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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!

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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 – KCd Mar 31 '12 at 3:53
@KCd, a thousand thanks! – Qiang Zhang Mar 31 '12 at 16:29

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