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If $H{H}^2(A,A)=0$, every formal associative deformation on the adic completion of a module $A[[t]]$ is trivial. After killing lower coefficients, associativity makes the order-$r$ coefficient a Hochschild cocycle ${\mu }_r=\delta g_r$. Transport by $\mathrm{id}+t^r{g}_r$ subtracts $\delta g_r$ and kills it. The successive transformations stabilize modulo each $t^n$ and converge to an invertible change of coordinates. Completeness, and equivalence congruent to the identity modulo $t$, are essential to this argument.

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.

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 $k[X,Y]$.

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 $H{H}^2(A,M)$.