Conjugates of an average of roots of unity

Let $m_i$ be the positive order of $z_i$, and set $N=m_1\cdots m_n>0$. Each $m_i$ divides $N$, hence $z_i^N=(z_i^{m_i}{)}^{N\mathrm{/}{m}_i}=1$. The nonconstant polynomial $q(t)=t^N-1$ splits in $\mathbb{C}$ by F5. Let $R$ be its set of roots there and $E=\mathbb{Q}(R)\subset \mathbb{C}$. The linear factorization has $N$ factors, so $R$ is finite. The polynomial splits over $E$, and $E$ is generated by these roots, so it is the cyclotomic splitting field of F4. All $z_i$ belong to $E$.

Each member of $R$ is algebraic over $\mathbb{Q}$, being a root of $q\in \mathbb{Q}[t]\setminus \{0\}$. F6 makes $E\mathrm{/}\mathbb{Q}$ finite, and F7 makes it algebraic. As $1\mathrm{/}n\in \mathbb{Q}$, repeated addition and multiplication in $E$ show that $a\in E$; F10 gives the same finite sum in $E$ and $\mathbb{C}$. In particular $a$ is algebraic by F7.

F8 applies to the algebraic splitting field $E\mathrm{/}\mathbb{Q}$ of the nonzero polynomial $q$, giving normality. Let $p$ be the monic minimal polynomial of $a$ over $\mathbb{Q}$. By F12, $p$ splits over $E$. If $b\in \mathbb{C}$ is a conjugate of $a$, then $p(b)=0$ by F3. Write $p(t)={\prod }_j(t-r_j)$ with $r_j\in E$. The field $\mathbb{C}$ has no zero divisors, so ${\prod }_j(b-r_j)=0$ forces $b=r_j$ for some $j$; thus $b\in E$.

Apply F9 to the identity isomorphism of $\mathbb{Q}$ and the monic irreducible polynomial $p$, with roots $a$ and $b$. It gives a $\mathbb{Q}$-isomorphism $\tau :\mathbb{Q}(a)\to \mathbb{Q}(b)$ satisfying $\tau (a)=b$. Both fields lie in $E$. Since $\mathbb{Q}\subseteq \mathbb{Q}(a)\subseteq E$, one has $\mathbb{Q}(a)(R)=E$; the two inclusions follow respectively because $E$ contains $a,R$ and because $\mathbb{Q}(R)=E$. The same argument gives $\mathbb{Q}(b)(R)=E$. Thus $E$ is a splitting field of $q$ over each of these base fields.

The coefficients of $q$ are rational and $\tau$ fixes them, so ${\tau }_{\ast }q=q\mathrm{\ne }0$. Apply F2 to the two splitting-field structures from step 4.1. It extends $\tau$ to a field isomorphism $\sigma :E\to E$. This is a $\mathbb{Q}$-automorphism and $\sigma (a)=b$.

A field homomorphism preserves finite sums: it sends the empty sum to $0$, and $\sigma (A+z)=\sigma (A)+\sigma (z)$ preserves the assertion when a term is appended; F11 proves this for every length in F10. Since $\sigma (1\mathrm{/}n)=1\mathrm{/}n$, it follows that $b=\sigma (a)=n^{-1}{\sum }_i\sigma (z_i)$.

Multiplicativity gives $\sigma (z_i{)}^{m_i}=\sigma (z_i^{m_i})=1$. If $\sigma (z_i{)}^d=1$ for a positive $d<m_i$, applying ${\sigma }^{-1}$ gives $z_i^d=1$, contrary to the definition of $m_i$ in F1. Thus the order is exactly $m_i$. This covers $m_i=1$ and repeated roots. When $n=1$, the construction still applies with $a=z_1$. When $a=0$, its minimal polynomial is $t$, so $b=0$; steps 4.1–6.1 still apply. No step imposed integrality on $a$.