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I am looking an explicit form for the ring class field of the order $\mathbb{Z}[\sqrt{-19}]$ in the quadratic field $\mathbb{Q}(\sqrt{-19})$. Does anyone know if there is some and how it is?

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    $\begingroup$ The ring class field of conductor...? $\endgroup$ – Keenan Kidwell Aug 26 '13 at 23:20
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Let $u = \frac{1+ \sqrt{-19}}2$, so that the ring of integers is $O = \mathbb Z[u]$.
The units of $O$ are $U=\{+1;-1\}$.

Your ring class field is related to the class field of conductor $(2)$. Modulo $(2)$, $U$ is trivial, and $(O/(2))^* = \{1;u;1+u\}$ is cyclic of order $3$.

To each class $c$ mod $(2)$, we get a corresponding lattice class $\Lambda_c = \langle (2), c \rangle$ whose ring of endomorphism is $\Bbb Z[\sqrt{-19}]$ : we get $\Lambda_1 = \langle 1,2u\rangle = \Bbb Z[\sqrt{-19}], \Lambda_u = \langle 2,u \rangle, \Lambda_{1+u} = \langle 2,1+u \rangle$.

In general, those are not distinct since $\Lambda_c = \Lambda_{xc}$ for $x$ prime with $n$, so this correspondence factors through $(\Bbb Z/n \Bbb Z)^*$. If $L_n$ is the class field of conductor $(n)$ and $K_n$ is the ring class field of $\Bbb Z[nu]$, then $Gal(L_n,K_n) \cong (\Bbb Z/n \Bbb Z)^*$ and actually $L_n = K_n (\zeta_n)$. When $n=2$, this group is trivial, so $L_2 = K_2$ and the correspondence is one-to-one.

Let $L$ be the class field of $K = \Bbb Q(\sqrt{-19})$ of conductor $(2)$. We know that $K \subset L$ is cyclic of degree $3$. Since the conjugation of $K$ over $\Bbb Q$ permutes $u$ with $1-u$, the Galois group of $L$ over $\Bbb Q$ is not abelian, so it is isomorphic to $S_3$.

Getting an explicit polynomial $P$ over $\Bbb Q$ of degree $3$ such that $L$ is its splitting field is hard but doable for degree $3$ extensions.

You can use the modular invariant $j$, and pick $P(X) = \left(X-j(\sqrt{-19})\right)\left(X-j(\frac{1+\sqrt{-19}}4)\right)\left(X-j(\frac{-1+\sqrt{-19}}4)\right)$, giving the rather unenlightening $P(X) = X^3 - 784074438864 X^2 + 1128678666363648 X - 827237892283232260$ (according to wolframalpha).

Let $\sigma$ be a generator of $Gal(L/K)$ and pick a generator $x$ such that $L=K(x)$.
Since $L/K$ is cyclic of order $3$ and $\zeta_3 \notin K$, $L(\zeta_3)/K$ is cyclic of order $6$.
Define $\alpha = (x + \zeta_3 x^\sigma + \zeta_3^2 x^{\sigma^2})$ and $\beta = (x + \zeta_3^2 x^\sigma + \zeta_3 x^{\sigma^2})$.
Some calculations show that $\alpha \beta \in K$, $\alpha^3,\beta^3 \in K(\zeta_3)$, that $Gal(L(\zeta_3)/K)$ is generated by a map $\tau : \alpha \mapsto \zeta_3 \beta \mapsto \zeta_3 \alpha \mapsto \beta \mapsto \zeta_3^2 \alpha \mapsto \zeta_3^2 \beta \mapsto \alpha$. Hence $L$ is determined by the choice of some $\alpha^3 \in K(\zeta_3)$ (up to cubes an inversion) whose norm $\alpha^3\beta^3$ is a cube $(\alpha\beta)^3$ in $K$.

Next, $L$ (and $L(\zeta_3)$ too) are supposed to be Galois over $\Bbb Q$, which means that $\bar{\alpha}$ must be in there too. After looking at when two cubic root gives the same extension, this is equivalent to one of $\alpha/\bar{\alpha}$ or $\alpha \bar \alpha$ being in $K(\zeta_3)$. In the first case, $L = K(\zeta_3,\alpha^2) = K(\zeta_3,\alpha \bar \alpha)$ so we can actually choose $\alpha^3 \in \Bbb Q(\sqrt {57})$. In the second case, $L = K(\zeta_3,\alpha) = K(\zeta_3,\alpha (\alpha\beta) (\alpha\bar{\alpha})) = K(\zeta_3,\beta\bar \alpha)$ so we can actually choose $\alpha^3 \in \Bbb Q(\sqrt{-3})$ with $\alpha\bar{\alpha} \in \Bbb Q$, so the whole thing has nothing to do with $\Bbb Q(\sqrt{-19})$ and $\Bbb Q(\zeta_3,\alpha)/\Bbb Q$ is abelian cyclic of degree $6$.

So now we are looking for suitable $\alpha^3 \in \Bbb Q(\sqrt{57})$ whose norm is a cube in $\Bbb Q$. Let $v = \frac {-1+\sqrt{57}}2$. We know that $L/K$ only ramifies at $(2)$, and so does $L(\zeta_3)/K(\zeta_3)$. In $\Bbb Q(\sqrt {57})$, $(2)$ splits into $(2,v)(2,v+1)$. Those are principal, as $-2 = (4+v)(3-v)$. Hence we must have $\alpha^3 = 8+2v$ up to cubes and units. The unit group is generated by $(171+40v)$ and $-1$, which leaves $3$ choices for $L(\zeta_3)$. Finally, if $\alpha^3 \neq \pm 1 \pmod 3$, then $K(\zeta_3) \subset L(\zeta_3)$ will further ramify at $(3)$. So we are left with exactly one possibility, $\alpha^3 = (8+2v)(171+40v) = 2197 + 291\sqrt{57}$.

Hence $L = K(x = \alpha-2/\alpha)$ with $\alpha^3 = 2197 + 291\sqrt{57}$, and the minimal polynomial over $K$ of this generator is $P(x) = x^3 + 6x - 4394$. After a few computations we get an explicit automorphism generating $Gal(L/K)$ with $x \mapsto - \frac x2 + \frac {(x^2+8)(x^2+2)}{584\sqrt{-19}}$, and we can experimentally check that it behaves appropriately for primes not dividing the discriminant of $P$:

  • $P$ splits into $3$ factors modulo $p$ iff ($-19$ is a square modulo $p$ and $(p)$ splits into $(p_1)(p_2)$ with $p_i = 1 \pmod 2$) iff $p = a^2 + 19b^2$
  • $P$ splits into $2$ factors modulo $p$ iff $-19$ is not a square modulo $p$
  • $P$ doesn't factor modulo $p$ iff ($-19$ is a square modulo $p$ and $(p)$ splits into $(p_1)(p_2)$ with $p_i \neq 1 \pmod 2$) iff $p = 4a^2 + 2ab + 5b^2$
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