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
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