Lemma 60.8.3. In Situation 60.7.5. Let

$\xymatrix{ (U_3, T_3, \delta _3) \ar[d] \ar[r] & (U_2, T_2, \delta _2) \ar[d] \\ (U_1, T_1, \delta _1) \ar[r] & (U, T, \delta ) }$

be a fibre square in the category of divided power thickenings of $X$ relative to $(S, \mathcal{I}, \gamma )$. If $T_2 \to T$ is flat and $U_2 = T_2 \times _ T U$, then $T_3 = T_1 \times _ T T_2$ (as schemes).

Proof. This is true because a divided power structure extends uniquely along a flat ring map. See Divided Power Algebra, Lemma 23.4.2. $\square$

Comment #2313 by Daxin Xu on

Should we add the condition $U_2\simeq T_2\times_T U$? (compare to Berthelot P, Breen L, Messing W. Théorie de Dieudonné cristalline. II, lemma 1.1.2)

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