The Stacks project

Lemma 60.6.6. Let $(A, I, \gamma )$ be a divided power ring. Let $A \to B$ be a ring map and let $IB \subset J \subset B$ be an ideal. Let $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$ be the divided power envelope. Then we have

\[ \Omega _{D/A, \bar\gamma } = \Omega _{B/A} \otimes _ B D \]

First proof. Let $M$ be a $D$-module. We claim that an $A$-derivation $\vartheta : B \to M$ is the same thing as a divided power $A$-derivation $\theta : D \to M$. The claim implies the statement by the Yoneda lemma.

Consider the square zero thickening $D \oplus M$ of $D$. There is a divided power structure $\delta $ on $\bar J \oplus M$ if we set the higher divided power operations zero on $M$. In other words, we set $\delta _ n(x + m) = \bar\gamma _ n(x) + \bar\gamma _{n - 1}(x)m$ for any $x \in \bar J$ and $m \in M$, see Lemma 60.3.1. Consider the $A$-algebra map $B \to D \oplus M$ whose first component is given by the map $B \to D$ and whose second component is $\vartheta $. By the universal property we get a corresponding homomorphism $D \to D \oplus M$ of divided power algebras whose second component is the divided power $A$-derivation $\theta $ corresponding to $\vartheta $. $\square$

Second proof. We will prove this first when $B$ is flat over $A$. In this case $\gamma $ extends to a divided power structure $\gamma '$ on $IB$, see Divided Power Algebra, Lemma 23.4.2. Hence $D = D_{B, \gamma '}(J)$ is equal to a quotient of the divided power ring $(D', J', \delta )$ where $D' = B\langle x_ t \rangle $ and $J' = IB\langle x_ t \rangle + B\langle x_ t \rangle _{+}$ by the elements $x_ t - f_ t$ and $\delta _ n(\sum r_ t x_ t - r_0)$, see Lemma 60.2.4 for notation and explanation. Write $\text{d} : D' \to \Omega _{D'/A, \delta }$ for the universal derivation. Note that

\[ \Omega _{D'/A, \delta } = \Omega _{B/A} \otimes _ B D' \oplus \bigoplus D' \text{d}x_ t, \]

see Lemma 60.6.2. We conclude that $\Omega _{D/A, \bar\gamma }$ is the quotient of $\Omega _{D'/A, \delta } \otimes _{D'} D$ by the submodule generated by $\text{d}$ applied to the generators of the kernel of $D' \to D$ listed above, see Lemma 60.6.2. Since $\text{d}(x_ t - f_ t) = - \text{d}f_ t + \text{d}x_ t$ we see that we have $\text{d}x_ t = \text{d}f_ t$ in the quotient. In particular we see that $\Omega _{B/A} \otimes _ B D \to \Omega _{D/A, \gamma }$ is surjective with kernel given by the images of $\text{d}$ applied to the elements $\delta _ n(\sum r_ t x_ t - r_0)$. However, given a relation $\sum r_ tf_ t - r_0 = 0$ in $B$ with $r_ t \in B$ and $r_0 \in IB$ we see that

\begin{align*} \text{d}\delta _ n(\sum r_ t x_ t - r_0) & = \delta _{n - 1}(\sum r_ t x_ t - r_0)\text{d}(\sum r_ t x_ t - r_0) \\ & = \delta _{n - 1}(\sum r_ t x_ t - r_0) \left( \sum r_ t\text{d}(x_ t - f_ t) + \sum (x_ t - f_ t)\text{d}r_ t \right) \end{align*}

because $\sum r_ tf_ t - r_0 = 0$ in $B$. Hence this is already zero in $\Omega _{B/A} \otimes _ A D$ and we win in the case that $B$ is flat over $A$.

In the general case we write $B$ as a quotient of a polynomial ring $P \to B$ and let $J' \subset P$ be the inverse image of $J$. Then $D = D'/K'$ with notation as in Lemma 60.2.3. By the case handled in the first paragraph of the proof we have $\Omega _{D'/A, \bar\gamma '} = \Omega _{P/A} \otimes _ P D'$. Then $\Omega _{D/A, \bar\gamma }$ is the quotient of $\Omega _{P/A} \otimes _ P D$ by the submodule generated by $\text{d}\bar\gamma _ n'(k)$ where $k$ is an element of the kernel of $P \to B$, see Lemma 60.6.2 and the description of $K'$ from Lemma 60.2.3. Since $\text{d}\bar\gamma _ n'(k) = \bar\gamma '_{n - 1}(k)\text{d}k$ we see again that it suffices to divided by the submodule generated by $\text{d}k$ with $k \in \mathop{\mathrm{Ker}}(P \to B)$ and since $\Omega _{B/A}$ is the quotient of $\Omega _{P/A} \otimes _ A B$ by these elements (Algebra, Lemma 10.131.9) we win. $\square$


Comments (4)

Comment #3445 by ZY on

I was trying to understand the proof of this lemma and got a bit confused. Suppose I am in the following (simplest) setup: let and let be as in the lemma. Suppose that is flat over , and let (D, ID, \gamma') be the extension of to . Then clearly and the lemma asserts that In order to prove this (i.e., the surjection is an isomorphism, we need to show that already in . The proof above suggests that in order to show this, we need to first write and obtain a surjective PD morphism , which gives rise to and then check by lifting it to . Is this the right way to think about the proof? Can we not check the equality directly in ? (Maybe using the fact that is a quotient of the PD symmetric polynomial ?)

Comment #3499 by on

OK, I have checked this proof and it seems OK to me.

But I think there should be another proof of this lemma as well. Namely, we should be able to show directly that given a -module an -derivation is the same thing as a divided power -derivation using the universal property of . To do this consider the square zero thickening of . There is a divided power structure on if we set the higher divided power operations zero on . Consider the -algebra map whose first component is the given map and second component is . By the universal property we get a corresponding map whose second component should be the map corresponding to .

I didn't check this completely, but this should work and we should replace the proof given in the Stacks project by this argument.

Comment #4555 by Tongmu He on

You should take care of the divided power structure on . The correct one is using 60.3.1 to set , for any and .

For example, take . Then the induced map will send to for any , which is indeed a divided power derivation.

Such a beautiful proof should not be disregarded for over one year :)

There are also:

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

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 07HW. Beware of the difference between the letter 'O' and the digit '0'.