Lemma 90.4.5. Let $S$ be an object of $\widehat{\mathcal{C}}_\Lambda$. Then $\dim _ k \text{Der}_\Lambda (S, k) < \infty$.

Proof. Let $x_1, \ldots , x_ n \in \mathfrak m_ S$ map to a $k$-basis for the relative cotangent space $\mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$. Choose $y_1, \ldots , y_ m \in S$ whose images in $k$ generate $k$ over $k'$. We claim that $\dim _ k \text{Der}_\Lambda (S, k) \leq n + m$. To see this it suffices to prove that if $D(x_ i) = 0$ and $D(y_ j) = 0$, then $D = 0$. Let $a \in S$. We can find a polynomial $P = \sum \lambda _ J y^ J$ with $\lambda _ J \in \Lambda$ whose image in $k$ is the same as the image of $a$ in $k$. Then we see that $D(a - P) = D(a) - D(P) = D(a)$ by our assumption that $D(y_ j) = 0$ for all $j$. Thus we may assume $a \in \mathfrak m_ S$. Write $a = \sum a_ i x_ i$ with $a_ i \in S$. By the Leibniz rule

$D(a) = \sum x_ iD(a_ i) + \sum a_ iD(x_ i) = \sum x_ iD(a_ i)$

as we assumed $D(x_ i) = 0$. We have $\sum x_ iD(a_ i) = 0$ as multiplication by $x_ i$ is zero on $k$. $\square$

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