Lemma 60.6.2. Let $A$ be a ring. Let $(B, J, \delta )$ be a divided power ring and $A \to B$ a ring map.

1. Consider $B[x]$ with divided power ideal $(JB[x], \delta ')$ where $\delta '$ is the extension of $\delta$ to $B[x]$. Then

$\Omega _{B[x]/A, \delta '} = \Omega _{B/A, \delta } \otimes _ B B[x] \oplus B[x]\text{d}x.$
2. Consider $B\langle x \rangle$ with divided power ideal $(JB\langle x \rangle + B\langle x \rangle _{+}, \delta ')$. Then

$\Omega _{B\langle x\rangle /A, \delta '} = \Omega _{B/A, \delta } \otimes _ B B\langle x \rangle \oplus B\langle x\rangle \text{d}x.$
3. Let $K \subset J$ be an ideal preserved by $\delta _ n$ for all $n > 0$. Set $B' = B/K$ and denote $\delta '$ the induced divided power on $J/K$. Then $\Omega _{B'/A, \delta '}$ is the quotient of $\Omega _{B/A, \delta } \otimes _ B B'$ by the $B'$-submodule generated by $\text{d}k$ for $k \in K$.

Proof. These are proved directly from the construction of $\Omega _{B/A, \delta }$ as the free $B$-module on the elements $\text{d}b$ modulo the relations

1. $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,

2. $\text{d}a = 0$, $a \in A$,

3. $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,

4. $\text{d}\delta _ n(f) = \delta _{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.

Note that the last relation explains why we get “the same” answer for the divided power polynomial algebra and the usual polynomial algebra: in the first case $x$ is an element of the divided power ideal and hence $\text{d}x^{[n]} = x^{[n - 1]}\text{d}x$. $\square$

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