext^{n+1}_R(M,N')=0$$ Since the middle term is not zero and the leftmost map is surjective, we see that $\Ext^n_R(M,N)\neq0$.
Remove now the hypothesis that $R$ be local and let again $M$ be finitely generated and of projective dimension $n$. There is a maximal ideal $\def\m{\mathfrak m}$ in $R$ such that $\def\pdim{\operatorname{pdim}}\pdim_RM=\pdim_{R_\m}M_\m$.
Then $\def\Ext{\operatorname{Ext}}\Ext^n_R(M,R)_\m\cong\Ext^n_{R_\m}(M_m,R_m)\neq0$, by the local case.
