# Dedekind's example of a cubic field $K$ for which $O_K$ does not have the form $\mathbb Z[\alpha]$

Let $\alpha$ be a root of $f(X)=X^3+X^2-2X+8$ and $\beta = \frac{4}{\alpha}$. It can be shown that $O_K=\mathbb{Z}[\alpha, \beta]$.

How does one then establish the following ring isomorphism $$\frac{O_K}{2O_K} \cong \mathbb{F}_2 \times \mathbb{F}_2 \times\mathbb{F}_2\ ?$$

We can't use the Dedekind Criterion here, since $2=|O_K : \mathbb{Z}[\alpha]|$ so the index and prime $2$ are not coprime.

This result is important, as it shows that $(2)$ is totally split and the result $O_K \neq \mathbb{Z}[\nu]$ for any $\nu$ would then follow by Dedekind.

I would be able to prove the existence of this isomorphism indirectly (after some calculation/manipulation of norms of ideals) if I know that $2$ is unramified, but I don't see why this should hold.

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If all you need is that $2$ is unramified, and you have that $\mathcal{O}_K=\mathbb{Z}[\alpha,\beta]$, then you should be able to show that $\Delta_K=-503$. The ramified primes are precisely those primes that divide the discriminant.
@Saha It would be easier to decompose the ideal $(2)$ into it's factors. The proof involves that fact that $p$ ramifies if and only if $\mathcal{O}_K/(p)$ has a non-zero nilpotent element, but isn't as quick as showing it directly for this example. If you can show what an integral basis for $K$ is, you should be able to decompose the ideal $(2)$ by looking at the minimal polynomial of $\beta$ and how it factors modulo $2$. – Warren Moore Mar 27 '14 at 21:57