I would like to run through the proof of [Dirichlet Unit Theorem](http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf) for a cubic field.  

Let's try $\mathbb{Q}[x]/(x^3 - x - 1)$. This has 1 real root and 2 complex roots (or embeddings). 


The units in the order $\mathbb{Z}[x]$ should be $\mathbb{Z}[x]^\times \simeq \mathbb{Z}$ so we can find a single unit. Since the field $\mathbb{Q}(\alpha)$ has a real embedding, the only roots of unity are $\pm 1$.



> 
> Dirichlet did not have Minkowski's theorem available; he proved the Unit Theorem in 1846 while Minkowski developed the geometry of numbers only near the 19th century. His substitute for the convex-body theorem was the pigeon-hole principle. Dirichlet did not state the unit theorem to all orders, but only those of the form $\mathbb{Z}[\alpha]$, since at the time those were the kinds of rings people considered
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Consider all three embeddings at once (two of them are conjugate):


* $ V = \{ (x, z) \in \mathbb{R} \times \mathbb{C} \}$ with norm $ |(x,z)| = x \cdot |z|^2$
* Restricted to algebraic numbers we just multiply both absolute values.
* Let $G = \{ (x,z) : x \, |z|^2 = x \, (z\_1^2 + z\_2^2) = 1 \} \subseteq V^\times $
* $U = \{ (x,z): x = x(\alpha)\in \mathbb{Z}[\alpha]^\times\subseteq \mathbb{R}^\times \text{ and } z = z(\alpha)\in \mathbb{Z}[\alpha]^\times\subseteq \mathbb{C}^\times \} \subseteq G$


This is a strange looking group action on a strange-looking 3D surface. With my mimimal knowledge of number theory I find another key sentence



> 
> the key to proving the theorem is showing the compactness of $G/U$ without knowing the structure of the unit group in advance
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I wanted to say that $G/U$ was a torus but I am not sure of that. Later he does show $\log U $ is a lattice of full rank.




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Using this answer from [Math.SE](https://math.stackexchange.com/a/207775/4997) I can outline a strategy quite similar in spirit. 


My order is $\mathbb{Z}[x] = \mathbb{Z}\cdot 1 \oplus \mathbb{Z}\cdot x \oplus \mathbb{Z}\cdot x^2$ which is a cubic lattice with a certain unit volume.


[**1**](https://math.stackexchange.com/a/56054/4997) - Using Pigeonhole Principle, I can find many integer ~~pairs~~ triples $(a,b,c)$ solving 
$$|a + b x + cx^2 | < \frac{1}{(2N)^2}$$ with $ |b|, |c| \leq 2N$. 


**EDIT** I am forgetting that obviously $N(x) = x|z|^2 =1$, in this particular case, since $x^3 - x - 1 = 0$ (with $x \in \mathbb{R}$).


If we continue, the way I have written it, this seems contradictory that $N(a + bx + cx^2) \in \mathbb{N}$ and yet I have just shown $|a+bx + cx^2| < 1$ in the real embedding. Obviously the other two factors $|a + bz + cz^2| > 1$... 


**2** - Show there are infinitely many $\alpha \in \mathbb{Z}[x]$ such that $|x(\alpha)| \, |z(\alpha)|^2| = M$ for some $1 \leq M \color{lightgray}{< x}$. In fact, here the real root is $x \approx 1.32$. Without a computer $1-1-1 < 0$ and $8-2-1 > 0$ so the real root is $x \in [1,2]$.


~~**3** - By Pigeonhole Principle (again) we find infinitely many $\alpha \in \mathbb{Z}[x]$ such that we conclude the Dirichlet Unit Theorem~~ There should be a step 3.




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Can someone help me fill in details?


* The main step seems to be showing that if we find that $|a + bx + cx^2 | < \frac{1}{N^2} $ then the triple $(a,b,c) \in \mathbb{Z}^3$ also has:
$$ |a + bz + cz^2 | < M \cdot N^2 $$
and conclude infinitely many numbers with $\mathrm{N}( a + bx + cx^2) < M$.
* KConrad's proof suggests we can use any convex set we like. Such as translates of:
$$ a^2 + b^2 + c^2 < r^2 \text{ with } \frac{4}{3}\pi r^3 > 2^3 \cdot \mathrm{Vol}( \mathbb{Z}[x] ) \geq\_? 8$$
using property of the field to show we can cover $G/U$ with only finitely many translates, which is a type of compactness, and do pigeon-hole on that.
* Dirichlet's [Lectures on Number Theory](https://archive.org/details/vorlesungenberz02dirigoog) may have a proof on Supplement XI - which is 150 pages long - and since I don't know enough German I can't pin the exact pages...



Can the Units of a Cubic Field be Proven from Pigeonhole Principle alone?
I didn't check the inequalities in (Step 1), but (Step 3) is straightforward. Let there be infinitely many solutions to $N(\alpha) = M$ for some $M \neq 0$. Then there are two solutions which are congruent modulo $M$: Say $N(\alpha) = N(\alpha') = M$ and $\alpha' = \alpha+M \beta$. Now, $\alpha$ divides $N(\alpha)$, so $\alpha$ divides $\alpha+M \beta = \alpha'$. Likewise, $\alpha'$ divides $\alpha$. So $\alpha/\alpha'$ is a unit.


