Ring of integers of every cyclotomic field

For $n=1$ one has ${\zeta }_1=1$, $\phi (1)=1$, $\mathbb{Q}({\zeta }_1)=\mathbb{Q}$ and ${\mathcal{O}}_{\mathbb{Q}}=\mathbb{Z}$, so ${\mathcal{O}}_{\mathbb{Q}({\zeta }_1)}=\mathbb{Z}[{\zeta }_1]=\mathbb{Z}$ and the single element $1={\zeta }_1^0$ is a $\mathbb{Z}$-basis. For $n>1$ we keep the notation of the Given; each ${\zeta }_i={\zeta }_n^{m_i}$ has order $p_i^{a_i}$, because ${\zeta }_n$ has order $n$; hence ${\zeta }_i$ is a primitive $p_i^{a_i}$-th root of unity and $K_i=\mathbb{Q}({\zeta }_i)$ is a prime-power cyclotomic field as in [F4].

By induction on $j$ the compositum is $M_j=\mathbb{Q}({\zeta }_{n_j})$ and $[M_j:\mathbb{Q}]=\phi (n_j)={\prod }_{i\le j}\phi (p_i^{a_i})$: for $j=1$ this is $K_1=\mathbb{Q}({\zeta }_{p_1^{a_1}})$ with degree $\phi (p_1^{a_1})$ by [F1]; and if it holds for $j-1$, then $M_j=M_{j-1}{K}_j=\mathbb{Q}({\mu }_{n_{j-1}})\mathbb{Q}({\mu }_{p_j^{a_j}})=\mathbb{Q}({\mu }_{n_j})$ by [F2] because $\mathrm{lcm}(n_{j-1},p_j^{a_j})=n_j$, while $[M_j:\mathbb{Q}]=\phi (n_j)=\phi (n_{j-1})\phi (p_j^{a_j})=[M_{j-1}:\mathbb{Q}][K_j:\mathbb{Q}]$ by [F1] and [F3]. In particular $[M_r:\mathbb{Q}]=\phi (n)=[\mathbb{Q}({\zeta }_n):\mathbb{Q}]$ and $M_r=\mathbb{Q}({\zeta }_n)$ inside $\mathrm{\Omega }$.

The two rings agree: $\mathbb{Z}[{\zeta }_1]\cdots \mathbb{Z}[{\zeta }_r]=\mathbb{Z}[{\zeta }_n]$. Indeed each ${\zeta }_i={\zeta }_n^{m_i}$ lies in $\mathbb{Z}[{\zeta }_n]$, which gives the inclusion $\subseteq$. Conversely, for each $i$ the numbers $m_i$ and $p_i^{a_i}$ are coprime, so [F6] provides $e_i\in \mathbb{Z}$ with $e_i{m}_i\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{p}_i^{a_i})$; since $p_i^{a_i}\mid m_j$ for $j\mathrm{\ne }i$, the integer ${\sum }_j{e}_j{m}_j-1$ is divisible by every $p_i^{a_i}$, and these are pairwise coprime with product $n$, so $n\mid {\sum }_j{e}_j{m}_j-1$ by [F7]. Hence ${\zeta }_n^{{\sum }_j{e}_j{m}_j}={\zeta }_n$, that is ${\zeta }_n={\prod }_j({\zeta }_n^{m_j}{)}^{e_j}={\prod }_j{\zeta }_j^{e_j}\in \mathbb{Z}[{\zeta }_1]\cdots \mathbb{Z}[{\zeta }_r]$, giving the reverse inclusion.

Applying step 1.2 and the compositum theorem [F5] inductively on $j$ gives ${\mathcal{O}}_{M_j}={\mathcal{O}}_{K_1}\cdots {\mathcal{O}}_{K_j}=\mathbb{Z}[{\zeta }_1]\cdots \mathbb{Z}[{\zeta }_j]$, with the products of the individual power bases as an integral basis, and $\mathrm{\mid }{d}_{M_j}\mathrm{\mid }={\prod }_{i\le j}{p}_i^{N_i\cdot [M_j:K_i]}$ is a product of powers of the distinct primes $p_1,\dots ,p_j$. Indeed, for $j=1$ this is [F4]; and for the induction step $M_j=M_{j-1}{K}_j$ satisfies the degree hypothesis by step 1.2, while $\gcd (d_{M_{j-1}},d_{K_j})=1$ because the first discriminant is $\pm$ a product of powers of $p_1,\dots ,p_{j-1}$ and the second is $\pm p_j^{N_j}$ by [F4], so [F5] converts ${\mathcal{O}}_{M_{j-1}}{\mathcal{O}}_{K_j}$ into ${\mathcal{O}}_{M_j}$ and preserves the basis statement.

By steps 1.2, 1.3 and 2.1 with $j=r$, ${\mathcal{O}}_{\mathbb{Q}({\zeta }_n)}={\mathcal{O}}_{M_r}=\mathbb{Z}[{\zeta }_1]\cdots \mathbb{Z}[{\zeta }_r]=\mathbb{Z}[{\zeta }_n]$. Every power ${\zeta }_n^k$ is a $\mathbb{Z}$-linear combination of $1,{\zeta }_n,\dots ,{\zeta }_n^{\phi (n)-1}$: this is clear for $k<\phi (n)$, and the monic relation ${\mathrm{\Phi }}_n({\zeta }_n)=0$ of degree $\phi (n)$ expresses ${\zeta }_n^{\phi (n)}$ as such a combination, after which induction on $k$ handles all larger powers; hence $1,{\zeta }_n,\dots ,{\zeta }_n^{\phi (n)-1}$ spans the $\mathbb{Z}$-module $\mathbb{Z}[{\zeta }_n]={\mathcal{O}}_{\mathbb{Q}({\zeta }_n)}$, and by [F1] these $\phi (n)$ elements are also linearly independent over $\mathbb{Q}$, hence over $\mathbb{Z}$. A linearly independent spanning set of a $\mathbb{Z}$-module is a $\mathbb{Z}$-basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), so $1,{\zeta }_n,\dots ,{\zeta }_n^{\phi (n)-1}$ is an ordered $\mathbb{Z}$-basis of ${\mathcal{O}}_{\mathbb{Q}({\zeta }_n)}$, that is, an integral basis by [F8].