Loading [MathJax]/extensions/tex2jax.js

The Stacks project

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. $\square$


Comments (0)

There are also:

  • 9 comment(s) on Section 60.6: Module of differentials

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.