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 :)

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

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