Lemma 60.6.10. Let $A \to B$ be a ring map and let $(J, \delta )$ be a divided power structure on $B$. Let $p$ be a prime number. Assume that $A$ is a $\mathbf{Z}_{(p)}$-algebra and that $p$ is nilpotent in $B/J$. Then we have

\[ \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B_ e/A, \bar\delta } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B/A, \delta }/p^ e\Omega _{B/A, \delta } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B^\wedge /A, \delta ^\wedge }/p^ e \Omega _{B^\wedge /A, \delta ^\wedge } \]

see proof for notation and explanation.

**Proof.**
By Divided Power Algebra, Lemma 23.4.5 we see that $\delta $ extends to $B_ e = B/p^ eB$ for all sufficiently large $e$. Hence the first limit make sense. The lemma also produces a divided power structure $\delta ^\wedge $ on the completion $B^\wedge = \mathop{\mathrm{lim}}\nolimits _ e B_ e$, hence the last limit makes sense. By Lemma 60.6.2 and the fact that $\text{d}p^ e = 0$ (always) we see that the surjection $\Omega _{B/A, \delta } \to \Omega _{B_ e/A, \bar\delta }$ has kernel $p^ e\Omega _{B/A, \delta }$. Similarly for the kernel of $\Omega _{B^\wedge /A, \delta ^\wedge } \to \Omega _{B_ e/A, \bar\delta }$. Hence the lemma is clear.
$\square$

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