# Class number/ quadratic field/divisibility-$2$

As I am poor in construction of mathematical problem, I am not getting good answers from MSE members. However, this time I constructed the following problem in best possible way. So, I hope I will get some good answers from readers!

Let $n$ be a positive integer, $k$ be an odd prime number, and $e$ a nonnegative integer. If $n\ne4$ and $e\ne0$, then $n$ divides the class number of the imaginary quadratic field $\Bbb Q\left[\sqrt{1-4(2k^e)^n}\right]$.

If anyone can justify the above statement with clear proof, I am so thankful to them.

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A decent LaTeX reference: ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf –  Todd Wilcox Dec 20 '12 at 6:11
Dear vmrfdu, As far as I can tell, this is a special case of your previous question math.stackexchange.com/questions/258485/…, and copycat's answer is the same as Bullwinkle's answer to your previous question. Why are you posting (a special case of) the same question again? Regards, –  Matt E Dec 22 '12 at 4:36

Let $l = 2 k^e$.
Let $\alpha = \sqrt{1 - 4 l^n}$, 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 $$\theta^2 - \theta + l^n = 0.$$ Indeed, we even have $\mathbf{Z}[\alpha]$ is the ring of integers of $K$. Let $p$ be a prime dividing $l$. 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 $l$. Since $\theta$ and $1 - \theta$ are co-prime, and since $$\theta \overline{\theta} = \theta (1 - \theta) = l^n,$$ it follows that the exponent of $\mathfrak{p}$ in $\theta$ is $n$-times the exponent of $\mathfrak{p}$ in $l$. In particular, there exists an ideal $\mathfrak{a}$ of norm $l$ such that $\mathfrak{a}^n = (\theta)$. Suppose that $\mathfrak{a}^m$ is principal. Then $$l^m = N(\mathfrak{a}) = a^2 + ab + b^2 l^n = (a + b/2)^2 + b^2(l^n - 1/4) \ge l^n,$$ as long as $b \ne 0$. Yet if $b = 0$, then $(\theta^m) = \mathfrak{a}^{mn} = (a^n)$, and then $\theta^m = \pm a^n$ (the only units in $K$ are $\pm 1$), which is nonsense. Hence $l^m \ge l^n$, and thus $m \ge n$ (using the fact that $l > 1$), 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 $n$.