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An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For central $t_1,\dots ,t_n$, define the Koszul complex on central ring elements using formal exterior symbols:with zero terms outside $0\le k\le n$ and differentialThe terms are free bimodules with the formal symbols commuting with coefficients. Centrality makes $d$ a bimodule map, and the terms in $d^2$ cancel in pairs because the $t_i$ commute. This defines the Koszul complex over a possibly noncommutative ring without using an undefined exterior algebra of arbitrary one-sided modules. In particular $K(\mathbf{t}){\otimes }_{R}M$ is a well-defined chain complex of left modules.

A regular sequence on a module means that multiplication by $t_i$ is injective on $M/(t_1,\dots ,t_{i-1})M$ for each $i$, usually with the additional convention that the final quotient is nonzero. The homology vanishing below uses only the injectivity conditions. For one element, the chain complex is $M\stackrel{t_1}{\to }M$, with zero first homology and zeroth homology $M/t_{1}M$.

For the induction let $C=K(t_1,\dots ,t_{n-1}){\otimes }_{R}M$. Adjoining the last generator identifies the new Koszul complex with the mapping cone of multiplication by $t_n$ on $C$. In the convention $\mathrm{Cone}(t_n{)}_k=C_k\oplus C_{k-1}$ its differential is $(a,b)\mapsto (da+t_{n}b,-db)$; the identification is $a+e_n\wedge b$. The long exact sequence in homology of this mapping cone givesThe induction hypothesis is $H_k(C)=0$ for $k>0$ and $H_0(C)=N=M/(t_1,\dots ,t_{n-1})M$. Regularity makes $t_n:N\to N$ injective. Hence the new positive homology is zero and its zeroth homology is $N/t_{n}N$. The augmentation to this quotient induces these homology isomorphisms, giving the quasi-isomorphismwhere the right side is placed in degree zero. The results used are the long exact sequence in homology of a degreewise short exact sequence of complexes and its mapping cone form, together with the stated induction; no unmentioned acyclicity criterion is needed.

For $R=\mathbb{Z}[X]$, $(p,X)$ is a regular sequence: $p$ is a non-zero-divisor in $R$, and multiplication by $X$ is injective in $(\mathbb{Z}/p\mathbb{Z})[X]$, even when $p$ is composite. The Koszul resolution of $R/(p,X)$ isWith $N=\mathbb{Z}/q\mathbb{Z}$ and $X$ acting as zero, applying the Hom functor gives the cochain complexThus, more generally, the Ext groups between polynomial-ring residue modules areand vanish in all degrees above two. For the equal-modulus case,These are $R$-modules through $X=0$ and reduction modulo $p$. If the multiplicative self-Ext functor is desired, the Koszul self-Ext algebra is the exterior algebra on two degree-one generators over $\mathbb{Z}/p\mathbb{Z}$. The two contractions on the Koszul resolution lift these classes, square to zero and anticommute, so this description also holds for composite $p$ and characteristic two. For coprime $p,q$, multiplication by $p$ is invertible on $N$, so both its kernel and cokernel vanish:

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

For the $R$-$R$-bimodule $M$, the Hochschild chain complex iswith boundaryThenThe Hochschild cochain complex is $C^n(R,M)={\mathrm{Hom}}_k(R^{{\otimes }_{k}n},M)$ withand $H{H}^n(R,M)=H^n(C^{\bullet }(R,M),\delta )$.

A derivation into a bimodule is a $k$-linear map $d:R\to M$ satisfyingIt is an inner derivation when $d=d_m$ for some $m\in M$, where $d_m(r)=rm-mr$. The degree-one cocycle equation is precisely the Leibniz rule, and the degree-one coboundaries are the inner derivations. Therefore the First Hochschild cohomology as outer derivations is

Let $\mu :R{\otimes }_{k}R\to R$ be multiplication and $\Omega =\ker \mu$, equipped with the outer bimodule structure. The universal bimodule derivationsatisfies $D(rs)=rD(s)+D(r)s$. Composition with $D$ definesFor a derivation $d$, its inverse image under this map isIf ${\sum }_i{r}_i{s}_i=0$, the Leibniz rule shows that this formula is both left and right $R$-linear. Moreover every element of $\Omega$ is ${\sum }_{i}D(r_i)s_i$, proving existence and uniqueness.

For $d_m(r)=rm-mr$, the corresponding map isThese are exactly the maps $\Omega \to M$ which extend to bimodule maps $R{\otimes }_{k}R\to M$.

For $a,b\in I$, the square-zero ideal condition givesThus the square-zero unit subgroup $1+I$ is abelian, andis a group isomorphism from the additive group of $I$.

Use the specified ring isomorphism $R/I\cong \mathbb{Z}G$. For $x\in \mathbb{Z}G$, choose a lift $r\in R$ and define $x\cdot a=ra$ for $a\in I$. Two lifts differ by an element of $I$, whose product with $a$ vanishes, so this is well defined. Right multiplication is handled identically. The two actions commute by associativity, making $I$ a bimodule. If $u\in R$ lifts $g\in G$, then $u$ is a unit: a lift $v$ of $g^{-1}$ makes both $uv$ and $vu$ elements of $1+I$, hence units, and a ring element with both a left and a right inverse is invertible. Conjugation therefore definesChanging $u$ by an element of $I$ does not change this expression because $I^2=0$. Moreover,so $a\mapsto 1+a$ is an isomorphism of $\mathbb{Z}G$-modules for these conjugation actions.

Let $R^{\times }\to (R/I{)}^{\times }$ be reduction on unit groups, and define $U$ as the inverse image of the distinguished subgroup $G\subseteq (\mathbb{Z}G{)}^{\times }$. Every $g\in G$ has a unit lift by the preceding argument, and the kernel consists exactly of the units congruent to $1$, namely $1+I$. Multiplication in $R$ therefore gives the group extensionChoose a set-theoretic section $s:G\to U$ with $s(1)=1$. Its extension cocyclesatisfies the two-cocycle identity by associativity. A different section changes $c$ by a group coboundary, so second group cohomology classifies group extensions gives a well-defined classThe same construction for $(R_1,I_1)$ gives $U_1$, an extension cocycle $c_1$, and $x_1=[c_1]\in H^2(G,1+I_1)$.

The answer to the final question is no. An abstract ring isomorphism $R_1\cong R$ need not carry the distinguished ideal $I_1$ to $I$, need not induce the identity under the two chosen identifications of the quotient rings with $\mathbb{Z}G$, and need not induce the prescribed $\mathbb{Z}G$-module isomorphism $\theta$. Hence it need not give an isomorphism of the two displayed group extensions, so it imposes no equality $\varphi (x_1)=x$. That equality does hold if the ring isomorphism has all three compatibility properties, because it then carries one extension cocycle to the other up to a group coboundary.

For a block idempotent $b$ with defect $D$, the Brauer quotient of a module of its modular block bimodule at $\Delta D$ is nonzero: on the fixed basis $C_G(D)$, the block projection acts by ${\mathrm{Br}}_D(b)$. Restriction to $D\times D$ splits into transitive permutation modules; their stabilizers have order at most $|D|$. A nonzero quotient at $\Delta D$ therefore forces a summand with stabilizer exactly $\Delta D$, which is the regular $kD$ bimodule. Tensoring on the right with any $kD$-module $U$ gives $U\mid {\mathrm{Res}}_D^G(b{\mathrm{Ind}}_D^{G}U)$. Consequently finite representation type of the modular block forces finite representation type of $kD$.

Fix an $R$-bimodule $M$. A unital ground-ring-linear split extension with square-zero kernel $M$ can be identified with $R\oplus M$, with multiplication $(a,u)(b,v)=(ab,av+ub+f(a,b))$. Associativity is exactly $\delta f=0$. Changing a unital linear section by $g:R\to M$ changes $f$ by $\delta g$. Thus equivalence classes of these split extensions are $H{H}^2(R,M)$. Over a field every such vector-space extension splits; over a general ground ring a splitting hypothesis is essential for this ordinary cochain classification.

For a unital associative algebra $A$ and prescribed $A$-bimodule $M$, an extension is $0\to M\to E\to A\to 0$ where $E$ is unital, $E\to A$ is unital, $M$ is a square-zero ideal, and the induced actions are the prescribed ones. Equivalences induce the identity on $A$ and $M$. A unital linear section produces a normalized Hochschild cocycle $\mu (a,b)=s(a)s(b)-s(ab)$. Conversely, $(a,m)(b,n)=(ab,an+mb+\mu (a,b))$ on $A\oplus M$ defines the extension. Changing section changes $\mu$ by a coboundary, giving classification by $H{H}^2(A,M)$.