Lemma 15.115.13 (09F2)—The Stacks project

Lemma 15.115.13. Let $A$ be a ring which contains a primitive $p$ th root of unity $\zeta$ . Set $w = 1 - \zeta$ . Then

$P(z) = \frac{(1 + wz)^ p - 1}{w^ p} = z^ p - z + \sum \nolimits _{0 < i < p} a_ i z^ i$

is an element of $A[z]$ and in fact $a_ i \in (w)$ . Moreover, we have

$P(z_1 + z_2 + w z_1 z_2) = P(z_1) + P(z_2) + w^ p P(z_1) P(z_2)$

in the polynomial ring $A[z_1, z_2]$ .

Proof. It suffices to prove this when

$A = \mathbf{Z}[\zeta ] = \mathbf{Z}[x]/(x^{p - 1} + \ldots + x + 1)$

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

$a_ i = \frac{1}{p} {p \choose i} u w^{i - 1}$

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. $\square$

The Stacks project

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