Compute the Unit and Class Number of a pure cubic field $\mathbb{Q}(\sqrt[3]{6})$

Find a unit in $\mathbb{Q}(\sqrt[3]{6})$ and show that this field has class number $h=1$.

I am done with the first part which is relatively simple:

Suppose that $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$. Then we have $\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2}$, since the integral base of $\mathbb{Q}(\sqrt[3]{6})$ can be written as $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$. Thus, $$\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2},$$ $$\sqrt[3]{6}\varepsilon=6a+c\sqrt[3]{6}+b\sqrt[3]{6^2},$$ $$\sqrt[3]{6^2}\varepsilon=6b+6a\sqrt[3]{6}+c\sqrt[3]{6^2}.$$

As we see it, the system of equations with variable $\varepsilon$ has only zero solution, since $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$ is a base. Then $$\det\left( \begin{array}{ccc} c-\varepsilon & b & a \\ 6a & c-\varepsilon & b \\ 6b & 6a & c-\varepsilon \\ \end{array} \right)$$ is the minimal polynomial of $\varepsilon$. Since $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$ if and only if $N(\varepsilon)=\pm1$, we take $\varepsilon=0$ in the above polynomial, and $$\det\left( \begin{array}{ccc} c & b & a \\ 6a & c & b \\ 6b & 6a & c \\ \end{array} \right)=\pm1.$$ Compute the determinant we find that $a=33,~b=60,~c=109$ is one of the solutions. Hence a unit in $\mathbb{Q}(\sqrt[3]{6})$ is of the form $\varepsilon=109+60\sqrt[3]{6}+33\sqrt[3]{6^2}$.

For the second part of the problem, I have no idea how to show that $\mathbb{Q}(\sqrt[3]{6})$ is a principal ideal domain.

Any comment will be appreciated!

• Use the Minkowski bound. – KCd Mar 28 '12 at 1:44
• @KCd: Thanks a lot! The bound is approximately 8, but I am confused with the prime decomposition in the field. – Qiang Zhang Mar 28 '12 at 1:49
• See page 1 of www.math.uconn.edu/~kconrad/blurbs/gradnumthy/Qw6.pdf. – KCd Mar 31 '12 at 3:53
• @KCd, a thousand thanks! – Qiang Zhang Mar 31 '12 at 16:29

The first idea for computing units in such fields is finding a generator of a purely ramified prime. Here $2 - \sqrt[3]{6}$ has norm $2$, hence $$(2 - \sqrt[3]{6})^3 = 2(1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2)$$ is $2$ times a unit.
Finding an element generating the prime ideal above $3$ is more difficult, but it turns out that $\beta = 3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2$ is such an element with norm $3$. As above you now get $$(3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2)^3 = 3(109+60\sqrt[3]{6} +33\sqrt[3]{6}^2),$$ and you get the unit you mentioned in your question. Actually we have $$\frac1{1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2} = 109+60\sqrt[3]{6} +33\sqrt[3]{6}^2 .$$ Finding elements of norms $5$ and $7$ is rather easy, which then shows that the ring ${\mathbb Z}[\sqrt[3]{6}]$ is principal.