**Proof.**
Write $A$ as the quotient of a polynomial algebra $P$ over $R^ G$. Then $(P \otimes _{R^ G} R)^ G = P$ because $P$ is free as an $R^ G$-module. Hence part (1) follows from Lemma 15.110.1.

Let $J = \mathop{\mathrm{Ker}}(P \to A)$. Lift $a$ as in (2) to an element $f \in P$. Then $f \otimes 1$ maps to zero in $A \otimes _{R^ G} R$. Hence $f \otimes 1$ is in $(J')^ G$ where $J' \subset P \otimes _{R^ G} R$ is the image of the map $J \otimes _{R^ G} R \to P \otimes _{R^ G} R$. Apply Lemma 15.110.1 to $f \otimes 1$ and the surjective ring map

\[ \text{Sym}^*_{R^ G}(J) \otimes _{R^ G} R \longrightarrow A' \subset \text{Sym}^*_{R^ G}(P) \otimes _{R^ G} R \]

which defines $A'$. We obtain $P \in (\text{Sym}^*_{R^ G}(J) \otimes _{R^ G} R)^ G[T]$ mapping to $(T - f \otimes 1)^ n$ in $A'[T]$. Apply part (1) to see that there exists a $P' \in \text{Sym}^*_{R^ G}(J)[T, T']$ whose image is $(T' - P)^ n$. Since $\text{Sym}_{R^ G}^*(P)$ is still free over $R^ G$ we conclude that $P'$ maps to $(T' - (T - f)^ n)^ n$ in $\text{Sym}_{R^ G}^*(P)$. On the other hand, tracing through the construction of the polynomials $P$ and $P'$ in Lemma 15.110.1 we see that $P'$ is congruent to $(T' - T^ n)^ n$ modulo the irrelevant ideal of the graded ring $\text{Sym}^*_{R^ G}(J)$. It follows that

\[ (T' - (T - a)^ n)^ n = (T' - T^ n)^ n \]

in $A[T', T]$. Setting $T' = 0$ for example we obtain the statement of the lemma.
$\square$

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