3
$\begingroup$

Let $ K=\mathbb Q(\sqrt[3]{m})$ be a pure cubic field with ring of integer as $\mathfrak{O}_{K}. $ Then by Dirichlet's unit theorem we know that group of units is generated by single unit which is known as fundamental unit. My question is if we are able to get a unit then\

(i) How to know whether it is fundamental?\ (ii) How to get fundamental unit from any unit in the ring?\

Thank you in advance.

$\endgroup$
2
$\begingroup$

Let $\sigma_1,\sigma_2$ be the real and comlex embeddings, and let $f = (\sigma_1,\sigma_2) : K \to \Bbb R \times \Bbb C$.

Then $f$ preserves addition and multiplication, and $N(x) = \sigma_1(x)|\sigma_2(x)|$.

So units are all on the surface $S = \{(y,z) \in \Bbb R \times \Bbb C \mid y|z|=1 \}$. Suppose that you have found a unit (of norm $1$) $u$ with $\sigma_1(u) > 1$. (If you got $\sigma_1(u)<1$, just pick $1/u$ instead).

Then any $n$th root of $u$ is sent by $f$ onto the compact piece of the surface $S$ given by the condition $1 \le y \le \sigma_1(u)$.

Then you only have to look for possible lattice points on that compact surface.
More precisely this gives you some bounds on the coordinates of the possible roots of $u$ in whatever basis for $\mathfrak O_K$ you have, then it is a finite computation to check all of them.

The elements of norm $-1$ are $(-1)$ times an element of norm $1$ so you don't need to check for a possible square root of $u$ of norm $-1$ once you have found the $u$ with the $y$-component closest to $1$.


Here's how to check @Lubin's comment : Let $u = 2^{1/3}$ and $v = u-1$. We want to be sure that $v$ is a fundamental unit.

we have $f(a+bu+cu^2) = (a+bu+cu^2, a-\frac12(bu+cu^2) + i \frac{\sqrt3}2(bu-cu^2))$
Then $f^{-1}(x,y+iz) = (\frac 13(x+2y), \frac 1{3u}(x-y+\sqrt3 z), \frac 1{3u^2}(x-y-\sqrt3 z))$

Very roughly, the surface $x|y+iz|=1$, restricted to $\sqrt v \le x \le 1$ is bounded by $|y|,|z| \le \frac 1{|x|} \le \frac 1 {\sqrt v}$.

Going back to $a,b,c$ we get the bounds
$-1.137698 = \frac 13 (\sqrt v -2/\sqrt v) \le a \le \frac13 (1 + 2/\sqrt v) = 1.640973$
$-1.282880 = \frac 1{3u}(\sqrt v-(1+\sqrt3)/\sqrt v) \le b \le \frac 1{3u}(1+(1+\sqrt 3)/\sqrt v) = 1.682329$
$-1.018222 = \frac 1{3u^2}(\sqrt v-(1+\sqrt3)/\sqrt v) \le c \le \frac 1{3u^2}(1+(1+\sqrt 3)/\sqrt v) = 1.335266$

Hence if $v$ has a norm $1$ root $a+bu+cu^2$, we have $a,b,c \in \{-1;0;1\}$ which leaves $27$ candidates. The elements of norm $1$ among those are $1,-1+u=v,1+u+u^2=1/v$.

As a reality check we can compute the first few roots in $\Bbb R \times \Bbb C$ then look at their coefficients in $\Bbb R \otimes_\Bbb Q K$

The square roots of norm $1$ are $(-0.101474-0.371471u+0.679930u^2),(-0.441357+0.641236u-0.465818u^2)$

The cube roots are $(0.480750-0.480750u+0.480750u^2), (-0.605707+0.605707u+0.302853u^2), (0.763143+0.381571u-0.381571u^2)$

$\endgroup$
  • $\begingroup$ Have you done any computations like this? $\endgroup$ – Lubin Sep 26 '15 at 3:38
  • $\begingroup$ yes, but only on real quadratic fields =s. Also you can restrict the search to $1 \le y \le \sqrt{\sigma_1(u)}$ $\endgroup$ – mercio Sep 26 '15 at 15:29
  • $\begingroup$ Let’s see. It’s “well known” that a primitive unit of $\Bbb Z[2^{1/3}]$ is $2^{1/3}-1$. Can the method you outline verify this? $\endgroup$ – Lubin Sep 26 '15 at 15:43
1
$\begingroup$

Since there are no rots of unity other than $\pm 1$ you only need to check whether your unit is a square, cube, fifth power etc. There do exist elementary lower bounds on the size of units in cubic fields that I have seen I think in Hasse's book on number theory; such lower bounds can be used to bound the possible exponents.

$\endgroup$
0
$\begingroup$

You may want to have a look at:

Jeans, N.S. and Hendy, M.D., 1978. Determining the fundamental unit of a pure cubic field given any unit. Mathematics of Computation, 32(143), pp.925-935.

Here is an image of the abstract:

enter image description here

$\endgroup$

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.