# Hilbert class field of $\mathbb{Q}(\sqrt{65})$

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?

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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 $K$.

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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. en.wikipedia.org/wiki/Different_ideal#Relative_different –  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.

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