Techniques for computing fundamental units in cubic extensions

Question

Are there any slick ways of computing the fundamental unit for the cubic polynomial of the form $X^3+aX+b$ over $\mathbb{Q}$? The simplest example would be $X^3+X-1$, where a root $\alpha$ is a unit with inverse $\alpha^2+1$. What about the general cubic of this form?

There's one example in Milne's Algebraic Number Theory notes, but the method only works for a some cubics of this form. Some people I talked to suggested finding the regulator, but that doesn't seem much easier either as I don't see how this would be accessible through anything but the zeta-function.

Answer 1

See for example, section V.3 of Frölich & Taylor's Algebraic number theory, or section 13.6 of Alaca & Wiliams' Introductory algebraic number theory. The general cubic case was done by Voronoi and is (allegedly, as I haven't checked myself) covered in Delone & Fadeev's The theory of irrationalities of the third degree.

Answer 2

Use Artin's inequality: if $K$ is a cubic field with one real embedding and $v > 1$ is a unit in $O_K$ then $|\text{disc}(K)| < 4v^3 + 24$. (I use inequalities on elements of $K$ via the one real embedding of $K$.) The proof of Artin's inequality is a pain. As a corollary, if $u > 1$ is a unit in $O_K$ and $4u^{3/2} + 24 < |\text{disc}(K)|$ then $u$ must be the fundamental unit of $O_K$. (Proof: Let $\varepsilon > 1$ be a fundamental unit, so $u = \varepsilon^n$ for some $n \geq 1$. If $n \geq 2$ then Artin's inequality with $v = \varepsilon$ implies $|\text{disc}(K)| < 4\varepsilon^3 + 24 = 4u^{3/n} + 24 \leq 4u^{3/2} + 24$, which contradicts the hypothesis of the corollary and also explains where the hypothesis comes from.)

Example: $K = {\mathbf Q}(\sqrt[3]{2})$, so $|\text{disc}(K)| = 108$. A unit in $O_K$ is $\sqrt[3]{2}-1$, which is between 0 and 1. Its reciprocal is $u := 1 + \sqrt[3]{2} + \sqrt[3]{4}$. Numerically, $4u^{3/2} + 24$ is around 54.1, which is less than 108, so $u$ is a fundamental unit of $O_K$.

Example: $K = {\mathbf Q}(\alpha)$ where $\alpha^3 +\alpha - 1 = 0$. Then $|\text{disc}(K)| = 31$. The unique real root of $X^3 + X - 1$ is around .68, so less than 1. Its reciprocal will be a unit greater than 1. If we identify the real root of $X^3 + X - 1$ with $\alpha$ then $u := 1/\alpha = \alpha^2 + 1$ is a unit greater than 1 in $O_K$. From a computer, $4u^{3/2} + 24$ is around 31.096, which is greater than $|\text{disc}(K)| = 31$, so it looks like we can't use Artin's inequality to deduce that $u$ is the fundamental unit. However, you could modify the corollary to show $u$ is no worse than the square of a fundamental unit and then check $u$ is not a square in $O_K$ by showing it isn't a square mod $\mathfrak p$ for some prime $\mathfrak p$ in $O_K$. That would prove $u$ is a fundamental unit of $O_K$.

Answer 3

If you want algorithms, check the sources by Voronoi and Buchmann already mentioned, or Cohen's books. If you're satisfied with a method, the following works very well in extensions of small degree and small discriminant. Let $\omega$ be a root of your polynomial and compute lots of norms of elements $a+b\omega$ as well as their prime factorization. Then use linear algebra to eliminate the prime ideals, and you will get a unit $\eta$ (possibly not fundamental). For example, if (say) $(1+3\omega) = {\mathfrak p}_1{\mathfrak q}_1$, $(3-\omega) = {\mathfrak p}_1$ and $(4+7\omega) = {\mathfrak p}_1{\mathfrak q}_1^2$, then the element $(1+3\omega)^2/(3-\omega)(4+7\omega)$ will be a unit.

If you can find a prime ideal such that $\eta \bmod {\mathfrak p}$ is not a square (cube, 5th power) etc, then $\eta$ will not be a power. In the last pages of Artin's notes on algebraic number theory you can find an estimate for cubic fields that shows you how far you will have to go.