We may assume \(p=2\) as otherwise this follows from ordinary commutativity; this will allow us to use the universal formula (for \(a,b\) in any \(\delta\)-ring)
\begin{equation}
\delta(a-b) = \delta(a) - \delta(b) + b(a-b).\tag{12.2}
\end{equation}
Using
Lemma 5.2.5, we may also reduce to the case where
\(I = (f)\) with
\(f \in A\) distinguished (this is mostly just to simplify notation).
We use the fact that
\((R/A)_{\Prism}\) contains a weakly final object
\((F, IF)\) which moreover is
\(f\)-torsion-free (
Proposition 11.6.5) to compute Hodge-Tate cohomology using the cosimplicial ring
\((F^\bullet, d^\bullet)\)as per
Remark 11.6.7. Lift
\(\eta(t) \in F/IF\) to
\(T \in F^0\text{.}\) Let
\(U,V \in F^1\) and
\(X,Y,Z \in F^2\) be the images of
\(T\) under the various maps
\(F^0 \to F^1\) and
\(F^0 \to F^2\) in the cosimplicial ring
\(F^\bullet\text{,}\) so that
\begin{equation*}
d^0(T) = U-V, \qquad d^1(U) = X - Y + Z.
\end{equation*}
Since \(U-V \in F^1\) vanishes modulo \(f\) (the reductions of \(U\) and \(V\) modulo \(f\) are the two images of \(t\)) and \(F^\bullet\) is \(f\)-torsion-free, the unique element \(\alpha \in F^1\) with \(U-V = f\alpha\) is also a cocycle. Tracing through the construction of the Bockstein differential, we see that \(\beta_I(t)\) equals the image of \(\alpha\) in \(H^1(F^\bullet/fF^\bullet)\text{,}\) so we need to check that the latter squares to zero.
Multiplying by \(f\) again, we may instead check that \(U-V\) squares to zero in \(H^2(f^2 F^\bullet/f^3 F^\bullet)\text{.}\) The square is represented by \((X-Y)(Y-Z) \in f^2 F^2\text{;}\) we will check that this is the boundary of \(f^2 \delta(\alpha) \in f^2 F^1\text{.}\) To begin, note that \(\phi(\alpha)\) is a cocycle because \(\phi\) commutes with the differential in \(F^\bullet\text{.}\) Hence on one hand,
\begin{equation*}
\delta(U-V) = \delta(f\alpha) = f^2 \delta(\alpha) + \phi(\alpha) \delta(f)
\end{equation*}
and so
\begin{equation*}
d^1(\delta(U-V)) = d^1(f^2 \delta(\alpha)).
\end{equation*}
\begin{equation*}
\delta(U-V) = \delta(U) - \delta(V) + V(U-V) = d^1(\delta(T)) + V(U-V)
\end{equation*}
and so
\begin{equation*}
d^1(\delta(U-V)) = d^1(V(U-V)) = Y(X-Y) + Z(Y-Z) - Z(X-Z) = (X-Y)(Y-Z).
\end{equation*}
(Compare
[18], Lecture V, Lemma 5.4.)