Most references on Koszul complexes seem to assume that the elements $x_1,\ldots, x_n$ live in a commutative ring or are central. It appears to me that the proof also works provided the weaker assumption that $x_1,\ldots, x_n$ commute with each other. Confirmation, a reference, or pointing out an error would be greatly appreciated.
Claim:Suppose $R$ is a ring which need-not be commutative and let $x_1,\ldots,x_n\in R$ be commuting elements such that whenever$\lambda x_{i+1} \in (x_{1},\ldots,x_i) $it follows that $\lambda $ is in the left ideal $(x_{1},\ldots,x_i)$. Then the Koszul complex defined by$$\phi: \wedge^q R^n \to \wedge^{q-1}R^n : \lambda e_{i_1} \wedge \cdots \wedge e_{i_q} \mapsto \lambda \sum_j (-1)^{j-1} x_{i_j} e_{i_1} \wedge \cdots \widehat{e_{i_j}} \cdots\wedge e_{i_q} $$yields a free resolution for $R/(x_1,\ldots,x_n)$.
Sketch:
- Observe the statement is obvious for $n = 1$. We proceed by induction.
- Consider the following short exact sequence in Koszul complexes$$0 \to \wedge^\bullet R^{n-1} \to \wedge^\bullet R^n \to \wedge^\bullet R^{n-1}[-1]\to 0.$$
- There is an induced long exact sequence in cohomology$$ H^q(\wedge^\bullet R^{n-1}) \to H^q(\wedge^\bullet R^{n}) \to H^{q-1}(\wedge^\bullet R^{n-1})$$
- Combined with the induction hypothesis this shows $H^q(\wedge^\bullet R^{n}) = 0$ whenever $q-1 \neq 0$.
- The remaining claim for $H^1(\wedge^\bullet R^{n})$ is immediate from the assumption on the $x_i$. QED.
