By rewriting
\begin{equation*}
M \otimes^L_A A^{\triangleright} \langle T \rangle
\cong M \otimes^L_A A^{\triangleright} \otimes^L_{\underline{\ZZ}\solid} \underline{\ZZ} \langle T \rangle
\end{equation*}
we reduce to the case
\(A = (\underline{\ZZ}, \Mod_{\underline{\ZZ} \solid})\text{.}\) Hereafter, we implicitly apply
Theorem 7.4.1 to transfer results between
\(\Ab_\solid\) and
\(\CAb_\solid\text{.}\)
We next pass through a sequence of reduction steps on
\(M\text{.}\) By
Proposition 2.4.8, we may assume that
\(M\) is finitely presented. By
Proposition 2.3.12, we have a splitting
\begin{equation*}
M = \left( \prod_I \underline{\ZZ} \right) \oplus \iExt^1_{\underline{\ZZ}}(\underline{Q}, \underline{\ZZ})
\end{equation*}
where
\(I\) is at most countable and
\(Q\) is an at most countable abelian group, and we may treat the two factors separately. For the first factor, using the compatibility of solid tensor product with countable products, we may reduce to the case where
\(I\) is a singleton, at which point the claim is apparent. For the second factor, using
Remark 2.3.11 we may further reduce to the cases where
\(Q\) is either a torsion group or a
\(\QQ\)-vector space.
Suppose that
\(M = \iExt^1_{\underline{\ZZ}}(\underline{Q}, \underline{\ZZ})\) where
\(Q\) is a torsion group. Using
Proposition 2.3.8 and the compatibility of solid tensor product with countable limits, we may reduce to the case where
\(Q\) is itself finite. By the usual structure theorem for finite abelian groups, we may further assume
\(Q\) is even cyclic, say
\(Q = \ZZ/m\ZZ\) and hence
\(M \cong \underline{\ZZ/m\ZZ}\text{.}\) At this point, we must check that for any at most countable set
\(I\text{,}\) an element of
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \langle T \rangle \right)\) is divisible by
\(m\) if its image in
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \llbracket T \llbracket \right)\) is divisible by
\(m\text{.}\) Now identify the elements of
\begin{equation*}
\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \llbracket T \rrbracket \right)
= \prod_\NN \bigoplus_I \ZZ
\end{equation*}
with \(\NN \times I\) matrices whose row vectors have finite support, and the elements of \(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \langle T \rangle \right)\) with \(\NN \times I\) matrices whose row and column vectors have finite support. From this description, it is clear that dividing out a factor of \(m\) does not disturb any support conditions; alternatively, since we already start with a support condition on row vectors, we may reduce to the case where \(I\) is a singleton, which is apparent.
Suppose that
\(M = \iExt^1_{\underline{\ZZ}}(\underline{Q}, \underline{\ZZ})\) where
\(Q\) is a
\(\QQ\)-vector space. By
Example 2.3.10, we then have
\(M \cong \prod_J \underline{\widehat{\ZZ}/\ZZ}\) for some
\(I\) with
\(|J| \leq \kappa\text{;}\) we again reduce to the case where
\(I\) is a singleton. At this point, we must check that for any at most countable set
\(I\text{,}\) an element of
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\widehat{\ZZ}} \langle T \rangle \right)\) factors through
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \langle T \rangle \right)\) if its image in
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\widehat{\ZZ}} \llbracket T \rrbracket \right)\) factors through
\(\Hom_{\underline{\ZZ}}\left( \prod_I \underline{\ZZ}, \underline{\ZZ} \llbracket T \rrbracket \right)\text{.}\) Using the matrix interpretation from the previous paragraph, we reduce to the case where
\(I\) is a singleton, which is apparent.