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A finite projective module is a finitely generated module that is a projective module. Equivalently it is a direct summand of a finite free module: a finite free surjection splits by projectivity. Its dual module is finite projective and its evaluation map to its double dual module is an isomorphism, since both properties hold for finite free modules and pass to direct summands. Finite generation is needed for this elementary duality argument.

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.

A finitely generated module $M$ over a commutative ring $A$ is reflexive when its evaluation homomorphism $M\to M^{\ast \ast }$ to the double dual module is an isomorphism.

For a nonzero left finitely generated module over a left Noetherian ring, take a finite projective resolution with finite projective modules. Vanishing of its top Ext functor would make the dual of its final injection surjective. These duals are right finite projective modules, so the surjection splits. Dualizing back and using finite projective double dual modules splits the original injection and shortens the resolution, contradicting the definition of projective dimension. When the dimension is zero, a coordinate map of a finite free module containing the nonzero projective summand supplies a nonzero map to $R$. The left/right distinction in the duals is essential for a noncommutative ring.