Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
A decent LaTeX reference: ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf –  Todd Wilcox Dec 20 '12 at 6:11
1  
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

1 Answer 1

up vote 4 down vote accepted
+50

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 $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.