Subgroup of a unit group Let $\alpha$ be a root of $x^{3}-6x-3$, let $K=\mathbb{Q}(\alpha)$ and let $u_{1}=\frac{\alpha^{3}}{3}=2\alpha+1$, $u_{2}=\alpha+2$.
Prove that $u_{1}$ and $u_{2}$ generate a subgroup of $\mathbb{Z}_{K}^{\times}$ of rank $2$.
Since $K\subset\mathbb{R}$ is an extension of degree $3$ we have that $\mathbb{Z}_{K}^{\times}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}^{2}$.
 A: We should start by noting the degree is not enough to conclude the rank is $2$, if $r=s=1$ then the rank is $1+1-1=1$. However, we can tell that all three roots are real from some basic calculus, so we'll pass that detail over.
To show the indicated units generate a subgroup of rank $2$ we use the logarithm map from the proof of Dirichlet's Unit Theorem and show that the two log vectors are linearly independent. Recall the embedding is given by

$$L(u) = (\log |u|_{\infty_1},\ldots, \log|u|_{\infty_{r+s-1}})$$

In your case you have $L(u_1)$ and $L(u_2)$ and since the infinite places are just the embeddings into $\Bbb R$ you can just write these as
$$L(u) = (\log |\sigma_1(u)|, \ldots , \log|\sigma_{r+s-1}(u)|)$$
for the assorted embeddings into $\Bbb R$. Using calculus we compute the three roots of the minimal polynomial for $\alpha$ are about -2.1451, -0.52398, and 2.6691. Plugging in for $u_1$ and $u_2$ we get

$$\begin{cases}
L(u_1)\approx (1.19095, -3.03739, 1.84659) \\
L(u_2) \approx (-1.93033, 0.389349, 1.54097)
\end{cases}$$

Note that these cannot be linearly dependent, as this would mean they are multiples of one another, however clearly we would need to multiply by a negative number to make the first two coordinates match but a positive one for the third. So they are linearly independent logs, which makes them multiplicatively independent as units. This is in fact the usual way one does computations in practice is via the map, $L:\mathcal{O}_K^\times\to \Bbb R^{r+s}$, and checking the linear independence there.
This is in fact the essence of both Dirichlet's Unit Theorem and the logarithm map $\log: \Bbb R^+\to\Bbb R$ since $\log(x^ay^b) = a\log x +b\log y$ we can see that multiplicative dependence $x^ay^b=1\iff a\log x+b\log y=0$. Dirichlet used $L$ to show that the rank of the unit group was maximal in the trace-zero subspace, but the idea goes beyond that and extends to general computations of ranks of units. If you have stretch goals to compute things like this in general, you'd follow a similar path.
