I would like to run through the proof of Dirichlet Unit Theorem 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

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

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.

Using this answer from Math.SE 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 - 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.

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 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...

  • $\begingroup$ @kconrad Can you outline what Dirichlet's proof might have been? $\endgroup$ Jun 12, 2016 at 1:20
  • $\begingroup$ I have not looked at Dirichlet's original argument. If you want to see a proof via the Pigeonhole Principle, see section 183 of Supplement XI (mostly pp. 593-596) of Dirichlet-Dedekind's Vorlesungen Ueber Zahlentheorie. $\endgroup$
    – KConrad
    Jun 12, 2016 at 1:41
  • 1
    $\begingroup$ Forget my previous comment. A proof via the pigeonhole principle in English is in Section 2.10 of Koch's "Number Theory: Algebraic Numbers and Functions" (AMS, 2000). Comparing the notation/equations appearing in Dirichlet-Dedekind and Koch, it's clear that the proofs in these two places are essentially the same. $\endgroup$
    – KConrad
    Jun 12, 2016 at 1:52
  • $\begingroup$ @KConrad in this modern time Dirichlet's lectures are on the internet archive.org/details/vorlesungenberz02dirigoog There's a chapter on Quadratic Forms, on Pell's Equation and the very last chapter is on Number Fields $\endgroup$ Jun 12, 2016 at 2:06
  • 1
    $\begingroup$ OK, I've updated the file you linked to so anyone looking at that in the future will see references to proofs via the pigeonhole principle if they are curious. $\endgroup$
    – KConrad
    Jun 12, 2016 at 2:40

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.