Step in proof of determining the algebraic integers of $\mathbb Q(\sqrt[3]5)$ I'm working through Example 2.22 in Steward and Tall's Algebraic Number Theory book. The goal is to determine the ring of integers of $\mathbb Q(\sqrt[3]5)$. Let $\theta\in\mathbb R$ such that $\theta^3=5$. Let $\omega=e^{2\pi i/3}$. I'm at the point where I need to check whether
$$
\alpha=\frac 13(1+\theta+\theta^2)
$$
or
$$
\beta=\frac 13(2+2\theta+2\theta^2)
$$
are algebraic integers (I've checked the other cases). Let's start with $\alpha$. I can think of two ways to do this. The first would be to consider the norm
$$
N(\alpha)=\frac 1{27}(1+\theta+\theta^2)(1+\omega\theta+\omega^2\theta^2)(1+\omega^2\theta+\omega\theta^2).
$$
If $\alpha$ is an algebaric integer, then $N(\alpha)\in\mathbb Z$. However, the expression of $N(\alpha)$ is a bit complicated (unless I'm overlooking something). The second would be to determine the minimum polynomial $\mu$ of $\alpha$ over $\mathbb Q$, and then invoke that $\alpha$ is an algebraic integer iff $\mu\in\mathbb Z[X]$.
In both cases I'm not too hopeful. Should I still proceed in this way, or is there something obvious I'm missing?
 A: $\alpha$ is not an algebraic integer over $\Bbb Q(\theta),$ since
$$
\begin{align}27\,N(\alpha)&=(1+\theta+\theta^2)(1+\omega\theta+\omega^2\theta^2)(1+\omega^2\theta+\omega\theta^2)\\
&=(1+\theta+\theta^2)(1+5\omega+5\omega^2+(\omega^2+\omega+5)\theta+(\omega+1+\omega^2)\theta^2)\\
&=(1+\theta+\theta^2)(-4+4\theta)\\
&=-4(1-\theta^3)=16
\end{align}
$$
A: I would bite the bullet and carry out the multiplication for $N(\alpha)$. Since the product is clearly invariant if $\theta$ is exchanged throughout for $\omega\theta$ or $\omega^2\theta$, only  terms containing $1$, $\theta^3$ or $\theta^6$, which are themselves invariant under the exchanges given above, will be nonvanishing. We find the following coefficients:
$1$: $1\cdot1\cdot1=\color{blue}{1}$
$\theta^3:$
$1\cdot\omega\theta\cdot\omega^2\theta^2+1\cdot\omega^2\theta\cdot\omega\theta^2+\theta\cdot1\cdot\omega\theta^2+...+\theta\cdot\omega\theta\cdot\omega^2\theta=(1+\omega+\omega^2+1)\theta^3=\color{blue}{\theta^3}$
$\theta^6:\theta^2\cdot\omega^2\theta^2\cdot\omega\theta^2=\color{blue}{\theta^6}$
So the product is $1+\theta^3+\theta^6$ (the ninth cyclotomic polynomial), which is not a multiple of $27$ for $\theta^3=5$ or any other integer value for $\theta^3$. Cyclotomic polynomials of an integer variable where the cyclotomic order is a multiple of $3$ can be divisible by only one power of $3$.
