Lemma 15.110.6. Let $R$ be a ring. Let $G$ be a finite group of order $n$ acting on $R$. Let $A$ be an $R^ G$-algebra.
for $b \in (A \otimes _{R^ G} R)^ G$ there exists a monic polynomial $P \in A[T]$ whose image in $(A \otimes _{R^ G} R)^ G[T]$ is $(T - b)^ n$,
for $a \in \mathop{\mathrm{Ker}}(A \to (A \otimes _{R^ G} R)^ G)$ we have $(T - a)^ n = T^ n$ in $A[T]$.
Proof. Choose a surjection $E \to A$ where $E$ is a polynomial algebra over $R^ G$. Then $(E \otimes _{R^ G} R)^ G = E$ because $E$ is free as an $R^ G$-module. Denote $J = \mathop{\mathrm{Ker}}(E \to A)$. Since tensor product is right exact we see that $A \otimes _{R^ G} R$ is the quotient of $E \otimes _{R^ G} R$ by the ideal generated by $J$. In this way we see that our lemma is a special case of Lemma 15.110.4. $\square$
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