Lemma 60.17.1. Let $D$ and $D(n)$ be as in (60.17.0.1) and (60.17.0.4). The coprojection $P \to P \otimes _ A \ldots \otimes _ A P$, $f \mapsto f \otimes 1 \otimes \ldots \otimes 1$ induces an isomorphism

60.17.1.1
$$\label{crystalline-equation-structure-Dn} D(n) = \mathop{\mathrm{lim}}\nolimits _ e D\langle \xi _ i(j) \rangle /p^ eD\langle \xi _ i(j) \rangle$$

of algebras over $D$ with

$\xi _ i(j) = x_ i \otimes 1 \otimes \ldots \otimes 1 - 1 \otimes \ldots \otimes 1 \otimes x_ i \otimes 1 \otimes \ldots \otimes 1$

for $j = 1, \ldots , n$ where the second $x_ i$ is placed in the $j + 1$st slot; recall that $D(n)$ is constructed starting with the $n + 1$-fold tensor product of $P$ over $A$.

Proof. We have

$P \otimes _ A \ldots \otimes _ A P = P[\xi _ i(j)]$

and $J(n)$ is generated by $J$ and the elements $\xi _ i(j)$. Hence the lemma follows from Lemma 60.2.5. $\square$

Comment #5442 by Hao on

In the definition of $\xi_i(j)$, is the second $x_i$ at the $j$-th place?

Comment #5444 by Hao on

In the definition of $\xi_i(j)$, is the second $x_i$ at the $j$-th place? If so, why is the equation in the proof true? For example, there is only one $i$ and $n=1$?

Comment #5666 by on

Aha, the confusion is that the numbering starts at $0$. Observe that $D = D(0)$ see just above this lemma in the text. Hence $D(1)$ has one more variable than $D$ for each $i$ and you can take it to be $x_i \otimes 1 - 1 \otimes x_i$. Please don't get me started on the signs! Anyway, I tried to clarify this here.

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