Lemma 15.110.3. Let $G$ be a finite group of order $n$ acting on a ring $R$. Let $J \subset R^ G$ be an ideal. For $x \in JR$ we have $\prod _{\sigma \in G} (T - \sigma (x)) = T^ n + a_1 T^{n - 1} + \ldots + a_ n$ with $a_ i \in J$.
Proof. Observe that the polynomial is indeed monic and has coefficients in $R^ G$. We can write $x = f_1 b_1 + \ldots + f_ m b_ m$ with $f_ j \in J$ and $b_ j \in R$. Thus, arguing by induction on $m$, we may assume that $x = y - fb$ with $f \in J$, $b \in R$, and $y \in JR$ such that the result holds for $y$. Then we see that
\[ \prod \nolimits _{\sigma \in G} (T - \sigma (x)) = \prod \nolimits _{\sigma \in G} (T - \sigma (y) + f\sigma (b)) = \prod \nolimits _{\sigma \in G} (T - \sigma (y)) + \sum _{i = 1, \ldots , n} f^ i a_ i \]
where we have
\[ a_ i = \sum \nolimits _{S \subset G,\ |S| = i} \prod \nolimits _{\sigma \in S} \sigma (b) \prod \nolimits _{\sigma \not\in S} (T - \sigma (y)) \]
A computation we omit shows that $a_ i \in R^ G$ (hint: the given expression is symmetric). Thus the polynomial of the statement of the lemma for $x$ is congruent modulo $J$ to the polynomial for $y$ and this proves the induction step. $\square$
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