## 60.2 Divided power envelope

The construction of the following lemma will be dubbed the divided power envelope. It will play an important role later.

Lemma 60.2.1. 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

\[ \mathop{\mathrm{Hom}}\nolimits _{(A, I, \gamma )}((D, \bar J, \bar\gamma ), (C, K, \delta )) = \mathop{\mathrm{Hom}}\nolimits _{(A, I)}((B, J), (C, K)) \]

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 $(A, I)$.

**Proof.**
Denote $\mathcal{C}$ the category of divided power rings $(C, K, \delta )$. Consider the functor $F : \mathcal{C} \longrightarrow \textit{Sets}$ defined by

\[ F(C, K, \delta ) = \left\{ (\varphi , \psi ) \middle | \begin{matrix} \varphi : (A, I, \gamma ) \to (C, K, \delta ) \text{ homomorphism of divided power rings}
\\ \psi : (B, J) \to (C, K)\text{ an } A\text{-algebra homomorphism with }\psi (J) \subset K
\end{matrix} \right\} \]

We will show that Divided Power Algebra, Lemma 23.3.3 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 $B \to C' \to C$. This proves assumption (1) of Divided Power Algebra, Lemma 23.3.3 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 23.3.2 and its proof.
$\square$

Definition 60.2.2. 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 60.2.1 is called the *divided power envelope of $J$ in $B$ relative to $(A, I, \gamma )$* and is denoted $D_ B(J)$ or $D_{B, \gamma }(J)$.

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

60.2.2.1
\begin{equation} \label{crystalline-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.

Lemma 60.2.3. 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 $k \in K$.

**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 23.4.3. Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win.
$\square$

In the situation of Definition 60.2.2 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 23.4.2. 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.

Lemma 60.2.4. 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 $r_ t \in B$, $r_0 \in I$.

**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 23.5.2. 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 $K = \mathop{\mathrm{Ker}}(\Psi )$ 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 23.4.3. Let $K' \subset B\langle x_ t \rangle $ be the ideal generated by the elements

$\delta _ m(\sum r_ t x_ t - r_0)$ where $m > 0$ and $(r_0, r_ t) \in \mathcal{R}$,

$x_{t'}^{[m]}(x_ t - f_ t)$ where $m > 0$ and $t', t \in I$.

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 23.4.3 (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.
$\square$

Lemma 60.2.5. 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

\[ D_{B[x_ i], \gamma }(JB[x_ i] + (x_ i)) = D_{B, \gamma }(J) \langle x_ i \rangle \]

**Proof.**
One possible proof is to deduce this from Lemma 60.2.4 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.
$\square$

Conditions (1) and (2) of the following lemma hold if $B \to B'$ is flat at all primes of $V(IB') \subset \mathop{\mathrm{Spec}}(B')$ and is very closely related to that condition, see Algebra, Lemma 10.99.8. It in particular says that taking the divided power envelope commutes with localization.

Lemma 60.2.6. Let $(A, I, \gamma )$ be a divided power ring. Let $B \to B'$ be a homomorphism of $A$-algebras. Assume that

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

$\text{Tor}_1^ B(B', B/IB) = 0$.

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.

**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.

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 10.39.7. Then $\bar\gamma $ extends to $D \otimes _ B P$ by Divided Power Algebra, Lemma 23.4.2; we will denote this extension $\bar\gamma $ also. Set $\mathfrak a = \mathop{\mathrm{Ker}}(P \to B')$ 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 $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$, see Divided Power Algebra, Lemma 23.4.3. 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 10.39.12. Hence

\[ \mathop{\mathrm{Ker}}(B/J \otimes _{B/IB} \mathfrak a/I\mathfrak a \to B/J \otimes _{B/IB} P/IP) = \mathop{\mathrm{Ker}}(\mathfrak a/J\mathfrak a \to P/JP) \]

is zero. Thus $\beta $ is injective. It follows that $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$ 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.
$\square$

The following lemma is a special case of [Proposition 2.1.7, dJ-crystalline] which in turn is a generalization of [Proposition 2.8.2, Berthelot].

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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