Lemma 59.2.3. Let $(A, I, \gamma )$ be a divided power ring. Let $\varphi : B' \to B$ be a surjection of $A$-algebras with kernel $K$. Let $IB \subset J \subset B$ be an ideal. Let $J' \subset B'$ be the inverse image of $J$. Write $D_{B', \gamma }(J') = (D', \bar J', \bar\gamma )$. Then $D_{B, \gamma }(J) = (D'/K', \bar J'/K', \bar\gamma )$ where $K'$ is the ideal generated by the elements $\bar\gamma _ n(k)$ for $n \geq 1$ and $k \in K$.

Proof. Write $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$. The universal property of $D'$ gives us a homomorphism $D' \to D$ of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we see that $D' \to D$ is surjective (see remarks above). It is clear that $\bar\gamma _ n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e., we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided power structure on $\bar J'/K' \subset D'/K'$, see Divided Power Algebra, Lemma 23.4.3. Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win. $\square$

## Comments (0)

There are also:

• 2 comment(s) on Section 59.2: Divided power envelope

## Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07HB. Beware of the difference between the letter 'O' and the digit '0'.