Lemma 60.2.7. Let $(B, I, \gamma ) \to (B', I', \gamma ')$ be a homomorphism of divided power rings. Let $I \subset J \subset B$ and $I' \subset J' \subset B'$ be ideals. Assume

$B/I \to B'/I'$ is flat, and

$J' = JB' + I'$.

Then the canonical map

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

is an isomorphism.

**Proof.**
Set $D = D_{B, \gamma }(J)$. Choose elements $f_ t \in J$ which generate $J/I$. Set $\mathcal{R} = \{ (r_0, r_ t) \in I \oplus \bigoplus \nolimits _{t \in T} B \mid \sum r_ t f_ t = r_0 \text{ in }B\} $ as in the proof of Lemma 60.2.4. This lemma shows that

\[ D = B\langle x_ t \rangle / K \]

where $K$ is generated by the elements $x_ t - f_ t$ and $\delta _ n(\sum r_ t x_ t - r_0)$ for $(r_0, r_ t) \in \mathcal{R}$. Thus we see that

60.2.7.1
\begin{equation} \label{crystalline-equation-base-change} D \otimes _ B B' = B'\langle x_ t \rangle /K' \end{equation}

where $K'$ is generated by the images in $B'\langle x_ t \rangle $ of the generators of $K$ listed above. Let $f'_ t \in B'$ be the image of $f_ t$. By assumption (1) we see that the elements $f'_ t \in J'$ generate $J'/I'$ and we see that $x_ t - f'_ t \in K'$. Set

\[ \mathcal{R}' = \{ (r'_0, r'_ t) \in I' \oplus \bigoplus \nolimits _{t \in T} B' \mid \sum r'_ t f'_ t = r'_0 \text{ in }B'\} \]

To finish the proof we have to show that $\delta '_ n(\sum r'_ t x_ t - r'_0) \in K'$ for $(r'_0, r'_ t) \in \mathcal{R}'$, because then the presentation (60.2.7.1) of $D \otimes _ B B'$ is identical to the presentation of $D_{B', \gamma '}(J')$ obtain in Lemma 60.2.4 from the generators $f'_ t$. Suppose that $(r'_0, r'_ t) \in \mathcal{R}'$. Then $\sum r'_ t f'_ t = 0$ in $B'/I'$. As $B/I \to B'/I'$ is flat by assumption (1) we can apply the equational criterion of flatness (Algebra, Lemma 10.39.11) to see that there exist an $m > 0$ and $r_{jt} \in B$ and $c_ j \in B'$, $j = 1, \ldots , m$ such that

\[ r_{j0} = \sum \nolimits _ t r_{jt} f_ t \in I \text{ for } j = 1, \ldots , m \]

and

\[ i'_ t = r'_ t - \sum \nolimits _ j c_ j r_{jt} \in I' \text{ for all }t \]

Note that this also implies that $r'_0 = \sum _ t i'_ t f_ t + \sum _ j c_ j r_{j0}$. Then we have

\begin{align*} \delta '_ n(\sum \nolimits _ t r'_ t x_ t - r'_0) & = \delta '_ n( \sum \nolimits _ t i'_ t x_ t + \sum \nolimits _{t, j} c_ j r_{jt} x_ t - \sum \nolimits _ t i'_ t f_ t - \sum \nolimits _ j c_ j r_{j0}) \\ & = \delta '_ n( \sum \nolimits _ t i'_ t(x_ t - f_ t) + \sum \nolimits _ j c_ j (\sum \nolimits _ t r_{jt} x_ t - r_{j0})) \end{align*}

Since $\delta _ n(a + b) = \sum _{m = 0, \ldots , n} \delta _ m(a) \delta _{n - m}(b)$ and since $\delta _ m(\sum i'_ t(x_ t - f_ t))$ is in the ideal generated by $x_ t - f_ t \in K'$ for $m > 0$, it suffices to prove that $\delta _ n(\sum c_ j (\sum r_{jt} x_ t - r_{j0}))$ is in $K'$. For this we use

\[ \delta _ n(\sum \nolimits _ j c_ j (\sum \nolimits _ t r_{jt} x_ t - r_{j0})) = \sum c_1^{n_1} \ldots c_ m^{n_ m} \delta _{n_1}(\sum r_{1t} x_ t - r_{10}) \ldots \delta _{n_ m}(\sum r_{mt} x_ t - r_{m0}) \]

where the sum is over $n_1 + \ldots + n_ m = n$. This proves what we want.
$\square$

## Comments (3)

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