I am having trouble understanding a proof given in Marcus's Number Fields. The theorem is the following: Let $\omega=e^{2\pi i/m}$ where $m=p^r$ and $p$ is a prime. Then the ring of integers of $\mathbb{Q}[\omega]$ is $\mathbb{Z}[\omega]$. In the proof the following theorem is invoked:
Theorem 9: Let $\{\alpha_1,…,\alpha_n\}$ be a basis for $K$ over $\mathbb{Q}$ consisting entirely of algebraic integers, and set $d=\mathrm{disc}(\alpha_1,…,\alpha_m)$. Then every $\alpha\in R$ can be expressed in the form $\frac{m_1\alpha_1+\cdots m_n\alpha_n}{d}$ with all $m_j\in\mathbb{Z}$ and all $m_j^2$ divisible by $d$.
Now, the proof of the claim that the ring of integers of $\mathbb{Q}[\omega]$ is $\mathbb{Z}[\omega]$ goes as follows:
By Theorem $9$ every $\alpha\in R=\mathbb{A}\cap\mathbb{Q}[\omega]$ can be expressed in the form $\frac{m_1+m_2(1-\omega)+\cdots+ m_n(1-\omega)^{n-1}}{d}$ where $n=\varphi(p^r)$, all $m_i\in\mathbb{Z}$, and $d=\mathrm{disc}(1-\omega)=\mathrm{disc}(\omega)$. We have already seen that $\mathrm{disc}(\omega)$ is a divisor of $m^{\varphi(m)}$, hence $d$ is a power of $p$. We show that $R=\mathbb{Z}[1-\omega]$.
So far everything makes sense to me. The next line is the source of my confusion: "Suppose that $R\neq\mathbb{Z}[1-\omega]$. Then there must be some $\alpha$ for which not all the $m_i$ are divisible by $d$".
Why is this true? Doesn't Theorem $9$ ensure that the $m_i$ are divisible by $d$? What am I missing?
Best Answer
Nothing says $d^2 \mid m$ what it says is $d \mid m^2$. So if $d$ were square-free then $m$ is divisible by $d$. But if $d$ is not square free this doesn't need to hold.
For example: $d = 2^2 \cdot 3$ and $m = 2 \cdot 3$ then $m^2$ is divisible by $d$ but $m$ is not. (But observe that the square-free part of $d$ [the $3$] must divide $m$.)