Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 148 / 5 / Solution

For the $R$-$R$-bimodule $M$, the Hochschild chain complex iswith boundaryThenThe Hochschild cochain complex is $C^n(R,M)={\mathrm{Hom}}_k(R^{{\otimes }_{k}n},M)$ withand $H{H}^n(R,M)=H^n(C^{\bullet }(R,M),\delta )$.

A derivation into a bimodule is a $k$-linear map $d:R\to M$ satisfyingIt is an inner derivation when $d=d_m$ for some $m\in M$, where $d_m(r)=rm-mr$. The degree-one cocycle equation is precisely the Leibniz rule, and the degree-one coboundaries are the inner derivations. Therefore the First Hochschild cohomology as outer derivations is

Let $\mu :R{\otimes }_{k}R\to R$ be multiplication and $\Omega =\ker \mu$, equipped with the outer bimodule structure. The universal bimodule derivationsatisfies $D(rs)=rD(s)+D(r)s$. Composition with $D$ definesFor a derivation $d$, its inverse image under this map isIf ${\sum }_i{r}_i{s}_i=0$, the Leibniz rule shows that this formula is both left and right $R$-linear. Moreover every element of $\Omega$ is ${\sum }_{i}D(r_i)s_i$, proving existence and uniqueness.

For $d_m(r)=rm-mr$, the corresponding map isThese are exactly the maps $\Omega \to M$ which extend to bimodule maps $R{\otimes }_{k}R\to M$.