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?
 A: 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 $K$.
A: 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.
