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Choose a surjection $P_0=A^{r_0}\twoheadrightarrow M$. Since $A$ is a Noetherian ring, a finite free module is a Noetherian module, and its kernel $K_1$ is a finitely generated module. Choose a finite free surjection $P_1\twoheadrightarrow K_1$ and repeat. This constructs a resolution by finitely generated free moduleswhose terms are, in particular, finitely generated projective modules. It may be infinite; the request is finite generation of every term, not finite length.

Now suppose the nonzero finitely generated module $M$ has finite projective dimension $n$. For $n\ge 1$, its $n$th syzygy $K_n$ in this free resolution is projective. Indeed, dimension shifting along the successive kernel sequences givesfor every module $N$. The vanishing follows by computing the latter Ext functor from any length-$n$ projective resolution. Vanishing of ${\mathrm{Ext}}^1(K_n,-)$ makes every short exact sequence ending in $K_n$ split: apply the long exact sequence of Ext groups to a free surjection onto $K_n$ to lift its identity. Thus $K_n$ is a projective module. It is finitely generated by the preceding construction. Truncating giveswith every term finite projective.

Apply the Hom functor ${\mathrm{Hom}}_A(-,A)$ and write $P_i^{\ast }={\mathrm{Hom}}_A(P_i,A)$. At the top degree,If this group were zero, $d_n^{\ast }$ would be surjective. The dual module of a finite projective module is finite projective, so this surjection has a section $s:P_n^{\ast }\to P_{n-1}^{\ast }$. Finite projective modules identify with their double dual modules: the evaluation map is an isomorphism for a finite free module and passes to its direct summands. Dualizing $d_n^{\ast }s=\mathrm{id}$ therefore gives $s^{\ast }{d}_n={\mathrm{id}}_{P_n}$. Hence $d_n$ splits, and $P_{n-1}/d_n(P_n)$ is projective. Replacing the last two terms by this quotient shortens the projective resolution to length $n-1$, contradicting minimality. This proves top Ext detects finite projective dimension:For $n=0$, $M$ is a nonzero finite projective module and embeds as a summand of $A^r$. Some coordinate projection restricts nontrivially to $M$, proving ${\mathrm{Hom}}_A(M,A)\ne 0$ as required.

For infinite projective dimension, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and $M=k=A/(\epsilon )$. Multiplication by $\epsilon$ has image and kernel both equal to $(\epsilon )$, giving the periodic resolution over dual numbersApplying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. A finite projective resolution would force all sufficiently high Ext functors to vanish. Thus ${\mathrm{pd}}_{k[\epsilon ]/({\epsilon }^2)}k=\infty$.

The finite-generation hypothesis is essential if the middle sentence's “a module” is interpreted as an unrestricted new module. A concrete vanishing top Ext without finite generation occurs for $A=k[[t]]$ and $K=A[1/t]=k((t))$. This Noetherian ring has the exact free presentationInjectivity of $d$ follows by examining the largest index of a finitely supported input, since $t$ is a non-zero-divisor. For exactness in the middle, use the relations to move a finite sum to its largest index $N$; its remaining coefficient is ${\sum }_{i=0}^N{a}_i{t}^{N-i}$, which is zero precisely when its image in $K$ is zero. The map onto $K$ is surjective because every Laurent series has a lower bound on its exponents.

Applying ${\mathrm{Hom}}_A(-,A)$ gives on products the map $(b_i)\mapsto (b_i-t{b}_{i+1})$. It is surjective: for any $(c_i)$, set $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$. These are well-defined formal power series, because each coefficient receives only finitely many contributions, and $b_i-t{b}_{i+1}=c_i$. Consequently ${\mathrm{Ext}}_A^1(K,A)=0$. But ${\mathrm{Hom}}_A(K,A)=0$: the image of every $t^{-i}$ must lie in ${\bigcap }_{r\ge 0}{t}^{r}A=0$, since it is divisible by every power of $t$. A nonzero projective module is a summand of a free module and therefore has a nonzero coordinate map to $A$. Thus $K$ is not projective, while its displayed resolution has length one. In particular,The nonvanishing conclusion proved above retains the finitely generated $M$ introduced at the start of the question.

Put $A=k[[t]]$ and $K=A[1/t]$. A free resolution has countably many generators $e_i$ and relations $e_i-t{e}_{i+1}$. Applying the Hom functor with coefficients in $A$ gives $(b_i)\mapsto (b_i-t{b}_{i+1})$ on sequence products. This is surjective: for prescribed $(c_i)$ use $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$, a well-defined formal power series. Therefore the degree-one Ext functor vanishes. But ${\mathrm{Hom}}_A(K,A)=0$, since every image is infinitely divisible by $t$. A nonzero projective module has a nonzero coordinate map into $A$, so $K$ is not projective. Its displayed resolution proves that its projective dimension is exactly one. The finite-generation hypothesis in top Ext detects finite projective dimension is essential.