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For a complete Noetherian ring with ideal of definition $I$, completed localization is the completion of the localization of a ring $A_f$ in the induced adic topology. It gives the sections of the formal spectrum over $D(f)$. For $A=k[t][[q]]$, its completed localization at $t$ is $k[t,t^{-1}][[q]]$. Ordinary localization allows a bounded pole order in $t$ across all coefficients; the latter ring also contains ${\sum }_{n\ge 0}{q}^n{t}^{-n}$, whose pole orders are unbounded.

// Target: ringed-space.bigb

A formal associative deformation is a unital $k[[t]]$-bilinear product continuous for the adic topology on $A[[t]]$ of the form $a\ast b=ab+{\sum }_{r\ge 1}{t}^r{\mu }_r(a,b)$. Each coefficient ${\mu }_r$ is a $k$-bilinear map $A\otimes A\to A$, and the whole product satisfies associativity. For an infinite-dimensional algebra, $A[[t]]$ is the completed tensor product ${\underleftarrow{\lim }}_{n}A\otimes k[t]/(t^n)$ and generally differs from the ordinary $A\otimes k[[t]]$.

For a closed subscheme with ideal sheaf $\mathcal{I}$ in a locally Noetherian scheme, the completion is the formal scheme on $|Y|$ with topological structure sheaf ${\underleftarrow{\lim }}_n({\mathcal{O}}_X/{\mathcal{I}}^{n+1}){|}_{|Y|}$. On a Noetherian affine scheme $\mathrm{Spec}A$, it is $\mathrm{Spf}({\underleftarrow{\lim }}_{n}A/I^{n+1})$. Its adic topology remembers every infinitesimal neighbourhood.

// Target: commutative-algebra.bigb

For a complete ring with an adic topology defined by $I$, the formal spectrum has the prime ideals containing $I$ as its points and structure sheaf ${\underleftarrow{\lim }}_n{\mathcal{O}}_{\mathrm{Spec}(A/I^{n+1})}$. On a principal open subset $D(f)$ its sections form the completed localization of an adic ring, not generally the ordinary localization $A_f$.

// Target: ringed-space.bigb

Let $\mathcal{I}\subseteq {\mathcal{O}}_X$ be the ideal sheaf of $Y$. The formal completion of a scheme along $Y$ has underlying topological space $|Y|$ and topological structure sheafThe topology is the adic topology given by the kernels of the projections to the successive infinitesimal neighbourhoods $X_n=(|Y|,{\mathcal{O}}_X/{\mathcal{I}}^{n+1})$. In the locally Noetherian scheme setting of these examples this is a formal scheme. On a Noetherian affine scheme, with $Y=V(I)\subseteq \mathrm{Spec}A$, it is the formal spectrum $\mathrm{Spf}\hat{A}$, where $\hat{A}={\underleftarrow{\lim }}_{n}A/I^{n+1}$. Thus the completion retains all orders of infinitesimal information normal to $Y$.

Here $Z=D(t)=\mathrm{Spec}k[q,t,t^{-1}]$ is an open subscheme of $X$. The relevant affine completions arewith the $q$-adic topology. The complement of $t=0$ in the first formal scheme is its formal open $D(t)$. The sections on this open are the completed localization of an adic ring:The same calculation on every smaller principal open subset identifies the structure sheaves, not just their global sections. Equivalently, every infinitesimal neighbourhood of $Z\cap (q=0)$ is precisely the restriction to $D(t)$ of the corresponding neighbourhood in $X$. By formal completion commutes with restriction to an open subscheme,The isomorphism is canonical and respects the morphisms to $\mathrm{Spf}k[[q]]$.

It would give the wrong answer to replace the formal-open structure sheaf by the ordinary localization of a ring $k[t][[q]][t^{-1}]$. That ring has a uniform bound on the negative powers of $t$ across the coefficients of any one element, whereashas no such bound. This difference between ordinary and completed localization does not obstruct the isomorphism of formal schemes; it explains why the completed localization is essential.

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 topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

A formal associative deformation is trivial if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}(\mathrm{mod}t)$ fixing the identity transports the deformed multiplication to the original one. This is equivalence by a formal change of coordinates, stronger than merely an abstract algebra isomorphism between unspecified middle terms.