construct a unit on $\mathbb{Z}[\sqrt[3]{7}]$?

Question

how might I construct a unit on $\mathbb{Z}[\sqrt[3]{7}]$? Can it be done using pigeonhole principle as with square roots and Pell equation.

I had been reading about the Voronoi continued fraction or Delone-Fadeev algorithm.

In this case one has $2^3-1=7$ but I am hoping this can be done for general cubic fields. Perhaps I could construct a unit on $\mathbb{Z}[\sqrt[3]{5}]$ instead?

This may help: http://math.stackexchange.com/questions/1163995/what-is-the-norm-of-a-number-in-a-cubic-integer-ring. — lhf, Jun 11 '16 at 18:15
Related: https://math.stackexchange.com/questions/4614795/how-to-check-if-a-solution-to-the-cubic-pells-equation-is-a-fundamental-unit#4614957. — Oscar Lanzi, Jan 10 '23 at 16:42

score 3 · Accepted Answer · answered Jun 11 '16 at 18:12

3

Since $2^3-(\sqrt[3]{7})^3=1$, one unit is $2-\sqrt[3]{7}$

answered Jun 11 '16 at 18:12

Empy2

52,372

This relies on $a^3-b^3=(a-b)(a^2+ab+b^2)$. – lhf Jun 11 '16 at 18:29

construct a unit on $\mathbb{Z}[\sqrt[3]{7}]$?

1 Answers1