forked from stacks/stacks-project
-
Notifications
You must be signed in to change notification settings - Fork 0
/
crystalline.tex
5696 lines (5240 loc) · 208 KB
/
crystalline.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
\input{preamble}
% OK, start here.
%
\begin{document}
\title{Crystalline Cohomology}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
This chapter is based on a lecture series given by Johan de Jong
held in 2012 at Columbia University.
The goals of this chapter are to give a quick introduction to
crystalline cohomology. A reference is the book \cite{Berthelot}.
\medskip\noindent
We have moved the more elementary purely algebraic discussion of divided
power rings to a preliminary chapter as it is also useful
in discussing Tate resolutions in commutative algebra.
Please see Divided Power Algebra, Section \ref{dpa-section-introduction}.
\section{Divided power envelope}
\label{section-divided-power-envelope}
\noindent
The construction of the following lemma will be dubbed the
divided power envelope. It will play an important role later.
\begin{lemma}
\label{lemma-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal
with $IB \subset J$. There exists a homomorphism of
divided power rings
$$
(A, I, \gamma) \longrightarrow (D, \bar J, \bar \gamma)
$$
such that
$$
\Hom_{(A, I, \gamma)}((D, \bar J, \bar \gamma), (C, K, \delta)) =
\Hom_{(A, I)}((B, J), (C, K))
$$
functorially in the divided power algebra $(C, K, \delta)$ over
$(A, I, \gamma)$. Here the LHS is morphisms of divided
power rings over $(A, I, \gamma)$ and the RHS is morphisms of
(ring, ideal) pairs over $(A, I)$.
\end{lemma}
\begin{proof}
Denote $\mathcal{C}$ the category of divided power rings
$(C, K, \delta)$. Consider the functor
$F : \mathcal{C} \longrightarrow \textit{Sets}$ defined by
$$
F(C, K, \delta) =
\left\{
(\varphi, \psi)
\Big |
\begin{matrix}
\varphi : (A, I, \gamma) \to (C, K, \delta)
\text{ homomorphism of divided power rings} \\
\psi : (B, J) \to (C, K)\text{ an }
A\text{-algebra homomorphism with }\psi(J) \subset K
\end{matrix}
\right\}
$$
We will show that
Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown}
applies to this functor which will
prove the lemma. Suppose that $(\varphi, \psi) \in F(C, K, \delta)$.
Let $C' \subset C$ be the subring generated by $\varphi(A)$,
$\psi(B)$, and $\delta_n(\psi(f))$ for all $f \in J$.
Let $K' \subset K \cap C'$ be the ideal of $C'$ generated by
$\varphi(I)$ and $\delta_n(\psi(f))$ for $f \in J$.
Then $(C', K', \delta|_{K'})$ is a divided power ring and
$C'$ has cardinality bounded by the cardinal
$\kappa = |A| \otimes |B|^{\aleph_0}$.
Moreover, $\varphi$ factors as $A \to C' \to C$ and $\psi$ factors
as $B \to B' \to B$. This proves assumption (1) of
Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown}
holds. Assumption (2) is clear
as limits in the category of divided power rings commute with the
forgetful functor $(C, K, \delta) \mapsto (C, K)$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-limits} and its proof.
\end{proof}
\begin{definition}
\label{definition-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal
with $IB \subset J$. The divided power algebra $(D, \bar J, \bar\gamma)$
constructed in Lemma \ref{lemma-divided-power-envelope}
is called the {\it divided power envelope of $J$ in $B$
relative to $(A, I, \gamma)$} and is denoted $D_B(J)$ or $D_{B, \gamma}(J)$.
\end{definition}
\noindent
Let $(A, I, \gamma) \to (C, K, \delta)$ be a homomorphism of divided
power rings. The universal property of
$D_{B, \gamma}(J) = (D, \bar J, \bar \gamma)$ is
$$
\begin{matrix}
\text{ring maps }B \to C \\
\text{ which map }J\text{ into }K
\end{matrix}
\longleftrightarrow
\begin{matrix}
\text{divided power homomorphisms} \\
(D, \bar J, \bar \gamma) \to (C, K, \delta)
\end{matrix}
$$
and the correspondence is given by precomposing with the map $B \to D$
which corresponds to $\text{id}_D$. Here are some properties of
$(D, \bar J, \bar \gamma)$ which follow directly from the universal
property. There are $A$-algebra maps
\begin{equation}
\label{equation-divided-power-envelope}
B \longrightarrow D \longrightarrow B/J
\end{equation}
The first arrow maps $J$ into $\bar J$ and $\bar J$ is the kernel
of the second arrow. The elements $\bar\gamma_n(x)$ where $n > 0$
and $x$ is an element in the image of $J \to D$ generate $\bar J$
as an ideal in $D$ and generate $D$ as a $B$-algebra.
\begin{lemma}
\label{lemma-divided-power-envelop-quotient}
Let $(A, I, \gamma)$ be a divided power ring.
Let $\varphi : B' \to B$ be a surjection of $A$-algebras with kernel $K$.
Let $IB \subset J \subset B$ be an ideal. Let $J' \subset B'$
be the inverse image of $J$. Write
$D_{B', \gamma}(J') = (D', \bar J', \bar\gamma)$.
Then $D_{B, \gamma}(J) = (D'/K', \bar J'/K', \bar\gamma)$
where $K'$ is the ideal generated by the elements $\bar\gamma_n(k)$
for $n \geq 1$ and $k \in K$.
\end{lemma}
\begin{proof}
Write $D_{B, \gamma}(J) = (D, \bar J, \bar \gamma)$.
The universal property of $D'$ gives us a homomorphism $D' \to D$
of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we
see that $D' \to D$ is surjective (see remarks above). It is clear that
$\bar\gamma_n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e.,
we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided
power structure on $\bar J'/K' \subset D'/K'$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win.
\end{proof}
\noindent
In the situation of Definition \ref{definition-divided-power-envelope}
we can choose a surjection $P \to B$ where $P$ is a polynomial
algebra over $A$ and let $J' \subset P$ be the inverse image of $J$.
The previous lemma describes $D_{B, \gamma}(J)$ in terms of
$D_{P, \gamma}(J')$. Note that $\gamma$ extends to a divided power
structure $\gamma'$ on $IP$ by
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}. Hence
$D_{P, \gamma}(J') = D_{P, \gamma'}(J')$ is an example of a special
case of divided power envelopes we describe in the following lemma.
\begin{lemma}
\label{lemma-describe-divided-power-envelope}
Let $(B, I, \gamma)$ be a divided power algebra. Let $I \subset J \subset B$
be an ideal. Let $(D, \bar J, \bar \gamma)$ be the divided power envelope
of $J$ relative to $\gamma$. Choose elements $f_t \in J$, $t \in T$ such
that $J = I + (f_t)$. Then there exists a surjection
$$
\Psi : B\langle x_t \rangle \longrightarrow D
$$
of divided power rings mapping $x_t$ to the image of $f_t$ in $D$.
The kernel of $\Psi$ is generated by the elements $x_t - f_t$ and
all
$$
\delta_n\left(\sum r_t x_t - r_0\right)
$$
whenever $\sum r_t f_t = r_0$ in $B$ for some $r_t \in B$, $r_0 \in I$.
\end{lemma}
\begin{proof}
In the statement of the lemma we think of $B\langle x_t \rangle$
as a divided power ring with ideal
$J' = IB\langle x_t \rangle + B\langle x_t \rangle_{+}$, see
Divided Power Algebra, Remark \ref{dpa-remark-divided-power-polynomial-algebra}.
The existence of $\Psi$ follows from the universal property of
divided power polynomial rings. Surjectivity of $\Psi$ follows from
the fact that its image is a divided power subring of $D$, hence equal to $D$
by the universal property of $D$. It is clear that
$x_t - f_t$ is in the kernel. Set
$$
\mathcal{R} = \{(r_0, r_t) \in I \oplus \bigoplus\nolimits_{t \in T} B
\mid \sum r_t f_t = r_0 \text{ in }B\}
$$
If $(r_0, r_t) \in \mathcal{R}$ then it is clear that
$\sum r_t x_t - r_0$ is in the kernel.
As $\Psi$ is a homomorphism of divided power rings
and $\sum r_tx_t = r_0 \in J'$
it follows that $\delta_n(\sum r_t x_t - r_0)$ is in the kernel as well.
Let $K \subset B\langle x_t \rangle$ be the ideal generated by
$x_t - f_t$ and the elements $\delta_n(\sum r_t x_t - r_0)$ for
$(r_0, r_t) \in \mathcal{R}$.
To show that $K = \text{Ker}(\Psi)$ it suffices to show that
$\delta$ extends to $B\langle x_t \rangle/K$. Namely, if so the universal
property of $D$ gives a map $D \to B\langle x_t \rangle/K$
inverse to $\Psi$. Hence we have to show that $K \cap J'$ is
preserved by $\delta_n$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Let $K' \subset B\langle x_t \rangle$ be the ideal
generated by the elements
\begin{enumerate}
\item $\delta_m(\sum r_t x_t - r_0)$ where $m > 0$ and
$(r_0, r_t) \in \mathcal{R}$,
\item $x_{t'}^{[m]}(x_t - f_t)$ where $m > 0$ and $t', t \in I$.
\end{enumerate}
We claim that $K' = K \cap J'$. The claim proves that $K \cap J'$
is preserved by $\delta_n$, $n > 0$ by the criterion of
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel} (2)(c)
and a computation of $\delta_n$
of the elements listed which we leave to the reader.
To prove the claim note that $K' \subset K \cap J'$.
Conversely, if $h \in K \cap J'$ then, modulo $K'$ we can write
$$
h = \sum r_t (x_t - f_t)
$$
for some $r_t \in B$. As $h \in K \cap J' \subset J'$
we see that $r_0 = \sum r_t f_t \in I$. Hence $(r_0, r_t) \in \mathcal{R}$
and we see that
$$
h = \sum r_t x_t - r_0
$$
is in $K'$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-divided-power-envelope-add-variables}
Let $(A, I, \gamma)$ be a divided power ring.
Let $B$ be an $A$-algebra and $IB \subset J \subset B$ an ideal.
Let $x_i$ be a set of variables. Then
$$
D_{B[x_i], \gamma}(JB[x_i] + (x_i)) = D_{B, \gamma}(J) \langle x_i \rangle
$$
\end{lemma}
\begin{proof}
One possible proof is to deduce this from
Lemma \ref{lemma-describe-divided-power-envelope}
as any relation between $x_i$ in $B[x_i]$ is trivial.
On the other hand, the lemma follows from the universal property
of the divided power polynomial algebra and the universal property of
divided power envelopes.
\end{proof}
\noindent
Conditions (1) and (2) of the following lemma hold if $B \to B'$ is flat
at all primes of $V(IB') \subset \Spec(B')$ and is very closely related
to that condition, see
Algebra, Lemma \ref{algebra-lemma-what-does-it-mean}.
It in particular says that taking the divided power
envelope commutes with localization.
\begin{lemma}
\label{lemma-flat-base-change-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $B \to B'$ be a homomorphism of $A$-algebras.
Assume that
\begin{enumerate}
\item $B/IB \to B'/IB'$ is flat, and
\item $\text{Tor}_1^B(B', B/IB) = 0$.
\end{enumerate}
Then for any ideal $IB \subset J \subset B$ the canonical map
$$
D_B(J) \otimes_B B' \longrightarrow D_{B'}(JB')
$$
is an isomorphism.
\end{lemma}
\begin{proof}
Set $D = D_B(J)$ and denote $\bar J \subset D$ its divided power ideal
with divided power structure $\bar\gamma$. The universal property of
$D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in
the lemma. It suffices to show that
the divided powers $\bar\gamma$ extend to $D \otimes_B B'$ since then
the universal property of $D_{B'}(JB')$ will produce a map
$D_{B'}(JB') \to D \otimes_B B'$ inverse to the one in the lemma.
\medskip\noindent
Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$.
In particular $B \to P$ is flat, hence $D \to D \otimes_B P$ is flat by
Algebra, Lemma \ref{algebra-lemma-flat-base-change}.
Then $\bar\gamma$ extends to $D \otimes_B P$ by
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}; we
will denote this extension
$\bar\gamma$ also. Set $\mathfrak a = \text{Ker}(P \to B')$ so that
we have the short exact sequence
$$
0 \to \mathfrak a \to P \to B' \to 0
$$
Thus $\text{Tor}_1^B(B', B/IB) = 0$ implies that
$\mathfrak a \cap IP = I\mathfrak a$.
Now we have the following commutative diagram
$$
\xymatrix{
B/J \otimes_B \mathfrak a \ar[r]_\beta &
B/J \otimes_B P \ar[r] &
B/J \otimes_B B' \\
D \otimes_B \mathfrak a \ar[r]^\alpha \ar[u] &
D \otimes_B P \ar[r] \ar[u] &
D \otimes_B B' \ar[u] \\
\bar J \otimes_B \mathfrak a \ar[r] \ar[u] &
\bar J \otimes_B P \ar[r] \ar[u] &
\bar J \otimes_B B' \ar[u]
}
$$
This diagram is exact even with $0$'s added at the top and the right.
We have to show the divided powers on the ideal
$\bar J \otimes_B P$ preserve the ideal
$\text{Im}(\alpha) \cap \bar J \otimes_B P$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Consider the exact sequence
$$
0 \to \mathfrak a/I\mathfrak a \to P/IP \to B'/IB' \to 0
$$
(which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above).
As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after
applying $B/J \otimes_{B/IB} -$, see
Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}. Hence
$$
\text{Ker}(B/J \otimes_{B/IB} \mathfrak a/I\mathfrak a \to
B/J \otimes_{B/IB} P/IP) =
\text{Ker}(\mathfrak a/J\mathfrak a \to P/JP)
$$
is zero. Thus $\beta$ is injective. It follows that
$\text{Im}(\alpha) \cap \bar J \otimes_B P$ is the
image of $\bar J \otimes \mathfrak a$. Now if
$f \in \bar J$ and $a \in \mathfrak a$, then
$\bar\gamma_n(f \otimes a) = \bar\gamma_n(f) \otimes a^n$
hence the result is clear.
\end{proof}
\noindent
The following lemma is a special case of
\cite[Proposition 2.1.7]{dJ-crystalline} which in turn is a
generalization of \cite[Proposition 2.8.2]{Berthelot}.
\begin{lemma}
\label{lemma-flat-extension-divided-power-envelope}
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
\begin{enumerate}
\item $B/I \to B'/I'$ is flat, and
\item $J' = JB' + I'$.
\end{enumerate}
Then the canonical map
$$
D_{B, \gamma}(J) \otimes_B B' \longrightarrow D_{B', \gamma'}(J')
$$
is an isomorphism.
\end{lemma}
\begin{proof}
Set $D = D_B(J)$ and denote $\bar J \subset D$ its divided power ideal
with divided power structure $\bar\gamma$. The universal property of
$D$ produces a homomorphism of divided power rings $D \to D_{B'}(J')$,
whence a map as in the lemma. It suffices to show that
there exist divided powers on the image of
$D \otimes_B I' + \bar J \otimes_B B' \to D \otimes_B B'$
compatible with $\bar \gamma$ and $\gamma'$ since then
the universal property of $D_{B'}(J')$ will produce a map
$D_{B'}(J') \to D \otimes_B B'$ inverse to the one in the lemma.
\medskip\noindent
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 \ref{lemma-describe-divided-power-envelope}. 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
\begin{equation}
\label{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
(\ref{equation-base-change}) of $D \otimes_B B'$ is identical
to the presentation of $D_{B', \gamma'}(J')$ obtain in
Lemma \ref{lemma-describe-divided-power-envelope} 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 \ref{algebra-lemma-flat-eq}) to see
that there exist an $m > 0$ and
$r_{jt} \in B$ and $c_j \in B'$, $j = 1, \ldots, m$ such
that
$$
r_{j0} = \sum r_{jt} f_t \in I \text{ for } j = 1, \ldots, m,
\quad\text{and}\quad
r'_t = \sum c_j r_{jt}.
$$
Note that this also implies that $r'_0 = \sum c_j r_{j0}$.
Then we have
\begin{align*}
\delta'_n(\sum r'_t x_t - r'_0)
& =
\delta'_n(\sum c_j (\sum 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})
\end{align*}
where the sum is over $n_1 + \ldots + n_m = n$. This proves what we want.
\end{proof}
\section{Some explicit divided power thickenings}
\label{section-explicit-thickenings}
\noindent
The constructions in this section will help us to define the connection
on a crystal in modules on the crystalline site.
\begin{lemma}
\label{lemma-divided-power-first-order-thickening}
Let $(A, I, \gamma)$ be a divided power ring. Let $M$ be an $A$-module.
Let $B = A \oplus M$ as an $A$-algebra where $M$ is an ideal of square zero
and set $J = I \oplus M$. Set
$$
\delta_n(x + z) = \gamma_n(x) + \gamma_{n - 1}(x)z
$$
for $x \in I$ and $z \in M$.
Then $\delta$ is a divided power structure and
$A \to B$ is a homomorphism of divided power rings from
$(A, I, \gamma)$ to $(B, J, \delta)$.
\end{lemma}
\begin{proof}
We have to check conditions (1) -- (5) of
Divided Power Algebra, Definition \ref{dpa-definition-divided-powers}.
We will prove this directly for this case, but please see the proof of
the next lemma for a method which avoids calculations.
Conditions (1) and (3) are clear. Condition (2) follows from
\begin{align*}
\delta_n(x + z)\delta_m(x + z)
& =
(\gamma_n(x) + \gamma_{n - 1}(x)z)(\gamma_m(x) + \gamma_{m - 1}(x)z) \\
& = \gamma_n(x)\gamma_m(x) + \gamma_n(x)\gamma_{m - 1}(x)z +
\gamma_{n - 1}(x)\gamma_m(x)z \\
& =
\frac{(n + m)!}{n!m!} \gamma_{n + m}(x) +
\left(\frac{(n + m - 1)!}{n!(m - 1)!} +
\frac{(n + m - 1)!}{(n - 1)!m!}\right)
\gamma_{n + m - 1}(x) z \\
& =
\frac{(n + m)!}{n!m!}\delta_{n + m}(x + z)
\end{align*}
Condition (5) follows from
\begin{align*}
\delta_n(\delta_m(x + z))
& =
\delta_n(\gamma_m(x) + \gamma_{m - 1}(x)z) \\
& =
\gamma_n(\gamma_m(x)) + \gamma_{n - 1}(\gamma_m(x))\gamma_{m - 1}(x)z \\
& =
\frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x) +
\frac{((n - 1)m)!}{(n - 1)! (m!)^{n - 1}}
\gamma_{(n - 1)m}(x) \gamma_{m - 1}(x) z \\
& = \frac{(nm)!}{n! (m!)^n}(\gamma_{nm}(x) + \gamma_{nm - 1}(x) z)
\end{align*}
by elementary number theory. To prove (4) we have to see that
$$
\delta_n(x + x' + z + z')
=
\gamma_n(x + x') + \gamma_{n - 1}(x + x')(z + z')
$$
is equal to
$$
\sum\nolimits_{i = 0}^n
(\gamma_i(x) + \gamma_{i - 1}(x)z)
(\gamma_{n - i}(x') + \gamma_{n - i - 1}(x')z')
$$
This follows easily on collecting the coefficients of
$1$, $z$, and $z'$ and using condition (4) for $\gamma$.
\end{proof}
\begin{lemma}
\label{lemma-divided-power-second-order-thickening}
Let $(A, I, \gamma)$ be a divided power ring. Let $M$, $N$ be $A$-modules.
Let $q : M \times M \to N$ be an $A$-bilinear map.
Let $B = A \oplus M \oplus N$ as an $A$-algebra with multiplication
$$
(x, z, w)\cdot (x', z', w') = (xx', xz' + x'z, xw' + x'w + q(z, z') + q(z', z))
$$
and set $J = I \oplus M \oplus N$. Set
$$
\delta_n(x, z, w) = (\gamma_n(x), \gamma_{n - 1}(x)z,
\gamma_{n - 1}(z)w + \gamma_{n - 2}(x)q(z, z))
$$
for $(a, m, n) \in J$.
Then $\delta$ is a divided power structure and
$A \to B$ is a homomorphism of divided power rings from
$(A, I, \gamma)$ to $(B, J, \delta)$.
\end{lemma}
\begin{proof}
Suppose we want to prove that property (4) of
Divided Power Algebra, Definition \ref{dpa-definition-divided-powers}
is satisfied. Pick $(x, z, w)$ and $(x', z', w')$ in $J$.
Pick a map
$$
A_0 = \mathbf{Z}\langle s, s'\rangle \longrightarrow A,\quad
s \longmapsto x,
s' \longmapsto x'
$$
which is possible by the universal property of divided power
polynomial rings. Set $M_0 = A_0 \oplus A_0$ and
$N_0 = A_0 \oplus A_0 \oplus M_0 \otimes_{A_0} M_0$.
Let $q_0 : M_0 \times M_0 \to N_0$ be the obvious map.
Define $M_0 \to M$ as the $A_0$-linear map which sends
the basis vectors of $M_0$ to $z$ and $z'$. Define $N_0 \to N$
as the $A_0$ linear map which sends the first two basis vectors
of $N_0$ to $w$ and $w'$ and uses
$M_0 \otimes_{A_0} M_0 \to M \otimes_A M \xrightarrow{q} N$
on the last summand. Then we see that it suffices to prove the
identity (4) for the situation $(A_0, M_0, N_0, q_0)$.
Similarly for the other identities. This reduces us to the case
of a $\mathbf{Z}$-torsion free ring $A$ and $A$-torsion free modules.
In this case all we have to do is show that
$$
n! \delta_n(x, z, w) = (x, z, w)^n
$$
in the ring $A$, see Divided Power Algebra, Lemma \ref{dpa-lemma-silly}.
To see this note that
$$
(x, z, w)^2 = (x^2, 2xz, 2xw + 2q(z, z))
$$
and by induction
$$
(x, z, w)^n = (x^n, nx^{n - 1}z, nx^{n - 1}w + n(n - 1)x^{n - 2}q(z, z))
$$
On the other hand,
$$
n! \delta_n(x, z, w) = (n!\gamma_n(x), n!\gamma_{n - 1}(x)z,
n!\gamma_{n - 1}(x)w + n!\gamma_{n - 2}(x) q(z, z))
$$
which matches. This finishes the proof.
\end{proof}
\section{Compatibility}
\label{section-compatibility}
\noindent
This section isn't required reading; it explains how our discussion
fits with that of \cite{Berthelot}.
Consider the following technical notion.
\begin{definition}
\label{definition-compatible}
Let $(A, I, \gamma)$ and $(B, J, \delta)$ be divided power rings.
Let $A \to B$ be a ring map. We say
{\it $\delta$ is compatible with $\gamma$}
if there exists a divided power structure $\bar\gamma$ on
$J + IB$ such that
$$
(A, I, \gamma) \to (B, J + IB, \bar \gamma)\quad\text{and}\quad
(B, J, \delta) \to (B, J + IB, \bar \gamma)
$$
are homomorphisms of divided power rings.
\end{definition}
\noindent
Let $p$ be a prime number. Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to C$ be a ring map with $p$ nilpotent in $C$.
Assume that $\gamma$ extends to $IC$ (see
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}).
In this situation, the (big affine) crystalline site of
$\Spec(C)$ over $\Spec(A)$
as defined in \cite{Berthelot}
is the opposite of the category of systems
$$
(B, J, \delta, A \to B, C \to B/J)
$$
where
\begin{enumerate}
\item $(B, J, \delta)$ is a divided power ring with $p$ nilpotent in $B$,
\item $\delta$ is compatible with $\gamma$, and
\item the diagram
$$
\xymatrix{
B \ar[r] & B/J \\
A \ar[u] \ar[r] & C \ar[u]
}
$$
is commutative.
\end{enumerate}
The conditions
``$\gamma$ extends to $C$ and $\delta$ compatible with $\gamma$''
are used in \cite{Berthelot} to insure that
the crystalline cohomology of $\Spec(C)$ is the same as the crystalline
cohomology of $\Spec(C/IC)$. We will avoid this issue
by working exclusively with $C$ such that $IC = 0$\footnote{Of course there
will be a price to pay.}. In this case,
for a system $(B, J, \delta, A \to B, C \to B/J)$ as above,
the commutativity of the displayed diagram above implies $IB \subset J$ and
compatibility is equivalent to the condition that
$(A, I, \gamma) \to (B, J, \delta)$ is a homomorphism of divided
power rings.
\section{Affine crystalline site}
\label{section-affine-site}
\noindent
In this section we discuss the algebraic variant of the crystalline site.
Our basic situation in which we discuss this material will be as
follows.
\begin{situation}
\label{situation-affine}
Here $p$ is a prime number, $(A, I, \gamma)$ is a divided power
ring such that $A$ is a $\mathbf{Z}_{(p)}$-algebra, and $A \to C$ is a
ring map such that $IC = 0$ and such that $p$ is nilpotent in $C$.
\end{situation}
\noindent
Usually the prime number $p$ will be contained in the
divided power ideal $I$.
\begin{definition}
\label{definition-affine-thickening}
In Situation \ref{situation-affine}.
\begin{enumerate}
\item A {\it divided power thickening} of $C$ over $(A, I, \gamma)$
is a homomorphism of divided power algebras $(A, I, \gamma) \to (B, J, \delta)$
such that $p$ is nilpotent in $B$ and a ring map $C \to B/J$ such that
$$
\xymatrix{
B \ar[r] & B/J \\
& C \ar[u] \\
A \ar[uu] \ar[r] & A/I \ar[u]
}
$$
is commutative.
\item A {\it homomorphism of divided power thickenings}
$$
(B, J, \delta, C \to B/J) \longrightarrow (B', J', \delta', C \to B'/J')
$$
is a homomorphism $\varphi : B \to B'$ of divided power $A$-algebras such
that $C \to B/J \to B'/J'$ is the given map $C \to B'/J'$.
\item We denote $\text{CRIS}(C/A, I, \gamma)$ or simply $\text{CRIS}(C/A)$
the category of divided power thickenings of $C$ over $(A, I, \gamma)$.
\item We denote $\text{Cris}(C/A, I, \gamma)$ or simply $\text{Cris}(C/A)$
the full subcategory consisting of $(B, J, \delta, C \to B/J)$ such that
$C \to B/J$ is an isomorphism. We often denote such an object
$(B \to C, \delta)$ with $J = \text{Ker}(B \to C)$ being understood.
\end{enumerate}
\end{definition}
\noindent
Note that for a divided power thickening $(B, J, \delta)$ as above
the ideal $J$ is locally nilpotent, see
Divided Power Algebra, Lemma \ref{dpa-lemma-nil}.
There is a canonical functor
\begin{equation}
\label{equation-forget-affine}
\text{CRIS}(C/A) \longrightarrow C\text{-algebras},\quad
(B, J, \delta) \longmapsto B/J
\end{equation}
This category does not have equalizers or fibre products in general.
It also doesn't have an initial object ($=$ empty colimit) in general.
\begin{lemma}
\label{lemma-affine-thickenings-colimits}
In Situation \ref{situation-affine}.
\begin{enumerate}
\item $\text{CRIS}(C/A)$ has products,
\item $\text{CRIS}(C/A)$ has all finite nonempty colimits and
(\ref{equation-forget-affine}) commutes with these, and
\item $\text{Cris}(C/A)$ has all finite nonempty colimits and
$\text{Cris}(C/A) \to \text{CRIS}(C/A)$ commutes with them.
\end{enumerate}
\end{lemma}
\begin{proof}
The empty product is $(C, 0, \emptyset)$. If $(B_t, J_t, \delta_t)$ is a
family of objects of $\text{CRIS}(C/A)$ then we can form the product
$(\prod B_t, \prod J_t, \prod \delta_t)$ as in
Divided Power Algebra, Lemma \ref{dpa-lemma-colimits}.
The map $C \to \prod B_t/\prod J_t = \prod B_t/J_t$ is clear.
\medskip\noindent
Given two objects $(B, J, \gamma)$ and $(B', J', \gamma')$ of
$\text{CRIS}(C/A)$ we can form a cocartesian diagram
$$
\xymatrix{
(B, J, \delta) \ar[r] & (B'', J'', \delta'') \\
(A, I, \gamma) \ar[r] \ar[u] & (B', J', \delta') \ar[u]
}
$$
in the category of divided power rings. Then we see that we have
$$
B''/J'' = B/J \otimes_{A/I} B'/J' \longleftarrow C \otimes_{A/I} C
$$
see Divided Power Algebra, Remark \ref{dpa-remark-forgetful}.
Denote $J'' \subset K \subset B''$
the ideal such that
$$
\xymatrix{
B''/J'' \ar[r] & B''/K \\
C \otimes_{A/I} C \ar[r] \ar[u] & C \ar[u]
}
$$
is a pushout, i.e., $B''/K \cong B/J \otimes_C B'/J'$.
Let $D_{B''}(K) = (D, \bar K, \bar \delta)$
be the divided power envelope of $K$ in $B''$ relative to
$(B'', J'', \delta'')$. Then it is easily verified that
$(D, \bar K, \bar \delta)$ is a coproduct of $(B, J, \delta)$ and
$(B', J', \delta')$ in $\text{CRIS}(C/A)$.
\medskip\noindent
Next, we come to coequalizers. Let
$\alpha, \beta : (B, J, \delta) \to (B', J', \delta')$ be morphisms of
$\text{CRIS}(C/A)$. Consider $B'' = B'/ (\alpha(b) - \beta(b))$. Let
$J'' \subset B''$ be the image of $J'$. Let
$D_{B''}(J'') = (D, \bar J, \bar\delta)$ be the divided power envelope of
$J''$ in $B''$ relative to $(B', J', \delta')$. Then it is easily verified
that $(D, \bar J, \bar \delta)$ is the coequalizer of $(B, J, \delta)$ and
$(B', J', \delta')$ in $\text{CRIS}(C/A)$.
\medskip\noindent
By Categories, Lemma \ref{categories-lemma-almost-finite-colimits-exist}
we have all finite nonempty colimits in $\text{CRIS}(C/A)$. The constructions
above shows that (\ref{equation-forget-affine}) commutes with them.
This formally implies part (3) as $\text{Cris}(C/A)$ is the fibre category
of (\ref{equation-forget-affine}) over $C$.
\end{proof}
\begin{remark}
\label{remark-completed-affine-site}
In Situation \ref{situation-affine} we denote
$\text{Cris}^\wedge(C/A)$ the category whose objects are
pairs $(B \to C, \delta)$ such that
\begin{enumerate}
\item $B$ is a $p$-adically complete $A$-algebra,
\item $B \to C$ is a surjection of $A$-algebras,
\item $\delta$ is a divided power structure on $\text{Ker}(B \to C)$,
\item $A \to B$ is a homomorphism of divided power rings.
\end{enumerate}
Morphisms are defined as in Definition \ref{definition-affine-thickening}.
Then $\text{Cris}(C/A) \subset \text{Cris}^\wedge(C/A)$ is the full
subcategory consisting of those $B$ such that $p$ is nilpotent in $B$.
Conversely, any object $(B \to C, \delta)$ of $\text{Cris}^\wedge(C/A)$
is equal to the limit
$$
(B \to C, \delta) = \lim_e (B/p^eB \to C, \delta)
$$
where for $e \gg 0$ the object $(B/p^eB \to C, \delta)$ lies
in $\text{Cris}(C/A)$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
In particular, we see that $\text{Cris}^\wedge(C/A)$ is a full subcategory
of the category of pro-objects of $\text{Cris}(C/A)$, see
Categories, Remark \ref{categories-remark-pro-category}.
\end{remark}
\begin{lemma}
\label{lemma-list-properties}
In Situation \ref{situation-affine}.
Let $P \to C$ be a surjection of $A$-algebras with kernel $J$.
Write $D_{P, \gamma}(J) = (D, \bar J, \bar\gamma)$.
Let $(D^\wedge, J^\wedge, \bar\gamma^\wedge)$ be the
$p$-adic completion of $D$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
For every $e \geq 1$ set $P_e = P/p^eP$ and $J_e \subset P_e$
the image of $J$ and write
$D_{P_e, \gamma}(J_e) = (D_e, \bar J_e, \bar\gamma)$.
Then for all $e$ large enough we have
\begin{enumerate}
\item $p^eD \subset \bar J$ and $p^eD^\wedge \subset \bar J^\wedge$
are preserved by divided powers,
\item $D^\wedge/p^eD^\wedge = D/p^eD = D_e$ as divided power rings,
\item $(D_e, \bar J_e, \bar\gamma)$ is an object of $\text{Cris}(C/A)$,
\item $(D^\wedge, \bar J^\wedge, \bar\gamma^\wedge)$ is equal to
$\lim_e (D_e, \bar J_e, \bar\gamma)$, and
\item $(D^\wedge, \bar J^\wedge, \bar\gamma^\wedge)$ is an object of
$\text{Cris}^\wedge(C/A)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) follows from
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
It is a general property of $p$-adic completion that
$D/p^eD = D^\wedge/p^eD^\wedge$. Since $D/p^eD$ is a divided power ring
and since $P \to D/p^eD$ factors through $P_e$, the universal property of
$D_e$ produces a map $D_e \to D/p^eD$. Conversely, the universal property
of $D$ produces a map $D \to D_e$ which factors through $D/p^eD$. We omit
the verification that these maps are mutually inverse. This proves (2).
If $e$ is large enough, then $p^eC = 0$, hence we see (3) holds.
Part (4) follows from
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
Part (5) is clear from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-set-generators}
In Situation \ref{situation-affine}.
Let $P$ be a polynomial algebra over $A$ and let
$P \to C$ be a surjection of $A$-algebras with kernel $J$.
With $(D_e, \bar J_e, \bar\gamma)$ as in Lemma \ref{lemma-list-properties}:
for every object $(B, J_B, \delta)$ of $\text{CRIS}(C/A)$ there
exists an $e$ and a morphism $D_e \to B$ of $\text{CRIS}(C/A)$.
\end{lemma}
\begin{proof}
We can find an $A$-algebra homomorphism $P \to B$
lifting the map $C \to B/J_B$. By our definition of
$\text{CRIS}(C/A)$ we see that $p^eB = 0$ for
some $e$ hence $P \to B$ factors as $P \to P_e \to B$.
By the universal property of the divided power envelope we
conclude that $P_e \to B$ factors through $D_e$.
\end{proof}
\begin{lemma}
\label{lemma-generator-completion}
In Situation \ref{situation-affine}.
Let $P$ be a polynomial algebra over $A$ and let
$P \to C$ be a surjection of $A$-algebras with kernel $J$.
Let $(D, \bar J, \bar\gamma)$ be the $p$-adic completion of
$D_{P, \gamma}(J)$. For every object $(B \to C, \delta)$ of
$\text{Cris}^\wedge(C/A)$ there
exists a morphism $D \to B$ of $\text{Cris}^\wedge(C/A)$.
\end{lemma}
\begin{proof}
We can find an $A$-algebra homomorphism $P \to B$ compatible
with maps to $C$. By our definition of
$\text{Cris}(C/A)$ we see that $P \to B$ factors as
$P \to D_{P, \gamma}(J) \to B$. As $B$ is $p$-adically complete
we can factor this map through $D$.
\end{proof}
\section{Module of differentials}
\label{section-differentials}
\noindent
In this section we develop a theory of modules of differentials
for divided power rings.
\begin{definition}
\label{definition-derivation}
Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $M$ be an $B$-module.
A {\it divided power $A$-derivation} into $M$ is a map
$\theta : B \to M$ which is additive, annihilates the elements
of $A$, satisfies the Leibniz rule
$\theta(bb') = b\theta(b') + b'\theta(b)$ and satisfies
$$
\theta(\gamma_n(x)) = \gamma_{n - 1}(x)\theta(x)
$$
for all $n \geq 1$ and all $x \in J$.
\end{definition}
\noindent
In the situation of the definition, just as in the case of usual
derivations, there exists a {\it universal divided power $A$-derivation}
$$
\text{d}_{B/A, \delta} : B \to \Omega_{B/A, \delta}
$$
such that any divided power $A$-derivation $\theta : B \to M$ is equal to
$\theta = \xi \circ d_{B/A, \delta}$ for some $B$-linear map
$\Omega_{B/A, \delta} \to M$. If $(A, I, \gamma) \to (B, J, \delta)$
is a homomorphism of divided power rings, then we can forget the
divided powers on $A$ and consider the divided power derivations of
$B$ over $A$. Here are some basic properties of the divided power
module of differentials.
\begin{lemma}
\label{lemma-omega}
Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring and
$A \to B$ a ring map.
\begin{enumerate}
\item Consider $B[x]$ with divided power ideal $(JB[x], \delta')$
where $\delta'$ is the extension of $\delta$ to $B[x]$. Then
$$
\Omega_{B[x]/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B[x] \oplus B[x]\text{d}x.
$$
\item Consider $B\langle x \rangle$ with divided power ideal
$(JB\langle x \rangle + B\langle x \rangle_{+}, \delta')$. Then
$$
\Omega_{B\langle x\rangle/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B\langle x \rangle \oplus
B\langle x\rangle \text{d}x.
$$
\item 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 $k \in K$.
\end{enumerate}
\end{lemma}
\begin{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
\begin{enumerate}
\item $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,
\item $\text{d}a = 0$, $a \in A$,
\item $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,
\item $\text{d}\delta_n(f) = \delta_{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.
\end{enumerate}
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
$\text{d}x^{[n]} = x^{[n - 1]}\text{d}x$.
\end{proof}
\noindent
Let $(A, I, \gamma)$ be a divided power ring. In this setting the
correct version of the powers of $I$ is given by the divided powers
$$
I^{[n]} = \text{ideal generate by }
\gamma_{e_1}(x_1) \ldots \gamma_{e_t}(x_t)
\text{ with }\sum e_j \geq n\text{ and }x_j \in I.
$$
Of course we have $I^n \subset I^{[n]}$. Note that $I^{[1]} = I$.
Sometimes we also set $I^{[0]} = A$.
\begin{lemma}
\label{lemma-diagonal-and-differentials}
Let $(A, I, \gamma) \to (B, J, \delta)$ be a homomorphism
of divided power rings. Let $(B(1), J(1), \delta(1))$ be the coproduct
of $(B, J, \delta)$ with itself over $(A, I, \gamma)$, i.e.,
such that
$$
\xymatrix{
(B, J, \delta) \ar[r] & (B(1), J(1), \delta(1)) \\
(A, I, \gamma) \ar[r] \ar[u] & (B, J, \delta) \ar[u]
}
$$
is cocartesian. Denote $K = \text{Ker}(B(1) \to B)$.
Then $K \cap J(1) \subset J(1)$ is preserved by the divided power
structure and
$$
\Omega_{B/A, \delta} = K/ \left(K^2 + (K \cap J(1))^{[2]}\right)
$$
canonically.
\end{lemma}
\begin{proof}
The fact that $K \cap J(1) \subset J(1)$ is preserved by the divided power
structure follows from the fact that $B(1) \to B$ is a homomorphism of
divided power rings.
\medskip\noindent
Recall that $K/K^2$ has a canonical $B$-module structure.