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We prove $(ii)\Rightarrow (i)$. The central block idempotents split every finite-dimensional module into its block components. A nonzero indecomposable representation, in particular an indecomposable projective module, has only one such component; otherwise these give a nontrivial direct-sum decomposition.

Every Jordan–Hölder factor of a module in a block belongs to that block, since its central idempotent acts as identity on submodules and quotients. Thus any two simple factors of the same projective indecomposable are in the same block. Applying this along the proposed chain yields $\boxed{S,T\text{ belong to the same block}}$.

There is a natural isomorphismobtained by evaluating a homomorphism against a covector. Since $P$ is a direct summand of a free module and tensoring a free $kG$-module with $U^{\ast }$ using the diagonal action again gives a free module, $P\otimes U^{\ast }$ is a projective module.

Lift this projective module to an $\mathcal{O}G$-lattice. Projectivity makes the dimension of the homomorphism space equal to the multiplicity of the trivial representation after extending scalars to $K$. The lifted ordinary character vanishes on p-singular elements, while on p-regular elements it is the productOrdinary character orthogonality and the substitution $g\mapsto g^{-1}$ therefore giveEquivalently, this is the Brauer character inner product between the projective character of $P$ and the Brauer character of $U$.

True. Fix an isomorphism $\varphi :M{\otimes }_{R}N\to R^n$, and write the inverse image of the first standard basis vector asDefineSince $\psi ((n_j{)}_j)=1$, this is a split surjection. Tensor its splitting with $M$. The resulting split surjection has the formThus $M$ is a direct summand of a finite free module, so it is a projective module. Symmetry gives the same conclusion for $N$. This is projectivity of factors of a nonzero finite free tensor product.

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.

The finite complex computing cohomology in a proper flat family gives a bounded complex $K^{\bullet }$ of finite locally free $A$-modules such that, for every $A$-module $M$,In particular, $M=A$ computes $H^p(X,\mathcal{F})$ and $M=\kappa (s)$ computes $H^p(X_s,\mathcal{F}(s))$.

Since $H^p(K^{\bullet })=0$ for $p>n$, the finite exact tail above degree $n$ can be split successively: its last differential is surjective onto a projective module, hence splits, and induction moves left. Removing the resulting contractible summands leaves a finite locally free complex ending in degree $n$. Thereforeand after tensoring with $\kappa (s)$ the same formula computes the fiber cohomology.

If $H^n(X,\mathcal{F})=0$, the last differential is surjective, and remains so after every base change; hence every $H^n(X_s,\mathcal{F}(s))$ vanishes. Conversely, if all fiber groups vanish, the finitely generated cokernel $C$ satisfies $C{\otimes }_A\kappa (s)=0$ for every $s\in \mathrm{Spec}A$. Localizing and applying Nakayama lemma gives $C_s=0$ for every $s$, so $C=0$. Thus

Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its conjugacy classes. Hence $N$ is the direct sum of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its centralizer. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb{Z}$. Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

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.

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 ${\mathrm{Ext}}^1$.

A short exact sequence presenting a module as a quotient of a projective module. The kernel need not be projective. Continuing a projective presentation gives a projective resolution.

A projective resolution of an $R$-module $M$ is an exact sequencein which every $P_i$ is a projective module.

If $M{\otimes }_{R}N\cong R^n$ for a positive integer $n$, then both $M$ and $N$ are projective modules. Choose a finite expression for the inverse image of one basis vector. It produces a split surjection $N^r\to R$; tensoring the splitting with $M$ exhibits $M$ as a direct summand of the finite free module $(M{\otimes }_{R}N{)}^r$. Symmetry gives the result for $N$.

Let $W,W^{\prime }$ be choices of integral form of a group representation over a complete p-modular system. Then $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$ is finite free, and $K{\otimes }_{\mathcal{O}}I={\mathrm{Hom}}_{KG}(K\otimes W,K\otimes W^{\prime })$: clearing denominators proves the spanning assertion. Also ${\mathrm{Hom}}_{\mathcal{O}G}(W,\pi W^{\prime })=\pi I$. If $W$ is a projective module, applying the Hom functor to $0\to \pi W^{\prime }\to W^{\prime }\to W^{\prime }/\pi W^{\prime }\to 0$ yieldsConsequently these ordinary and modular Hom spaces have the same dimension of a vector space. Projectivity is essential: for $G=C_2$ over ${\mathbb{Z}}_2$, the trivial and sign lattices have zero Hom between them, whereas their reductions in characteristic $2$ coincide and have a one-dimensional Hom space.

Shapiro's lemma gives a natural isomorphismIt follows by applying the Hom functor adjunction for a coinduced module to a projective resolution and observing that restriction from $G$ to $K$ preserves projective modules.

Set $A=kQ$, $P_i=A{e}_i$, $X_i=e_{i}X$. There is an exact sequencewhere $d(p\otimes x)=p\rho \otimes x-p\otimes f_{\rho }(x)$ in the source and target summands, and the augmentation is $p\otimes x\mapsto px$. Each $P_i$ is a summand of $A$, so the two terms are projective modules. This also works for infinite-dimensional modules.

For a nonzero finitely generated indecomposable module $P$ that is a projective module over a right Artinian ring $A$, its top $P/PJ(A)$ is a simple module. The Jacobson radical $J(A)$ is a nilpotent ideal, so the top is nonzero; the quotient is a semisimple module over the semisimple ring $A/J(A)$. If it decomposed, a nontrivial idempotent of its endomorphism ring would lift by projectivity to an endomorphism of $P$. The Hopkins-Levitzki theorem gives finite composition length, and the Fitting lemma makes the lift either invertible or nilpotent. Its induced map on the top would have the same property, impossible for that nontrivial idempotent.

The vertex projective module of a path algebra $P(i)=kQ{e}_i$ has at vertex $j$ the span of paths from $i$ to $j$. Each arrow appends itself to such paths. It is a projective module because it is a summand of the free module $kQ$. Its endomorphism ring is $(e_{i}kQ{e}_i{)}^{\mathrm{op}}$, acting by right multiplication. The closed-path corner of a path algebra is a domain, so this ring has no nontrivial idempotents and $P(i)$ is indecomposable. Directed cycles may make these projectives infinite-dimensional.