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The order-$\epsilon$ part of associativity is $af(b,c)-f(ab,c)+f(a,bc)-f(a,b)c=0$, so $f$ is a Hochschild cocycle. A change of coordinates $T=1+\epsilon g$ changes $f$ to $f+\delta g$. Therefore equivalence classes of first-order associative deformations of a fixed underlying algebra are its second Hochschild cohomology. Unital deformations use normalized cochains.

A Gerstenhaber algebra has an associative product making it a graded commutative algebra and a degree-minus-one graded Lie bracket. In the left convention, $[u,vw]=[u,v]w+(-1{)}^{(|u|-1)|v|}v[u,w]$. The shifted degrees $|u|-1$ determine the signs of antisymmetry and the Jacobi identity. Hochschild cohomology and the exterior algebra of polynomial derivations are examples.

Write $f\circ g={\sum }_{i=0}^{p-1}(-1{)}^{i(q-1)}f{\circ }_{i}g$ for insertion of a degree-$q$ cochain into a degree-$p$ one. With the unsigned Hochschild cup product and the left graded Leibniz rule, take $[f,g]=(-1{)}^{(p-1)(q-1)}f\circ g-g\circ f$. It descends to Hochschild cohomology and gives a Gerstenhaber algebra. The alternative insertion bracket $f\circ g-(-1{)}^{(p-1)(q-1)}g\circ f$ differs by the displayed degree sign and obeys the corresponding right rule. Degree-one brackets are commutators of derivations in either convention.

A graded algebra is a graded commutative algebra if $uv=(-1{)}^{pq}vu$ for homogeneous elements of a graded algebra of degrees $p,q$. In characteristic not equal to two, every odd-degree element then squares to zero. The exterior algebra and Hochschild cohomology with the Hochschild cup product are examples.

On a graded algebra, a left derivation of degree $r$ satisfies the displayed rule for homogeneous $a$. A right derivation of degree $r$ satisfies $R(ab)=aR(b)+(-1{)}^{r|b|}R(a)b$. For odd $r$, these reduce to the familiar parity sign rules. The left Gerstenhaber bracket makes $[u,-]$ a left derivation of degree $|u|-1$ on Hochschild cohomology; this rule holds for the induced cohomology operations, not necessarily for their cochain representatives.

The ordinary Hochschild cochain complex is defined using a chosen commutative ground ring $k$. The bar complex is exact via insertion of $1$, but its terms are projective over the enveloping algebra when $R$ is projective over $k$, in particular when $k$ is a field. Without that hypothesis the ordinary cochain groups compute relative Hochschild cohomology and cannot automatically be identified with absolute Ext. Likewise, ordinary second cohomology classifies ground-ring-linearly split square-zero extensions, not arbitrary nonsplit ring extensions.

// Destination: module-theory.bigb

Over a field, the bimodule resolution $0\to k[x]\otimes k[x]\stackrel{x\otimes 1-1\otimes x}{\to }k[x]\otimes k[x]\to k[x]\to 0$ is exact: multiplication identifies the quotient by $x\otimes 1-1\otimes x$, and that element is a non-zero-divisor in the two-variable polynomial ring. Applying the bimodule Hom functor gives zero differential, proving the displayed Hochschild cohomology. Degree-one classes are the derivations $f(x){\partial }_x$.

For $f\in C^p(A,A)$ and $g\in C^q(A,A)$, set $(f⌣g)(a_1,\dots ,a_{p+q})=f(a_1,\dots ,a_p)g(a_{p+1},\dots ,a_{p+q})$. This associative cochain multiplication descends to Hochschild cohomology, where it gives a graded commutative algebra. It is one operation in the Gerstenhaber algebra structure.

For a unital associative algebra, normalized positive-degree cochains vanish whenever an argument is $1$. They form a subcomplex of the Hochschild cochain complex computing the same Hochschild cohomology. The unit-insertion contracting homotopy on degenerate bar terms proves the normalization equivalence. A normalized degree-two Hochschild cocycle defines a unital square-zero extension of an algebra.

Specify a commutative ground ring $k$ and a unital associative $k$-algebra $R$. For an $R$-bimodule $M$, the Hochschild cochain complex hasand coboundaryThe bimodule identities and associativity cancel the terms of ${\delta }^2$ in pairs. The Hochschild cohomology is $H{H}_k^q(R,M)=\ker {\delta }_q/\mathrm{im}{\delta }_{q-1}$; the paper's $H^q(R,R)$ uses the regular bimodule as coefficients.

Over a field, the bar resolution of an associative algebra explains its homological meaning. Put $R^e=R{\otimes }_k{R}^{\mathrm{op}}$. The degree-$q$ bar term is $R^{{\otimes }_k(q+2)}$, with the outer factors supplying the $R^e$ action, boundary obtained by multiplying adjacent factors with alternating signs, and augmentation multiplication $R\otimes R\to R$. Inserting $1$ at the left gives a $k$-linear contracting homotopy, proving exactness. The terms are free $R^e$-modules after choosing a $k$-basis for the inner tensor factors. Applying ${\mathrm{Hom}}_{R^e}(-,M)$ gives precisely the displayed cochain complex and thereforeThis ground-ring qualification in Hochschild cohomology matters for an arbitrary ring: one can take $k=\mathbb{Z}$, but the bar terms need not be projective over $R^e$. The ordinary cochain complex then computes relative Hochschild cohomology; the absolute Ext identification needs, for example, $R$ projective over $k$. Likewise the extension classification below applies to ground-ring-linearly split extensions. For algebras over a field these qualifications are automatic.

In degree zero, $(\delta m)(a)=am-ma$, so $H{H}^0(R,R)=Z(R)$. In degree one the cocycle equation isexactly the Leibniz rule for a derivation of an algebra. The coboundaries are the inner derivations $a\mapsto am-ma$. Thus the First Hochschild cohomology as outer derivations isIts significance is infinitesimal automorphisms modulo inner ones. Over the dual numbers $k[\epsilon ]/({\epsilon }^2)$, the map $a\mapsto a+\epsilon d(a)$ is multiplicative precisely when $d$ satisfies the Leibniz rule. Conjugation by $1+\epsilon m$ gives $a\mapsto a+\epsilon (ma-am)$, an inner derivation with the opposite sign convention. The quotient therefore measures infinitesimal symmetry not arising from conjugation.

Degree two controls extensions and changes of multiplication. A $k$-linear split square-zero extension of an algebra with prescribed kernel bimodule $M$ can be written as $E=R\oplus M$. Choose a unital section and let $f(a,b)$ be the difference between its product and its value on $ab$. ThenComputing both parenthesizations of a triple gives the associativity conditionnamely $\delta f=0$. A unital product has $f(1,a)=f(a,1)=0$, so one uses the normalized Hochschild cochain complex. Changing section by $g:R\to M$ gives $f\mapsto f+\delta g$. Conversely the displayed product for any normalized two-cocycle constructs such an extension, and a section change gives the corresponding extension isomorphism. This proves that second Hochschild cohomology classifies split square-zero extensions:The zero class is the split algebra extension with no extra product term. With $M=R$, this gives one meaning of the requested $H^2(R,R)$.

For a first-order associative deformation, retain the underlying $k$-module and put $a\ast b=ab+\epsilon f(a,b)$. Expanding $(a\ast b)\ast c=a\ast (b\ast c)$ modulo ${\epsilon }^2$ gives exactly $\delta f=0$. A coordinate change $T=1+\epsilon g$ transports the product to $T^{-1}(T(a)\ast T(b))$, whose first-order coefficient is $f+\delta g$. HenceThis classification is of infinitesimal deformations, not a guarantee that all classes extend to every order. For $a\ast b=ab+tf(a,b)+t^2{\mu }_2(a,b)$ modulo $t^3$, the next associativity equation isSubstitution of $\delta f=0$ shows the right-hand side is a three-cocycle. Its class in $H{H}^3(R,R)$ is the second-order obstruction to an associative deformation. It must vanish for a second-order extension to exist. At higher orders the same expansion gives further obstruction cocycles. If $H{H}^2(R,R)=0$, every first nonzero coefficient of a formal associative deformation is a coboundary and can be removed by a coordinate change. Repeating this order by order gives formal rigidity from vanishing second Hochschild cohomology, with the transformations converging in the $t$-adic topology. Vanishing of $H{H}^3$ instead removes these extension obstructions; these are distinct roles.

Examples make the two low degrees concrete. For $R=k[x]$ over a field, inner derivations vanish and every derivation is $h(x){\partial }_x$, determined by its value on $x$. The bimodule resolutionis exact: the middle tensor ring is the polynomial integral domain $k[u,v]$, multiplication identifies its quotient by $u-v$, and multiplication by $u-v$ is injective. Applying bimodule Hom gives zero differential. Consequently the Hochschild cohomology of a polynomial algebra in one variable isThis algebra has many infinitesimal automorphisms, while its ordinary associative multiplication is formally rigid.

For $R=k[x,y]$ in characteristic zero, the bilinear map $f(a,b)=({\partial }_{y}a)({\partial }_{x}b)$ is a two-cocycle: substitution in $\delta f$ and the two product rules cancel its four terms. It is not a coboundary, because every $\delta g$ on a commutative algebra is symmetric in its two inputs, whereas $f(y,x)-f(x,y)=1$. This class integrates to the Weyl deformation of a polynomial algebra in two variables, with $yx-xy=t$. The differential Ore extension construction supplies the ordered basis $x^i{y}^j$ over $k[t]$, so the $t$-adic completion has the fixed underlying module $k[x,y][[t]]$. Its first multiplication correction is exactly this $f$. This displays a genuinely noncommutative formal associative deformation detected by $H{H}^2$.

Finally, the Hochschild cup product $(f⌣g)(a_1,\dots ,a_{p+q})=f(a_1,\dots ,a_p)g(a_{p+1},\dots ,a_{p+q})$ descends to the graded cohomology algebra, and the Gerstenhaber bracket organizes the interaction of derivations and formal associative deformation cocycles. The calculations above explain why the first two positive degrees are especially useful: $H{H}^1$ measures outer infinitesimal symmetries, while $H{H}^2$ measures split extensions and infinitesimal associative structures, with degree three detecting the first obstruction to continuing them.

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.

A derivation of an algebra is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. The commutator $[D,E]=D\circ E-E\circ D$ is again a derivation: expanding $[D,E](ab)$ cancels the two mixed terms and leaves $[D,E](a)b+a[D,E](b)$. The commutator on endomorphisms is bilinear, antisymmetric, and satisfies the Jacobi identity by cancellation of its twelve triple-composition terms. Therefore ${\mathrm{Der}}_k(A)$ is a Lie algebra.

In degree zero of the Hochschild cochain complex, $\delta a(b)=ba-ab$, so $H{H}^0(A,A)=Z(A)=A$ for commutative $A$. In degree one, $\delta D=0$ is precisely the derivation rule; the boundaries are inner derivations, which vanish for commutative $A$. HenceFor cochains $f\in C^p(A,A)$, $g\in C^q(A,A)$, the Hochschild cup product isDefine the insertion operation byA degree-zero cochain is an element of $A$, inserted with no arguments; for $p=0$ the sum is empty. For the Gerstenhaber bracket we use the left graded Leibniz rule convention, compatible with the unsigned Hochschild cup product just displayed:Both degree-zero inputs have bracket zero. Another common insertion convention writes $f\circ g-(-1{)}^{(p-1)(q-1)}g\circ f$; the two brackets differ by $(-1{)}^{(p-1)(q-1)}$. With an unsigned Hochschild cup product, that convention uses the corresponding right graded Leibniz rule. The distinction matters for a degree-two cochain bracketed with a function. Either consistent convention gives the same degree-one Lie bracket and the same derivation action on functions.

If $m(a,b)=ab$, this convention gives $\delta f=[m,f]$. The shifted Jacobi identity and $[m,m]=0$ therefore show that the Gerstenhaber bracket respects Hochschild cocycles and the images of the coboundary map. The Hochschild cup product and Gerstenhaber bracket induce operations on Hochschild cohomology. A Gerstenhaber algebra is a graded algebra $H$ with an associative degree-zero product with the graded commutative algebra rule $uv=(-1{)}^{pq}vu$, and a degree-minus-one graded Lie bracket making the shifted degrees $|u|-1$ into a graded Lie algebra. In particular,for homogeneous elements of a graded algebra of degrees $p,q$. The shifted Jacobi identity isThe Hochschild cup product does not make the cochains a graded commutative algebra in general, but does make their cohomology a graded commutative algebra; the insertion operation supplies the homotopy for this assertion and for the graded Leibniz rule. Thus these axioms describe the induced Gerstenhaber algebra, not a claim of a graded commutative algebra structure on the cochain multiplication itself.

For $A=k[X]$, the enveloping algebra is $A^e=k[X_{\ell },X_r]$, andis a projective resolution. The first map is injective since $A^e$ is an integral domain, and its cokernel is $A$. Applying ${\mathrm{Hom}}_{A^e}(-,A)$ gives a zero coboundary map. ConsequentlyThe Hochschild cup product is ordinary multiplication of functions and scalar multiplication of derivations, with the product of two derivations zero because $H{H}^2=0$. Every derivation is $f{\partial }_X$, since it is determined by its value on $X$. The Gerstenhaber bracket iswith all other orders fixed by graded antisymmetry. These formulas fully determine the Gerstenhaber algebra.

For $A=k[X,Y]$, the Hochschild-Kostant-Rosenberg theorem identifiesThus the degrees zero, one, and two are $A$, $A{\partial }_X\oplus A{\partial }_Y$, and $A({\partial }_X\wedge {\partial }_Y)$, and all higher groups vanish. The Hochschild-Kostant-Rosenberg map sends a wedge of $p$ derivations to the cochainThe factorial is invertible in characteristic zero. Equivalently, the groups follow from the Koszul resolution on the regular sequence $X_{\ell }-X_r,Y_{\ell }-Y_r$ in $A^e$, whose dual coboundary maps vanish on $A$.

The Hochschild cup product becomes the exterior product, and our Gerstenhaber bracket becomes the left Schouten-Nijenhuis bracket. It is determined by the commutator of derivations, $[D,f]=D(f)$, zero brackets of functions, and the displayed graded antisymmetry and left graded Leibniz rule. For explicit signs, put $\Pi ={\partial }_X\wedge {\partial }_Y$ and $D=F{\partial }_X+G{\partial }_Y$. ThenThe last bracket has degree three, whose exterior power is zero. For two derivations, the coefficient functions of their commutator give the remaining formula. This specifies the entire Gerstenhaber algebra; under the alternate insertion convention mentioned above, the first displayed bracket changes sign, together with the Leibniz convention. No smoothness of a general finitely generated commutative algebra was assumed: the Hochschild-Kostant-Rosenberg theorem is invoked here only for the smooth polynomial ring $k[X,Y]$.