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On Serge Lang's book Algebra, he wrote on page 748, after defining the module of differentials $\Omega^1_{A/R}$ of an $R$-algebra $A$ and higher differentials $\Omega^i_{A/R}=\wedge^i \Omega^1_{A/R}$, that there exists a unique sequence of $R$-homomorphisms $d_i:\Omega^i_{A/R}\rightarrow\Omega^{i+1}_{A/R}$ such that for $\omega \in\Omega^i_{A/R}$ and $\eta\in\Omega^j_{A/R}$ we have $d(\omega\wedge\eta)=d\omega\wedge\eta+(-1)^i\omega\wedge d\eta$. He then pointed out $d^2=0$ as a consequence of the definition.

The question is, as I saw elsewhere, people usually impose one more axiom, that $d^2r=0$ for $r\in R$, so that we can define $d$ inductively. And I think Lang is missing something here. So, is it unnecessary to require $d^2r=0$ here?

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I checked with the third edition, and I agree that some conditions are missing.

  1. $d_0: A\to \Omega_{A/R}^1$ must be the universal derivation (otherwise take $d_i=0$ for all $i$, there is no uniqueness).

  2. One must assume $d^2=0$. Example: Let $A=R[X,Y]$. Then $\Omega^1_{A/R}=AdX\oplus AdY$, $\Omega^2_{A/R}=A(dX\wedge dY)$. Let $d_0$ be the universal derivation and let $d_1 : \Omega^1_{A/R}\to \Omega^2_{A/R}$ be any $A$-linear map (e.g. mapping $dX, dY$ to $dX\wedge dY$). The identity on $d(\omega\wedge \eta)$ is always satisfied but $d^2=d_1\circ d_0$ is not always zero.