Let \(D\) be the divided power envelope; it is the smallest subring of \(R \otimes_{\ZZ} \QQ\) containing \(R\) and \(\gamma_n(\delta^m(x))\) for all \(m \geq 0\text{,}\) \(n \geq 1\text{.}\) The maximal ideal on which \(D\) admits divided powers includes \(\delta^m(x)\) for all \(m \geq 0\) by construction, and hence also \(\phi(\delta^m(x)) = \delta^m(x)^p + p \delta^{m+1}(x)\) for all \(m \geq 0\text{;}\) consequently, for all \(m \geq 0\text{,}\) \(n \geq 1\text{,}\)
\begin{equation*}
\phi(\gamma_n(\delta^m(x))) = \gamma_n(\phi(\delta^m(x))) \in D.
\end{equation*}
Hence \(\phi\) induces an endomorphism of \(D\text{.}\)
We next check that \(\phi\) induces a Frobenius lift on \(D\text{;}\) this amounts to checking that for all \(m \geq 0\text{,}\) \(n \geq 1\text{,}\)
\begin{equation*}
\phi(\gamma_n(\delta^m(x))) \equiv \gamma_n(\delta^m(x))^p \pmod{pD}.
\end{equation*}
We will see that in fact both sides are divisible by \(p\text{.}\) For \(\phi(\gamma_n(\delta^m(x))) = \gamma_n(\phi(\delta^m(x)))\text{,}\) this holds by writing
\begin{equation*}
\phi(\delta^m(x))) = p(\delta^m(x)^p/p) + p \delta^{m+1}(x) \in pD
\end{equation*}
and
\begin{equation*}
\phi(\gamma_n(\delta^m(x))) = \gamma_n(\phi(\delta^m(x))) = p^n \gamma_n(\phi(\delta^m(x))/p).
\end{equation*}
For
\(\gamma_n(\delta^m(x))^p\text{,}\) this holds by writing
\(\gamma_n(\delta^m(x))^p = p! \gamma_p(\gamma_n(\delta^m(x)))\) and applying
(14.4).
Since
\(D\) is
\(p\)-torsion-free, by
Lemma 2.1.3 we obtain a
\(\delta\)-structure compatible with
\(R\text{,}\) as desired.