Suppose that we have a field $K = \mathbb{Q}(\theta)$ such that \begin{equation} \theta^3 - 18\theta - 6 = 0. \tag{*} \end{equation}
- Show that $[K:\mathbb{Q}] = 3.$
- Hence prove that $\{1$, $\theta$, $\theta^2\}$ is an integral basis of $\mathcal{O}_K = \mathbb{Z}[\theta]$.
How would I go about solving this problem? Surely part 1 holds because the basis in part 2 has three elements, hence the dimension of $K/\mathbb{Q}$ as a vector space is 3. But the question is written in such a way that part 1 seems to be needed for part 2. How would I use (*) to prove part 1, and then how would part 2 follow?
(1) The polynomial $f = x^3 - 18x + 6$ is Eisenstein at $2$ (and also at $3$), so it is irreducible. This shows that $[K:\mathbb{Q}] = 3$.
(2) A priori all we know is that $\newcommand{\Z}{\mathbb{Z}} \mathbb{Z}[\theta]$ is a subring of $\newcommand{\O}{\mathcal{O}} \O_K$. We will show that $\newcommand{\disc}{\operatorname{disc}} \Z[\theta] = \O_K$ by showing that $[\O_K : \Z[\theta]] = 1$. By properties of discriminants, we have $$ \disc(\Z[\theta]) = [\O_K : \Z[\theta]]^2 \disc(\O_K) \, . $$ We compute $$ \disc(\Z[\theta]) = \disc(f) = 22356 = 2^2 \cdot 3^5 \cdot 23 \, . $$
Since $f$ is Eisenstein at $2$ and $3$, then the following result implies that $2 \nmid [\O_K : \Z[\theta]]$ and $3 \nmid [\O_K : \Z[\theta]]$. (See Theorem 2.3 of Keith Conrad's Totally ramified primes and Eisenstein polynomials. In fact, this exact field is discussed in Example 2.12 of these notes.)
Theorem. Let $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of an Eisenstein polynomial at $p$. Then $p \nmid [\mathcal{O}_K : \mathbb{Z}[\alpha]]$.
The only other prime factor of $\disc(\Z[\theta])$ is $23$, but since that doesn't appear to a power $\geq 2$, then $23 \nmid [\O_K : \Z[\theta]]$, either. Thus $[\O_K : \Z[\theta]]=1$, so $\O_K = \Z[\theta]$.