Projective dimension (${\mathrm{pd}}_{R}M$)

For a nonzero module, the projective dimension ${\mathrm{pd}}_{R}M$ is the shortest length of a projective resolution of the module $M$, or infinity if no finite projective resolution exists. Equivalently, it is the supremum of degrees with nonzero ${\mathrm{Ext}}_R^n(M,N)$ as $N$ ranges over all modules. This equivalence follows by dimension shifting and the characterization of projective modules by vanishing ${\mathrm{Ext}}^1$.

Let $M$ be a nonzero finitely generated module over a commutative Noetherian ring. Take a length-$n$ resolution with finite projective terms. If its top Ext functor with coefficients in $A$ 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 $n$. When $n=0$, a nonzero coordinate of a finite free module containing $M$ as a summand gives a nonzero map $M\to A$.

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.

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.

The global dimension of a ring is the supremum of the projective dimensions of all its modules, allowing infinity. It measures the longest projective resolution needed by any module. For a field it is zero, since every module is a vector space and hence free.

If global dimension is zero, every module is a projective module. Each short exact sequence splits because its quotient is projective. Conversely, splitting every short exact sequence makes each module a direct summand of a free module and therefore a projective module. Splitting also makes every module an injective module: for $A\hookrightarrow B$ choose a retraction $B\to A$ and compose it with any prescribed map out of $A$.