Lemma 15.115.13. Let $A$ be a ring which contains a primitive $p$th root of unity $\zeta $. Set $w = 1 - \zeta $. Then
is an element of $A[z]$ and in fact $a_ i \in (w)$. Moreover, we have
in the polynomial ring $A[z_1, z_2]$.
Lemma 15.115.13. Let $A$ be a ring which contains a primitive $p$th root of unity $\zeta $. Set $w = 1 - \zeta $. Then
is an element of $A[z]$ and in fact $a_ i \in (w)$. Moreover, we have
in the polynomial ring $A[z_1, z_2]$.
Proof. It suffices to prove this when
is the ring of integers of the cyclotomic field. The polynomial identity $t^ p - 1 = (t - 1)(t - \zeta ) \ldots (t - \zeta ^{p - 1})$ (which is proved by looking at the roots on both sides) shows that $t^{p - 1} + \ldots + t + 1 = (t - \zeta ) \ldots (t - \zeta ^{p - 1})$. Substituting $t = 1$ we obtain $p = (1 - \zeta )(1 - \zeta ^2) \ldots (1 - \zeta ^{p - 1})$. The maximal ideal $(p, w) = (w)$ is the unique prime ideal of $A$ lying over $p$ (as fields of characteristic $p$ do not have nontrivial $p$th roots of $1$). It follows that $p = u w^{p - 1}$ for some unit $u$. This implies that
for $p > i > 1$ and $- 1 + a_1 = pw/w^ p = u$. Since $P(-1) = 0$ we see that $0 = (-1)^ p - u$ modulo $(w)$. Hence $a_1 \in (w)$ and the proof if the first part is done. The second part follows from a direct computation we omit. $\square$
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