Primitive elements of number fields which span rings of integers My question is the following: given a number field $K$, does there exist a primitive element $\alpha$ of $K$ over $\mathbb{Q}$ such that the ring $\mathcal{O}_K$ of integers of $K$ is isomorphic to $\mathbb{Z}[\alpha]$?
For example, we have that the ring of integers of $\mathbb{Q}(\sqrt{d})$ (where $d$ is a square-free integer) the primitive element which satisfies the condition given above is $(1+\sqrt{d})/2$ or $\sqrt{d}$ whether $d$ is or not congruent to 1 modulo 4.
 A: Dedekind showed that there is not always a primitive element/power basis for the ring of integers of a N.F.
His counter-example was the cubic field $\mathbb Q(\theta)$ where $\theta^3 + \theta^2 - 2\theta + 8$.
It is mentioned as one of the examples on wikipedia and there is a writeup about this in the note by B. Conrad:


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*https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field

*http://math.stanford.edu/~conrad/154Page/handouts/nonprim.pdf
A: To elaborate on @Brennan.Tobias's answer, let $K = \mathbb Q(\alpha)$ where $\alpha$ is a root of $g(X) =X^3+X^2-2X+8$. We will show that there is no $\gamma\in\mathbb Q$ such that $\mathcal O_K = \mathbb Z[\gamma]$.
Indeed, suppose a stronger condition: that $\exists \gamma\in K$ such that $2\nmid [\mathcal O_K:\mathbb Z[\gamma]]$. Let $f(X)$ be the minimal polynomial of $\gamma$; clearly, $f$ is a cubic polynomial.
It follows that we can use the Kummer-Dedekind theorem to factorise the rational prime $2$ in $\mathcal O_K$ by considering $$f(X)\pmod 2,$$
and since any polynomial modulo $2$ can have at most two distinct roots, $2$ does not split completely in $\mathcal O_K$. 
However, one can show using Hensel's Lemma that the polynomial $g(X)$ has three distinct roots in $\mathbb Q_2$ (for example, $|g(0)|_2<|g'(0)_2|_2$ and $|g(1)|_2<|g'(1)_2|_2$), from which it follows that $2$ splits completely in $\mathcal O_K$.
