Hilbert class field

In this context, the Hilbert class field of ${\displaystyle K}$ is not just unramified at the finite places (the classical ideal theoretic interpretation) but also at the infinite places of ${\displaystyle K}$. That is, every real embedding of ${\displaystyle K}$ extends to a real embedding of ${\displaystyle E}$, rather than to a complex embedding of ${\displaystyle E}$.

Examples

• If the ring of integers of ${\displaystyle K}$ is a unique factorization domain, in particular if ${\displaystyle K=\mathbb{Q}}$, then ${\displaystyle K}$ is its own Hilbert class field.

• Let ${\displaystyle K=\mathbb{Q}(\sqrt{-15})}$, of discriminant ${\displaystyle -15}$. The field ${\displaystyle L=\mathbb{Q}(\sqrt{-3},\sqrt{5})}$ has discriminant ${\displaystyle 225=(-15{)}^2}$ and so is an everywhere unramified extension of ${\displaystyle K}$, and it is abelian. Using the Minkowski bound, one can show that ${\displaystyle K}$ has class number 2. Hence, its Hilbert class field is ${\displaystyle L}$. A non-principal ideal of ${\displaystyle K}$ is ${\displaystyle (2,(1+\sqrt{-15})/2)}$, and in ${\displaystyle L}$ this becomes the principal ideal ${\displaystyle ((1+\sqrt{5})/2)}$.

• The field ${\displaystyle \mathbb{Q}(\sqrt{-23})}$ has class number 3. Its Hilbert class field can be formed by adjoining a root of ${\displaystyle x^3-x-1}$, which has discriminant ${\displaystyle -23}$.

• To see why ramification at the archimedean primes must be taken into account, consider the real quadratic field ${\displaystyle K=\mathbb{Q}(\sqrt{3})}$ obtained by adjoining the square root of 3 to ${\displaystyle \mathbb{Q}}$. This field has class number 1 and discriminant 12. Its extension ${\displaystyle K(i)}$ has discriminant ${\displaystyle 144={12}^2}$, and so is unramified at all prime ideals in ${\displaystyle K}$, which means that ${\displaystyle K}$ admits finite abelian extensions of degree greater than 1 in which all finite primes of ${\displaystyle K}$ are unramified. This doesn't contradict the Hilbert class field of ${\displaystyle K}$ being ${\displaystyle K}$ itself: every proper finite abelian extension of ${\displaystyle K}$ must ramify at some place, and in the extension ${\displaystyle K(i)/K}$ there is ramification at the archimedean places: the real embeddings of ${\displaystyle K}$ extend to complex (rather than real) embeddings of ${\displaystyle K(i)}$.

• Likewise, for ${\displaystyle K=\mathbb{Q}(\sqrt{6})}$ (class number 1, discriminant 24), the extension ${\displaystyle K(\sqrt{-2},\sqrt{-3})}$ has discriminant ${\displaystyle 576={24}^2}$, so is unramified at all prime ideals in ${\displaystyle K}$, but ramified at both real places.

• By the theory of complex multiplication, the Hilbert class field of an imaginary quadratic field is generated by the value of the elliptic modular function at a generator for the ring of integers (as a ${\displaystyle \mathbb{Z}}$-module).