Proving that an element generates $\mathcal{O}_K^*/ (\mathcal{O}_K^*)^3$ Let $K = \mathbb{Q}[x] / \langle x^3 + 2x^2 + 6x + 6\rangle$. The polynomial has a single real root and its discriminant is $-588$. Let $\alpha$ be the image of $x$ in $K$.
Then how would I show that $\alpha +1 $ generates $\mathcal{O}_K^*/ (\mathcal{O}_K^*)^3$, where $\mathcal{O}_K^*$ is the set of units in $\mathcal{O}_K$?
 A: First, by Dirichlet's theorem, you can show that the group $\mathcal{O}^{\times}_K/\mathcal{O}^{\times 3}_K$ is cyclic of order $3$. (The signature is $(1,1)$, so there are no roots of unity of order greater than $2$, and $r_1 + r_2 - 1 = 1$.) It then suffices to show that $\alpha + 1$ is not a perfect cube. 
An easy way to show this is to show that $\alpha + 1$ is \emph{not} a cube after reducing to some residue field of $\mathcal{O}^{\times}_K$. For example, if
$$\mathfrak{p} = (\alpha + 61, 97),$$
then $\mathcal{O}_K/\mathfrak{p} = \mathbf{F}_{97}$, and the image of $\alpha + 1$ is $-60$. But
$$(-60)^{(97 - 1)/3} \equiv 35 \not\equiv 1 \mod 97,$$
so $\alpha + 1$ is not a cube.
Remarks:


*

*Why did I choose $97$? In order to work over $\mathbf{F}_p$ instead of a larger finite field, I wanted a prime $\mathfrak{p}$ of prime norm $p$. In order to find something over $\mathbf{F}_p$ which was not a cube, I wanted something $1 \mod 3$, since else everything is a cube. The first two primes with this property (for this field) are $61$ and $97$, and $97$ was the first for which this argument works.

*Was this method guaranteed to work? That is, could an element be a cube modulo all primes $p$ but not be (globally) a cube? No, this would violated the Grunewald-Wang theorem, so this method was guaranteed (eventually) to work (providing $\alpha + 1$ was not actually a cube, of course!)
