Projective module

A module $P$ is projective when every map from $P$ through the target of a surjection lifts through that surjection.

A projective module is a direct summand of a free module. Free modules are flat because tensoring with a free module produces a direct sum of copies of the original module and map. If $F=P\oplus Q$ is flat, tensoring an injection with $F$ is the direct sum of its tensor maps with $P$ and $Q$, so each component is injective. Applying this to a free module containing $P$ proves that $P$ is flat.

Every module $P$ is a quotient of a free module $F={⨁}_{p\in P}A{e}_p$, via $e_p\mapsto p$. If $P$ is a projective module, lift $1_P$ through this surjection to obtain a section $s:P\to F$. The resulting split short exact sequence identifies $F$ with $\ker \pi \oplus P$. Conversely, if $P$ is a summand of a free module, extend a prescribed map out of $P$ to the free module using the summand projection, lift each basis image through the given surjection, and restrict the lift to $P$. This proves the equivalence with the lifting definition of a projective module, for arbitrary ranks.

For a finite-dimensional algebra $A$, a principal indecomposable module is an indecomposable direct summand of the left regular module, equivalently $Ae$ for a primitive idempotent $e$. Its head of a module is simple, and it is the projective cover of that simple module.

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.

An invertible module over a commutative ring $A$ is a finitely generated module that is a projective module and whose localization at a prime ideal is free of rank one at every prime. Its dual module $L^{\ast }={\mathrm{Hom}}_A(L,A)$ is also finite projective. Evaluation $L{\otimes }_A{L}^{\ast }\to A$ is an isomorphism after every prime localization, hence is an isomorphism globally by localization detects zero elements. The Picard group of a ring consists of the isomorphism classes of these modules under the tensor product of modules.

For a ring homomorphism $A\to B$, extension of scalars takes an invertible module to an invertible $B$-module. A direct-summand presentation in a finite free module remains a direct-summand presentation after tensoring. At a prime $\mathfrak{q}\subset B$ contracting to $\mathfrak{p}$, localization gives $B_{\mathfrak{q}}{\otimes }_{A_{\mathfrak{p}}}{L}_{\mathfrak{p}}\cong B_{\mathfrak{q}}$. Compatibility with tensor products, identities and composition makes the Picard group of a ring a covariant functor.

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 $A$.

Lift the image of each basis element independently through the given surjection, then extend the chosen lifts linearly.

A projective cover of a module $M$ is an essential surjection $P\twoheadrightarrow M$ from a projective module. For a finite-dimensional algebra, every finite-dimensional module has one; the projective cover of a simple module is indecomposable and has that simple module as its head.

The head, or top, of a finite-length module $M$ is its largest semisimple quotient. It is $M/J(M)$, where $J(M)$ is the radical of a module.