= Solution
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> is
$$
\boxed{HH^n(A,M)=\operatorname{Ext}_{A^e}^n(A,M).}
$$
Equivalently, the <Hochschild cochain complex> has $C^n(A,M)=\operatorname{Hom}_k(A^{\otimes n},M)$ and <coboundary map>
$$
(\delta f)(a_1,\ldots,a_{n+1})=a_1f(a_2,\ldots,a_{n+1})+\sum_{i=1}^n(-1)^if(a_1,\ldots,a_ia_{i+1},\ldots,a_{n+1})+(-1)^{n+1}f(a_1,\ldots,a_n)a_{n+1}.
$$
Its <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> is
$$
\boxed{\operatorname{Dim}(A)=\operatorname{pd}_{A^e}A=\sup\{n:HH^n(A,N)\ne0\text{ for some bimodule }N\}.}
$$
The 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> $0\to M\to E\to A\to0$, 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 defect
$$
\mu(a,b)=s(a)s(b)-s(ab)\in M
$$
is 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 define
$$
(a,m)(b,n)=(ab,an+mb+\mu(a,b)).
$$
The <Hochschild cocycle> equation is exactly <associativity>, and $(1,0)$ is the identity. If $\mu'=\mu+\delta g$, the map $E_{\mu'}\to E_\mu$, $(a,m)\mapsto(a,m+g(a))$, 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 \b[a bijection between $HH^2(A,M)$ 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 module>
$$
A[[t]]=\varprojlim_r A\otimes_k k[t]/(t^r).
$$
This 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, $\sum_{n\geq0}X^nt^n$ 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 form
$$
a*b=ab+\sum_{r\geq1}t^r\mu_r(a,b),\qquad \mu_r\in\operatorname{Hom}_k(A\otimes A,A),\quad \mu_r(1,a)=\mu_r(a,1)=0.
$$
It is a <trivial formal deformation> if a $k[[t]]$-linear <automorphism> continuous for the <adic topology> $T=\operatorname{id}+\sum_{r\geq1}t^rT_r$, with $T(1)=1$, satisfies $T(a*b)=T(a)T(b)$.
If $\operatorname{Dim}(A)\leq1$, then $HH^2(A,A)=0$. 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 $T_r=\operatorname{id}+t^rg_r$:
$$
a*'b=T_r\bigl(T_r^{-1}(a)*T_r^{-1}(b)\bigr).
$$
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 \b[every star product is trivial under the completed formal-series convention]. In fact, the argument only needs the vanishing of $HH^2(A,A)$, not all of <Hochschild cohomological dimension> at most one.
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