I want to show that the ring of integers of the cubic number field $K = \mathbb Q(\alpha)$, where $\alpha$ is a root of $f = X^3 – X – 2$, is equal to $\mathbb Z[\alpha]$.
$(1, \alpha, \alpha^2)$ forms a $\mathbb Q$-basis of $K$ consisting of integers. I know $\mathbb Z[\alpha] \subseteq \mathcal O_K$. To show equality, I'm guessing the following result is useful:
Let $K$ be a number field of degree $n$, $(\alpha_1, \dots, \alpha_n)$ a $\mathbb Q$-basis of $K$ consisting of integers. Suppose there exist $m \in \mathbb N, m > 2$ and $k_1, \dots, k_n \in \mathbb Z$ with $gcd(m, k_1, \dots, k_n) = 1$ such that $$\frac{k_1 \alpha_1 + \dots + k_n \alpha_n}{m} \in \mathcal O_K$$ then $m^2 \,\big\vert\, disc(\alpha_1, \dots, \alpha_n)$
I have $disc(\alpha_1, \dots, \alpha_n) = -104$, so $m = 2$, hence I want to show that I cannot have $$\frac{a + b \alpha + c \alpha^2}{2} \in \mathcal O_K$$ for integers $a, b, c \in \mathbb Z$ of which at least one is odd. But I'm getting stuck here. How do I show these aren't integral elements? Can anyone give me a hint on how to proceed?
Best Answer
Here a totally different (but quite simple) approach to the problem:
The submodule of $\mathcal{O}_K$ generated by $(1,\alpha,\alpha^2)$ is clearly $\mathbb{Z}[\alpha]$. You have already shown that $$\text{disc}(1,\alpha,\alpha^2)=-104$$ Now, note that the following formula is true (it can be found in most introductory textbook on algebraic number theory): $$\text{disc}(1,\alpha,\alpha^2)=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\text{disc}(\mathcal{O}_K)$$ It follows that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ equals $1$ or $2$. In the latter case, it follows that $\text{disc}(\mathcal{O}_K)=-26$. But this cannot be true, since $\text{disc}(\mathcal{O}_K) \equiv 0$ or $1 \text{ mod } 4$ (this fact is known as "Stickelberger's theorem on discriminants" which is also contained in a lot of textbooks on the topic). Hence $[\mathcal{O}_K:\mathbb{Z}[\alpha]]=1$ and therefore $\mathcal{O}_K=\mathbb{Z}[\alpha]$.