Cyclotomic polynomial formula: is it valid in an arbitrary field?

Question

Let $F$ be a field and $n \in \mathbb{N}$. Then an element $\varepsilon \in \overline{F}$, being $\overline{F}$ an algebraic closure of $F$, is called a $n$-th root of unity if it is a root of the polynomial $X^n - 1 \in F[X]$. If I denote by $W_n$ the set of the $n$-th roots of unity, then it is a cyclic group and after I define the $n$-th cyclotomic polynomial by $$ {\phi}_n(X) = \prod_{\varepsilon \in W_n^*} (X - \varepsilon)\mbox{,} $$ where $W_n^*$ is the set of the elements $\varepsilon \in W_n$ such that $\langle \varepsilon \rangle = W_n$ (it is the set of the $n$-th primitive roots of unity). And now I can make my question: is the formula $$ X^n - 1 = \prod_{d \mid n} {\Phi}_d(X) $$ valid for each field $F$ or I need to consider $F = \mathbb{C}$?

Answer 1

There are subtleties if $n\cdot1_F=0_F$. If $F$ has characteristic $p$, then we have $$X^p-1=(X-1)^p.$$ This implies $W_{p}^*=W_{p^2}^*=\{1_F\}$, and the formula fails.

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Other than this, it does hold. See a relatively recent answer by yours truly.