Let be a nonzero finitely generated module over a commutative Noetherian ring. Take a length- resolution with finite projective terms. If its top Ext functor with coefficients in vanished, the dual of its last injection would be surjective. The dual of a finite projective module is projective, so that surjection splits. Finite projective modules are isomorphic to their double duals, so dualizing gives a splitting of the original injection. The resolution then shortens, contradicting minimality of . When , a nonzero coordinate of a finite free module containing as a summand gives a nonzero map .
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 . The left/right distinction in the duals is essential for a noncommutative ring.
Put and . A free resolution has countably many generators and relations . Applying the Hom functor with coefficients in gives on sequence products. This is surjective: for prescribed use , a well-defined formal power series. Therefore the degree-one Ext functor vanishes. But , since every image is infinitely divisible by . A nonzero projective module has a nonzero coordinate map into , so 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.

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