Lemma 10.140.6. Let $R \to S$ be a $\mathbf{Q}$-algebra map. Let $f \in S$ be such that $\Omega _{S/R} = S \text{d}f \oplus C$ for some $S$-submodule $C$. Then
$f$ is not nilpotent, and
if $S$ is a Noetherian local ring, then $f$ is a nonzerodivisor in $S$.
Proof. For $a \in S$ write $\text{d}(a) = \theta (a)\text{d}f + c(a)$ for some $\theta (a) \in S$ and $c(a) \in C$. Consider the $R$-derivation $S \to S$, $a \mapsto \theta (a)$. Note that $\theta (f) = 1$.
If $f^ n = 0$ with $n > 1$ minimal, then $0 = \theta (f^ n) = n f^{n - 1}$ contradicting the minimality of $n$. We conclude that $f$ is not nilpotent.
Suppose $fa = 0$. If $f$ is a unit then $a = 0$ and we win. Assume $f$ is not a unit. Then $0 = \theta (fa) = f\theta (a) + a$ by the Leibniz rule and hence $a \in (f)$. By induction suppose we have shown $fa = 0 \Rightarrow a \in (f^ n)$. Then writing $a = f^ nb$ we get $0 = \theta (f^{n + 1}b) = (n + 1)f^ nb + f^{n + 1}\theta (b)$. Hence $a = f^ n b = -f^{n + 1}\theta (b)/(n + 1) \in (f^{n + 1})$. Since in the Noetherian local ring $S$ we have $\bigcap (f^ n) = 0$, see Lemma 10.51.4 we win. $\square$
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