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If we take $n$ an even integer and greater than $5$, then $\mathbb Q(\sqrt{1 - 4k^n})$ are divisible by $n$, other than for $k = 13$ and $n = 8$. Why this is happened? If we take $n$ less than 5 (I mean for $n = 2$ or $n = 4$) what will happen?

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Your first claim, how do you know? –  Will Jagy Dec 14 '12 at 6:23
    
@Will Jagy! I had some idea on quadratic fields. I believe the statement. I am not sure. If you can, you can disprove also. –  vmrfdu123456 Dec 14 '12 at 6:44
    
@vmrfdu123456 As I've seen during my edit you have an idea about using LaTeX. Then why don't do this? –  user26857 Dec 29 '12 at 0:06
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@Copycat Please explain the reason of this bounty. As far as I've seen the problem has an accepted answer. –  user26857 Dec 29 '12 at 0:08

1 Answer 1

up vote 6 down vote accepted

Let $\alpha = \sqrt{1 - 4 k^n}$, and let $K = \mathbf{Q}(\sqrt{\alpha})$. Suppose that $k > 1$. I understand your question to be: why is the class number of $K$ divisible by $n$?

The minimal polynomial of $\theta = \displaystyle{\frac{1 + \alpha}{2}}$ is $$\theta^2 - \theta + k^n = 0.$$ Indeed, we even have $\mathbf{Z}[\alpha]$ is the ring of integers of $K$. Let $p$ be a prime dividing $k$. Then $p$ splits in $K$, as can be seen by factoring the polynomial above modulo $p$. Indeed, there is a unique such ideal which divides $\theta$, namely $\mathfrak{p} = (p,\theta)$. Equally, every prime dividing $\theta$ also divides $k$. Since $\theta$ and $1 - \theta$ are co-prime, and since $$\theta \overline{\theta} = \theta (1 - \theta) = k^n,$$ it follows that the exponent of $\mathfrak{p}$ in $\theta$ is $n$-times the exponent of $\mathfrak{p}$ in $k$. In particular, there exists an ideal $\mathfrak{a}$ of norm $k$ such that $\mathfrak{a}^n = (\theta)$. Suppose that $\mathfrak{a}^m$ is principal. Then $$k^m = N(\mathfrak{a}) = a^2 + ab + b^2 k^n = (a + b/2)^2 + b^2(k^n - 1/4) \ge k^n,$$ as long as $b \ne 0$. Yet if $b = 0$, then $(\theta^m) = \mathfrak{a}^{mn} = (a^n)$, and then $\theta = \pm a^n$ (the only units in $K$ are $\pm 1$), which is nonsense. Hence $k^m \ge k^n$, and thus $m \ge n$ (using the fact that $k > 1$), and thus the order of $\mathfrak{a}$ in the class group is exactly $n$. It follows that the class number is divisible by $n$ for any $n$.

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J.Moose! your explanation is too good. But, my question is why the class number of k is not divisible by n, in the cases of n = 2 and n = 4? Please explain. –  vmrfdu123456 Dec 15 '12 at 8:16
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@vmrdu, the class number is divisible by $n$ for all $n$ (I just proved it), and is thus so also for $n=2$, $n=4$, $n=1597$, and any other natural number. –  Bullwinkle J. Moose Dec 16 '12 at 5:48
    
@vrmdu, you have 24 hours to ask an coherent follow up question or accept this answer, otherwise it gets deleted. –  Bullwinkle J. Moose Dec 16 '12 at 6:29
    
Time is running out! –  Bullwinkle J. Moose Dec 17 '12 at 5:43
    
J.Moose! please click on math.stackexchange.com/questions/260516/… and answer... –  vmrfdu123456 Dec 17 '12 at 9:36

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