I'm reading an article about noncommutative geometry and I'm trying to prove the following theorem
Let $R$ be a commutative ring and assume that $I$ is ideal generated by regular sequence $x=(x_1,...,x_m)\in R^m$, i.e. for every $i=1,...,m$ multiplication by $x_i$ is an injective map in $R/I$. Then there is the following isomorphism of graded algebras
$\mathrm{Tor}_{n}^R\left( R/I, R/I\right)\cong \bigwedge_{R/I}^{n}\left(I/I^2\right)$.
In this article is the following statement:
$\mathrm{Tor}_{n}^R \left( R/I, R/I\right)\cong \left(R/I\right)\otimes_R\bigwedge_{R}^{n}(R^m)$
I know that Koszul complex $K_*(x)$ is a projective resolution of $R/I$, so using this resolution we obtain the following complex
$\mathcal{C} :\ 0\rightarrow \left(R/I\right)\otimes_R\bigwedge_{R}^{m}(R^m)\stackrel{\mathrm{id}\otimes_R d_m}{\longrightarrow}\left(R/I\right)\otimes_R\bigwedge_{R}^{m-1}(R^m)\stackrel{\mathrm{id}\otimes_R d_{m-1}}{\longrightarrow}...\stackrel{\mathrm{id}\otimes_R d_{2}}{\longrightarrow}\\\stackrel{\mathrm{id}\otimes_R d_2}{\longrightarrow}\left(R/I\right)\otimes_R R^m\stackrel{\mathrm{id}\otimes_R d_1}{\longrightarrow}\left(R/I\right)\otimes_R R\longrightarrow 0$
where maps $d_k$ are given by Koszul complex.
By definition $\mathrm{Tor}_{n}^R \left( R/I, R/I\right)=H_n(\mathcal{C})$. Why $n$-th homology group of this complex is equal to $\left(R/I\right)\otimes_R\bigwedge_{R}^{n}(R^m)$ ?
