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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.

Put $A=R[[X]]$, and let $J$ be any ideal of this formal power series ring. For $d\ge 0$ define an ideal of $R$ byMultiplication by $X$ shows $L_d\subseteq L_{d+1}$. Since $R$ is Noetherian, this chain stabilizes at some $d=N$, and each $L_d$ is finitely generated. For every $0\le d\le N$, choose finitely many series $f_{d,j}\in J\cap X^{d}A$ whose degree-$d$ coefficients generate $L_d$.

Given $f\in J$, cancel its degree-zero coefficient using the $f_{0,j}$, then its degree-one coefficient using the $f_{1,j}$, and so on. At a degree $d\ge N$, use $X^{d-N}{f}_{N,j}$. Each residual remains in $J$ and has zero coefficients below the next cancellation degree. For each of the finitely many chosen $f_{d,j}$, accumulate its multiplier into a formal power series $h_{d,j}$. The higher-degree increments of these multipliers converge coefficientwise. At every degree the cancellation procedure gives the exact coefficient identity, soThus the finitely many lifts generate $J$ as an actual ideal. No assumption that $J$ is closed has been used. This proves Noetherianity of formal power series rings:Iterating the argument proves the assertion for $R[[X_1,\dots ,X_s]]$, which is naturally the iterated formal power series ring.

For a surjective module map $f:M\to N$, define the adic completion of a module byTake $\hat{n}\in \hat{N}$ and choose representatives $n_r\in N$ for its components. Compatibility says $n_{r+1}-n_r\in I^{r}N$. Choose $m_1$ with $f(m_1)=n_1$. If $f(m_r)=n_r$, surjectivity gives $f(I^{r}M)=I^{r}N$, so choose ${\delta }_r\in I^{r}M$ with $f({\delta }_r)=n_{r+1}-n_r$ and put $m_{r+1}=m_r+{\delta }_r$. The classes of $m_r$ modulo $I^{r}M$ are compatible and map to the chosen components of $\hat{n}$. HenceThis surjectivity on adic completions does not require finite generation of $M$ or $N$. The compatibility of the lifts is the point; independent lifts of the finite-level classes would not suffice.

Since $R$ is Noetherian, write $I=(a_1,\dots ,a_s)$. Define a ring homomorphismModulo $I^r$, evaluating a series requires only its finitely many terms of total degree below $r$. These evaluations commute with the reduction maps, and define $\Phi$ without needing the original ring to be complete or separated.

To prove surjectivity, begin with $\hat{b}\in \hat{R}$. Choose $b_0\in R$ matching its component modulo $I$. After terms of total degree below $r$ have been chosen, their evaluation differs from $\hat{b}$ at level $R/I^{r+1}$ by a class in $I^r/I^{r+1}$. This class is an $R$-linear combination of the monomials $a^{\alpha }$ with $|\alpha |=r$. Choose the corresponding homogeneous degree-$r$ polynomial $p_r(X)$ as the next term. Then the series $b_0+{\sum }_{r\ge 1}{p}_r(X)$ evaluates to $\hat{b}$ at every level. This proves the power-series presentation of an adic completion, and thereforeA quotient of a Noetherian ring is Noetherian, since its ideals correspond to ideals containing the kernel. The cases $I=0$ and $I=R$ give $\hat{R}=R$ and $\hat{R}=0$, respectively, and satisfy the conclusion as well.

Put $R={\mathcal{O}}_{X,x}$, $\mathfrak{m}={\mathfrak{m}}_x$ and $d=\dim R$. Smoothness makes $R$ a regular local ring, with residue field $k$. A list of local parameters $t_1,\dots ,t_d$ is a regular system of parameters: its images form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$. Set $J=(t_1,\dots ,t_d)$. The basis condition says $\mathfrak{m}=J+{\mathfrak{m}}^2$, so the finitely generated module $\mathfrak{m}/J$ equals $\mathfrak{m}(\mathfrak{m}/J)$. The Nakayama lemma gives

We now construct the requested formal power series, rather than assert that a local function has a convergent expansion. The associated graded ring of a regular local ring isIndeed these initial forms generate the graded ring, since the $t_i$ generate $\mathfrak{m}$. The Dimension theorem for Noetherian local rings gives $\dim {\mathrm{gr}}_{\mathfrak{m}}R=d$. A nonzero kernel in the surjection from the polynomial ring would make its quotient have dimension at most $d-1$, so that kernel is zero.

For $f\in R$, choose $P_0\in k$ to be its residue at $x$. Inductively suppose homogeneous polynomials $P_0,\dots ,P_{n-1}$ have been chosen such thatThe graded-ring isomorphism gives a unique homogeneous polynomial $P_n(T_1,\dots ,T_d)$ of degree $n$ representing the class of $r_n$ in ${\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}$. Subtracting $P_n(t)$ makes the next remainder lie in ${\mathfrak{m}}^{n+1}$. HenceThis equality is interpreted in the adic completion of a module $\hat{R}=\underleftarrow{\lim }R/{\mathfrak{m}}^n$. The same induction applies to every compatible sequence of residues in this inverse limit, so evaluation $T_i\mapsto t_i$ defines an isomorphism $k[[T_1,\dots ,T_d]]\cong \hat{R}$. It is injective because the lowest nonzero homogeneous term of a series has nonzero initial form. The Krull intersection theorem gives ${\bigcap }_n{\mathfrak{m}}^n=0$, so $R$ embeds in its completion and the series determines $f$. This is the formal expansion at a smooth point; it depends on the chosen regular system of parameters, and does not claim that the infinite sum converges inside $R$. If $d=0$, the local ring is just $k$, and only the constant term is needed.

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 any ideal $I$ and surjective module map $f:M\to N$, compatible representatives $n_r$ modulo $I^{r}N$ can be lifted recursively. Having chosen $f(m_r)=n_r$, lift $n_{r+1}-n_r\in I^{r}N$ to ${\delta }_r\in I^{r}M$ and put $m_{r+1}=m_r+{\delta }_r$. These define the required element of the adic completion of a module. No finite-generation hypothesis on the modules is required.