Lemma 10.133.8. Let $A \to B$ be a ring map. The differentials $\text{d} : \Omega ^ i_{B/A} \to \Omega ^{i + 1}_{B/A}$ are differential operators of order $1$.

Proof. Given $b \in B$ we have to show that $\text{d} \circ b - b \circ \text{d}$ is a linear operator. Thus we have to show that

$\text{d} \circ b \circ b' - b \circ \text{d} \circ b' - b' \circ \text{d} \circ b + b' \circ b \circ \text{d} = 0$

To see this it suffices to check this on additive generators for $\Omega ^ i_{B/A}$. Thus it suffices to show that

$\text{d}(bb'b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) - b\text{d}(b'b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) - b'\text{d}(bb_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) + bb'\text{d}(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i)$

is zero. This is a pleasant calculation using the Leibniz rule which is left to the reader. $\square$

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