A question in proof of a theorem in finite fields I am unable to think about an argument while studying the section Finite Fields from Algebra by Thomas Hungerford. 


My question is  - How $m_k(\sum Z_{m_i} ) $ = 0 implies that each u belonging to G is a root of polynomial $ x^{m_k} -1_F $  belonging to F[x] . 

Please tell. 
 A: $$u \in G \cong_{\varphi} \mathbb{Z}_{m_1} \oplus \cdots \oplus \mathbb{Z}_{m_k} \implies \varphi(u) \in \prod_{i=1}^{k}\mathbb{Z}_{m_i} \\ \implies \mathrm{ord}_G(u) = \mathrm{ord}_{\prod_{i}\mathbb{Z}_{m_i}}(\varphi(u)) \  |\ m_k \implies u^{m_k} = 1_F.$$
A: Alternative approach with minimal group theory (but some number theory instead):
If $G$ is a finite (abelian, but who cares?) group, then every element of $G$ has order dividing $|G|$. By definition, this means $a^{|G|}=1$ for all $a\in G$. I.e., each $a\in G$ is a root of the polynomial $X^{|G|}-1$. (end of group theory)
As there are at most $|G|$ distinct roots of that polynomial, the elements of $G$ are precisely the roots of $X^{|G|}-1$. 
For $n\in\Bbb N$, let $r(n)$ denote the number of roots of $X^n-1$ in $F$.
let $p(n)$ denote the number of primitive $n$th roots of unity in $F$, i.e., numbers that are roots of $X^n-1$, but not roots of $X^k-1$ for any smaller $k$
Then each root of $X^n-1$ is a primitive root for some divisor of $n$, i.e., 
 $$r(n)=\sum_{d\mid n}p(d).$$
By elementary number theory, it then follows that 
$$p(n)=\sum_{d\mid n}\mu(d)r(n/d).$$
From the first two paragraphs, we know that $r(|G|)=|G|$. Consequently, $r(d)=d$ for all divisors $d$ of $|G|$.
Then
$$p(|G|)= \sum_{d\mid |G|}\mu(d)\frac {|G|}d=\varphi(|G|)>0.$$
(Where $\varphi$ is the Euler totient function).
Therefore, among the $|G|$ roots of $X^n-1$ in $F$ that make up the set $G$, there is at least one root $g$ with $g^n=1$ and $g^k\ne1$ for all $0<k<n$. Then (back to group theory) $G$ is a cyclic group with $g$ as a generator.
A: This is just a switch from viewing the group's operation as an addition in the context of talking about the group $G$ on its own, to the fact that the group is multiplicative in the context of field $F$.
$G$ was represented as $G = \mathbb{Z}_{m_1} \oplus \cdots \oplus \mathbb{Z}_{m_k} = \sum \mathbb{Z}_{m_i}$, so $m_k\left(\sum \mathbb{Z}_{m_i}\right) = 0$ is saying that multiplying any element $u$ of $G$ by $m_k$ gives the identity element of the group. Multiplying a non-negative integer and an element of an additive group really means:
$$ \underbrace{u + u + \cdots + u}_{m_k} = \mathrm{id}_{(G,+)} $$
To go to the context of field $F$, we need to consider $G$ as a multiplicative group, as it was originally defined, so change the operator:
$$ \underbrace{u \times u \times \cdots \times u}_{m_k} = \mathrm{id}_{(G,\times)} $$
Of course, in $F$, we can represent that product as $u^{m_k}$, and the identity element of the multiplicative group is $\mathrm{id}_{(G,\times)}=1_F$. So $u^{m_k}=1_F$; $u$ is a root of polynomial $x^{m_k}-1 \in F[x]$.
