| Version 17 |
Version 16 |
| The Dirichlet's unit theorem gives all units of an algebraic number field $\mathbb{Q}(\theta)$ in the unique form |
The Dirichlet's unit theorem gives all units of an algebraic number field $\mathbb{Q}(\theta)$ in the unique form |
| $$\epsilon = \zeta^{n}\eta_1^{k_1}\eta_2^{k_2}...\eta_t^{k_t},$$ |
$$\epsilon = \zeta^{n}\eta_1^{k_1}\eta_2^{k_2}...\eta_t^{k_t},$$ |
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where $\zeta$ is a primitive $w$th root of unity in $\mathbb{Q}(\theta)$, the $\eta_j$'s are the so-called fundamental unit of $\mathbb{Q}(\theta)$, $0 \le n \le w-1$, $k_j \in \mathbb{Z}$ $\forall j$, $t = r+s-1$.
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where $\zeta$ is a primitive $w$th root of unity in $\mathbb{Q}(\theta)$, the $\eta_j$'s are the so-called fundamental units of $\mathbb{Q}(\theta)$, $0 \le n \le w-1$, $k_j \in \mathbb{Z}$ $\forall j$, $t = r+s-1$.
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| \begin{itemize} |
\begin{itemize} |
| \item The case of a real quadratic field $\mathbb{Q}(\sqrt{m})$, the square-free $m > 0$: $r = 2$, $s = 0$, $t = r+s-1 = 1$. So we obtain |
\item The case of a real quadratic field $\mathbb{Q}(\sqrt{m})$, the square-free $m > 0$: $r = 2$, $s = 0$, $t = r+s-1 = 1$. So we obtain |
| $$\epsilon = \zeta^{n}\eta^{k} = \pm\eta^{k},$$ |
$$\epsilon = \zeta^{n}\eta^{k} = \pm\eta^{k},$$ |
| because $\zeta= -1$ is the only primitive real root of unity ($w =2$). Thus, every real quadratic field has infinitely many units and a unique fundamental unit $\eta$. |
because $\zeta= -1$ is the only primitive real root of unity ($w =2$). Thus, every real quadratic field has infinitely many units and a unique fundamental unit $\eta$. |
| Examples: If $m = 3$, then $\eta = 2+\sqrt{3}$; if $m = 421$, then $\eta = \frac{444939+21685\sqrt{421}}{2}$. |
Examples: If $m = 3$, then $\eta = 2+\sqrt{3}$; if $m = 421$, then $\eta = \frac{444939+21685\sqrt{421}}{2}$. |
| \item The case of an imaginary quadratic field $\mathbb{Q}(\theta)$; here $\theta = \sqrt{m}$, $m < 0$: The conjugates of $\theta$ are imaginary numbers $\pm\sqrt{m}$, hence $r = 0$, $2s = 2$, $t = r+s-1 = 0$. Thus we see that all units are |
\item The case of an imaginary quadratic field $\mathbb{Q}(\theta)$; here $\theta = \sqrt{m}$, $m < 0$: The conjugates of $\theta$ are imaginary numbers $\pm\sqrt{m}$, hence $r = 0$, $2s = 2$, $t = r+s-1 = 0$. Thus we see that all units are |
| $$\epsilon = \zeta^{n}.$$ |
$$\epsilon = \zeta^{n}.$$ |
| 1) $m = -1$. The field contains the primitive fourth root of unity, e.g. $i$, and therefore all units in the Gaussian field $\mathbb{Q}(i)$ are $i^n$, where $n = 0, 1, 2, 3$. |
1) $m = -1$. The field contains the primitive fourth root of unity, e.g. $i$, and therefore all units in the Gaussian field $\mathbb{Q}(i)$ are $i^n$, where $n = 0, 1, 2, 3$. |
| 2) $m = -3$. The field in question is a cyclotomic field containing the primitive third root of unity and also the primitive sixth root of unity, namely |
2) $m = -3$. The field in question is a cyclotomic field containing the primitive third root of unity and also the primitive sixth root of unity, namely |
| $$\zeta = \cos{\frac{2\pi}{6}}+i\sin{\frac{2\pi}{6}};$$ |
$$\zeta = \cos{\frac{2\pi}{6}}+i\sin{\frac{2\pi}{6}};$$ |
| hence all units are |
hence all units are |
| $\epsilon = (\frac{1+\sqrt{-3}}{2})^{n}$, where $n = 0, 1, ..., 5$, |
$\epsilon = (\frac{1+\sqrt{-3}}{2})^{n}$, where $n = 0, 1, ..., 5$, |
| or, equivalently, |
or, equivalently, |
| $\epsilon = \pm(\frac{-1+\sqrt{-3}}{2})^{n}$, where $n = 0, 1, 2$. |
$\epsilon = \pm(\frac{-1+\sqrt{-3}}{2})^{n}$, where $n = 0, 1, 2$. |
| 3) $m = -2$, $m <-3$. The only roots of unity in the field are $\pm 1$; hence $\zeta = -1$, $w = 2$, and the units of the field are simply |
3) $m = -2$, $m <-3$. The only roots of unity in the field are $\pm 1$; hence $\zeta = -1$, $w = 2$, and the units of the field are simply |
| $\epsilon = (-1)^{n}$, where $n = 0, 1$. |
$\epsilon = (-1)^{n}$, where $n = 0, 1$. |
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| \end{itemize} |
\end{itemize} |
