Diagonal Hochschild homology of a polynomial ring

Apply the polynomial Hochschild theorem with the regular coefficient bimodule. [F1, F3, given] Since $R$ is $k$-central, the theorem applies to $M=R$. Under AC it gives $H{H}_j(R,R)\cong H_j({\mathcal{K}}_{\bullet }(R))$, where the coefficient differential deletes ${\theta }_i$ with coefficient $x_{i}m-m{x}_i$. Its naturality in $M$ will also identify the coefficient $R$-module structure below.

The Hochschild-chain boundary is $R$-linear on the regular coefficient chains. [F4, step 1.1, algebra] For $r\in R$, let $r$ act on $C_q(R,R)=R{\otimes }_k{R}^{{\otimes }_{k}q}$ by multiplication on the first factor. This action commutes with every face: the first face uses $(rm)a_1=r(m{a}_1)$, internal faces leave the coefficient unchanged, and the last face uses $a_q(rm)=r(a_{q}m)$ because $R$ is commutative. Thus each boundary is $R$-linear and the homology inherits this $R$-action. For every $r\in R$, coefficient multiplication ${\mu }_r(m)=rm$ is an $R$-bimodule endomorphism of $R$. Naturality in [F1] shows the comparison with $H_j({\mathcal{K}}_{\bullet }(R))$ commutes with ${\mu }_r$, so it is $R$-linear.

Every differential in the coefficient Koszul complex for $M=R$ is zero. [F1, step 1.1, algebra] For every $m\in R$ and generator $x_i$, commutativity gives $x_{i}m-m{x}_i=0$. Hence each coefficient of every Koszul deletion map is zero, so $d_q=0$ for all $q$.

Read homology from the exterior basis. [F1, F2, step 2.2, algebra] Since all differentials vanish, $H_j({\mathcal{K}}_{\bullet }(R))={\mathcal{K}}_j(R)$. For $0\le j\le n$, the increasing wedges ${\theta }_I$ with $\mathrm{\mid }I\mathrm{\mid }=j$ form a basis, so ${\mathcal{K}}_j(R)={⨁}_{\mathrm{\mid }I\mathrm{\mid }=j}R{\theta }_I\cong R{\otimes }_k{\mathrm{\Lambda }}_k^j(k^n)$ with rank $\left(\genfrac{}{}{0px}{}{n}{j}\right)$. For $j>n$ the exterior power and Koszul term are zero by [F2], giving $H{H}_j(R,R)=0$. The isomorphism is $R$-linear by step 2.1.

Determine the internal grading and shift. [F1, F5, step 3.1, algebra] In the graded clause of [F1], each ${\theta }_i$ has degree $2$, so every ${\theta }_I$ with $\mathrm{\mid }I\mathrm{\mid }=j$ has degree $2j$. Thus each summand $R{\theta }_I$ is the internal shift $R\{2j\}$ under [F5], and the isomorphism in step 3.1 preserves internal degree.

Check the empty, one-variable, top, and degree-zero cases. [F1, F2, step 3.1, step 4.1, given] If $n=0$, there is only the empty wedge in degree zero and the complex is $k$ with zero differential, so $H{H}_0(k,k)=k$ and higher homology vanishes. If $n=1$, the two terms are $R$ and $R{\theta }_1$; the differential is zero, so $H{H}_0=R$, $H{H}_1=R\{2\}$, and higher groups vanish. In general, degree zero uses ${\mathrm{\Lambda }}^0(k^n)=k$ and has no outgoing differential; top degree $j=n$ has one basis wedge of degree $2n$ (the empty wedge when $n=0$); above top degree all terms vanish. The stated AC is used only through [F1]; the zero-differential calculation and exterior basis read-off make no further choices. There is no zero coefficient case because the coefficient is fixed to the nonzero regular module $R$ over a field. This claim is not an equivalence. [F1, F2, F3, step 1.1, step 2.2, step 3.1, step 4.1, algebra] $\mathrm{\square }$