Lemma 89.4.6. Let $f : R \to S$ be a morphism of $\widehat{\mathcal{C}}_\Lambda$. If $\text{Der}_\Lambda (S, k) \to \text{Der}_\Lambda (R, k)$ is injective, then $f$ is surjective.

Proof. If $f$ is not surjective, then $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero by Lemma 89.4.2. Then also $Q = S/(f(R) + \mathfrak m_ R S + \mathfrak m_ S^2)$ is nonzero. Note that $Q$ is a $k = R/\mathfrak m_ R$-vector space via $f$. We turn $Q$ into an $S$-module via $S \to k$. The quotient map $D : S \to Q$ is an $R$-derivation: if $a_1, a_2 \in S$, we can write $a_1 = f(b_1) + a_1'$ and $a_2 = f(b_2) + a_2'$ for some $b_1, b_2 \in R$ and $a_1', a_2' \in \mathfrak m_ S$. Then $b_ i$ and $a_ i$ have the same image in $k$ for $i = 1, 2$ and

\begin{align*} a_1a_2 & = (f(b_1) + a_1')(f(b_2) + a_2') \\ & = f(b_1)a_2' + f(b_2)a_1' \\ & = f(b_1)(f(b_2) + a_2') + f(b_2)(f(b_1) + a_1') \\ & = f(b_1)a_2 + f(b_2)a_1 \end{align*}

in $Q$ which proves the Leibniz rule. Hence $D : S \to Q$ is a $\Lambda$-derivation which is zero on composing with $R \to S$. Since $Q \not= 0$ there also exist derivations $D : S \to k$ which are zero on composing with $R \to S$, i.e., $\text{Der}_\Lambda (S, k) \to \text{Der}_\Lambda (R, k)$ is not injective. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).