Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K = \mathbb{Q}(\sqrt{65})$. Let $L = \mathbb{Q}(\sqrt{5}, \sqrt{13})$. Is $L$ the Hilbert class field of $K$? If yes, how would you prove this?

share|cite|improve this question
up vote 5 down vote accepted

First, compute the class number of $K$; the answer is $2$.

Now $L$ is a quadratic extension of $K$, which is unramified except possibly at primes above $5$ (write $L = K(\sqrt{5})$ ) and is also unramified except possibly at primes above $13$ (write $L = K(\sqrt{13})$). Thus $L/K$ is quadratic and unramified everywhere (including at infinity, since it is a totally real extension), and so must be the Hilbert Class Field of $K$.

share|cite|improve this answer
Dear Matt, Could you explain why $L$ is unramified except possibly at primes above 5? Regards, – Makoto Kato Aug 17 '12 at 5:37
@MakotoKato: Dear Makoto, Because this is true of $\mathbb Q(\sqrt{5})/\mathbb Q$, and $L$ is the compositum of $K$ with $\mathbb Q(\sqrt{5})$. Or, more concretely, the ring of integers of $L$ contains $\mathcal O_K[(1+\sqrt{5})/2]$, and the latter ring has relative discriminant $5$ over $\mathcal O_K$, so that the relative discriminant of $\mathcal O_L$ over $\mathcal O_K$ divides $5$. Regards, – Matt E Aug 17 '12 at 5:39
@SushmaPalimar Please read the section Ramification of the following article. – Makoto Kato Oct 7 '13 at 15:35

Since the class number of $K$ is 2, it suffices to prove that $L$ is unramified at every finite prime of $K$. First note that $L = K(\sqrt{5}) = K(\sqrt{13})$.

Let $\mathfrak{D}$ be the different of $L/K$. Let $f(X) = X^2 - 5$. $f'(\sqrt{5}) = 2\sqrt{5}$. Hence $2\sqrt{5} \in \mathfrak{D}$. Similarly $2\sqrt{13} \in \mathfrak{D}$.

Since 5 and 13 are relatively prime, $\sqrt{5}$ and $\sqrt{13}$ are relatively prime in $L$. Hence $1 = \alpha\sqrt{5} + \beta\sqrt{13}$ for some algebraic integers $\alpha, \beta \in L$. Hence $2 = \alpha 2\sqrt{5} + \beta 2\sqrt{13} \in \mathfrak{D}$.

Let $g(X) = X^2 + 3X + 1$. $\gamma = (-3 + \sqrt{5})/2$ is a root of $g(X)$. Hence $g'(\gamma) = \sqrt{5} \in \mathfrak{D}$. Hence $5 = (\sqrt{5})^2 \in \mathfrak{D}$. Since $2, 5 \in \mathfrak{D}$, $1 \in \mathfrak{D}$. Hence $L/K$ is unramified.

share|cite|improve this answer

Your Answer


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.