The Stacks project

Lemma 60.2.6. Let $(A, I, \gamma )$ be a divided power ring. Let $B \to B'$ be a homomorphism of $A$-algebras. Assume that

  1. $B/IB \to B'/IB'$ is flat, and

  2. $\text{Tor}_1^ B(B', B/IB) = 0$.

Then for any ideal $IB \subset J \subset B$ the canonical map

\[ D_ B(J) \otimes _ B B' \longrightarrow D_{B'}(JB') \]

is an isomorphism.

Proof. Set $D = D_ B(J)$ and denote $\bar J \subset D$ its divided power ideal with divided power structure $\bar\gamma $. The universal property of $D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in the lemma. It suffices to show that the divided powers $\bar\gamma $ extend to $D \otimes _ B B'$ since then the universal property of $D_{B'}(JB')$ will produce a map $D_{B'}(JB') \to D \otimes _ B B'$ inverse to the one in the lemma.

Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$. In particular $B \to P$ is flat, hence $D \to D \otimes _ B P$ is flat by Algebra, Lemma 10.39.7. Then $\bar\gamma $ extends to $D \otimes _ B P$ by Divided Power Algebra, Lemma 23.4.2; we will denote this extension $\bar\gamma $ also. Set $\mathfrak a = \mathop{\mathrm{Ker}}(P \to B')$ so that we have the short exact sequence

\[ 0 \to \mathfrak a \to P \to B' \to 0 \]

Thus $\text{Tor}_1^ B(B', B/IB) = 0$ implies that $\mathfrak a \cap IP = I\mathfrak a$. Now we have the following commutative diagram

\[ \xymatrix{ B/J \otimes _ B \mathfrak a \ar[r]_\beta & B/J \otimes _ B P \ar[r] & B/J \otimes _ B B' \\ D \otimes _ B \mathfrak a \ar[r]^\alpha \ar[u] & D \otimes _ B P \ar[r] \ar[u] & D \otimes _ B B' \ar[u] \\ \bar J \otimes _ B \mathfrak a \ar[r] \ar[u] & \bar J \otimes _ B P \ar[r] \ar[u] & \bar J \otimes _ B B' \ar[u] } \]

This diagram is exact even with $0$'s added at the top and the right. We have to show the divided powers on the ideal $\bar J \otimes _ B P$ preserve the ideal $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$, see Divided Power Algebra, Lemma 23.4.3. Consider the exact sequence

\[ 0 \to \mathfrak a/I\mathfrak a \to P/IP \to B'/IB' \to 0 \]

(which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above). As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after applying $B/J \otimes _{B/IB} -$, see Algebra, Lemma 10.39.12. Hence

\[ \mathop{\mathrm{Ker}}(B/J \otimes _{B/IB} \mathfrak a/I\mathfrak a \to B/J \otimes _{B/IB} P/IP) = \mathop{\mathrm{Ker}}(\mathfrak a/J\mathfrak a \to P/JP) \]

is zero. Thus $\beta $ is injective. It follows that $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$ is the image of $\bar J \otimes \mathfrak a$. Now if $f \in \bar J$ and $a \in \mathfrak a$, then $\bar\gamma _ n(f \otimes a) = \bar\gamma _ n(f) \otimes a^ n$ hence the result is clear. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 60.2: 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 07HD. Beware of the difference between the letter 'O' and the digit '0'.