# Hilbert class field of a quadratic field whose class number is 3

Is the following proposition true? If yes, how would you prove this?

Proposition Let $f(X) = X^3 + aX + b$ be an irreducible polynomial in $\mathbb{Z}[X]$. Let $d = -(4a^3 + 27b^2)$ be the discriminant of $f(X)$. Let $K = \mathbb{Q}(\sqrt{d})$. Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$. Suppose the following conditions hold.

(1) $|d| = |4a^3 + 27b^2|$ is a prime number.

(2) The class number of $K$ is 3.

(3) $f(X) \equiv (X - s)^2(X - t)$ (mod $d$), where $s$ and $t$ are distinct rational integers mod $d$.

Then $L$ is the Hilbert class field of $K$.

Examples Each of the following polynomials of negative discriminants satisfies the above conditions.

(1) $f(X) = X^3 - X + 1 \equiv (X - 13)^2(X - 20)$ (mod 23)

(2) $f(X) = X^3 + X + 1 \equiv (X-3)(X-14)^2$ (mod 31)

(3) $f(X) = X^3 + 2X + 1 \equiv (X - 14)^2(X - 31)$ (mod 59)

I could not find a polynomial of positive discriminant satisying the above conditions.

• Do you know the standard argument that the splitting field of example 1 is the Hilbert class field of $\mathbb{Q}(\sqrt{-23})$? You should try to mimic it in this generality and see if something goes wrong. Maybe negative discriminant comes up so that the real place of the conductor ramifies. I don't remember, but that seems to me the only thing that could go wrong. My guess is it is true. – Matt Aug 19 '12 at 23:27
• What you want is Acta Arithmetica, volume 57 (1991), pages 131-153, Kenneth S. Williams and Richard R. Hudson, Representation of primes by the principal form of discriminant $-D$ when the classnumber $h(-D)$ is $3.$ I have a pdf. – Will Jagy Aug 20 '12 at 0:17
• @WillJagy, I would appreciate it very much if you send me the pdf. Regards, – Makoto Kato Aug 20 '12 at 0:42
• Done. ${}{}{}$ – Will Jagy Aug 20 '12 at 0:47
• @WillJagy Thank you so much. – Makoto Kato Aug 20 '12 at 1:35

The above proposition is a special case of the following proposition.

Proposition Let $f(X)$ be a monic irreducible polynomial of degree 3 in $\mathbb{Z}[X]$. Let $d$ be the discriminant of $f(X)$. Let $K = \mathbb{Q}(\sqrt{d})$. Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$. Suppose the following conditions hold.

(1) $|d|$ is a prime number.

(2) The class number of $K$ is 3.

(3) $f(X) \equiv (X - s)^2(X - t)$ (mod $d$), where $s$ and $t$ are rational integers and $s$ and $t$ are distinct mod $d$.

Then $L$ is the Hilbert class field of $K$.

Proof: Since the degree of $f(X)$ is 3, $[L : \mathbb{Q}] \leq 6$. Since $f(X)$ is irreducible, $3|[L : \mathbb{Q}]$. By (1), $[K : \mathbb{Q}] = 2$. Hence $2|[L : \mathbb{Q}]$. Hence $[L : \mathbb{Q}] = 6$. Hence $[L : K] = 3$. Therefore, by (2), it suffices to prove that every prime ideal of $K$ is unramified in $L$.

Let $Q$ be a prime ideal of $K$ lying over a prime number $q \neq p$, where $p = |d|$. By the application of the proposition of this question, $q$ is unramified in $L$. Hence $Q$ is unramified in $L$.

Let $P$ be a prime ideal of $K$ lying over $p$. It remains to prove that $P$ is unramified in $L$.

Let $\theta$ be a root of $f(X)$ in $L$. Let $M = \mathbb{Q}(\theta)$. Since $f(X)$ is irreducible, $[M : \mathbb{Q}] = 3$. Hence $[L : M] = 2$.

We denote by $\mathcal{O}_K, \mathcal{O}_M, \mathcal{O}_L$ the rings of integers in $K$, $M$, $L$ respectively.

Let $D_M$ be the discriminant of $M$. It is well known that $d = k^2 D_M$ for some rational integer $k$. Since $k^2 = 1$ by (1), $d = D_M$. Hence $\mathcal{O}_M = \mathbb{Z}[\theta]$. It is is well known(e.g. Milne's online course note) that $p\mathcal{O}_M = \mathfrak{p}^2\mathfrak{q}$ by (3), where $\mathfrak{p}$ and $\mathfrak{q}$ are distinct prime ideals of $\mathcal{O}_M$.

Since $[L : M] = 2$, We have the following patterns of the prime decompositions in $L$.

(1) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}$.

(2) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P_1}\mathfrak{P_2}$, where $\mathfrak{P_1} \neq \mathfrak{P_2}$.

(3) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}^2$.

(1)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q}$.

(2)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q_1}\mathfrak{Q_2}$, where $\mathfrak{Q_1} \neq \mathfrak{Q_2}$.

(3)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q}^2$.

Since $L/\mathbb{Q}$ is Galois, each ramification index of prime ideals of $L$ lying over $p$ is the same. Hence only the combination of (2) and (3)' is possible. Hence $p\mathfrak{O}_L = \mathfrak{P_1}^2\mathfrak{P_2}^2\mathfrak{Q}^2$.

Since $p$ ramifies in $K$, $p\mathfrak{O}_K = P^2$. Hence by the above result, $P$ is unramified in $L$. QED

• As a reminder, you should specify that $p=|d|$, in order not to confuse the reader. Regards. – awllower May 1 '13 at 13:56
• @awllower I edited the answer. Thanks. – Makoto Kato May 2 '13 at 6:12
• Good to see that. :D – awllower May 2 '13 at 6:57

You can be even more lenient on the assumptions.

Proposition.
If $K$ is a cubic number field for which $\Delta_K=\Delta(\mathcal{O}_K)$ is squarefree then $3|h(\mathbb{Q}(\sqrt{\Delta_K)})$.

Remark: note that when $K=\mathbb{Q}(\alpha)$ with $\alpha$ a root of an irreducible monic cubic polynomial $f\in\mathbb{Z}[X]$ with $\Delta(f)$ squarefree, then the above hypotheses are satisfied: we have

$$\Delta(f)=\Delta(\mathbb{Z}[\alpha])=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\cdot\Delta_K,$$ from which we see that $\mathcal{O}_K=\mathbb{Z}[\alpha]$ and $\Delta_K=\Delta(f)$.

Proof. As $\Delta_K$ is not a square in $\mathbb{Q}$ it follows that the normal closure $N$ of $K/\mathbb{Q}$ has degree $6$ over $\mathbb{Q}$ and that $[\mathbb{Q}(\sqrt{\Delta_K}):\mathbb{Q}]=2$. Then $N/\mathbb{Q}(\sqrt{\Delta_K})$ is an Abelian extension of degree $3$. I claim that it is also unramified, so that it is contained in the Hilbert class field of $\mathbb{Q}(\sqrt{\Delta_K})$, proving the proposition.

To see this, let $(r,s)$ be the signature of $K$. Since $r+2s=3$ we have either $(r,s)=(1,1)$ or $(r,s)=(3,0)$. When $(r,s)=(1,1)$ we have that $\mathbb{Q}(\sqrt{\Delta_K})$ is complex as $\mathrm{sign}(\Delta_K)=(-1)^s$ so it is unramified at the infinite primes. When $(r,s)=(3,0)$ we see that $K$ and $\mathbb{Q}(\sqrt{\Delta_K})$ are both totally real, hence their compositum $N$ is also totally real. Thus $\mathbb{Q}(\sqrt{\Delta_K})$ is unramified at the infinite primes in that case as well.

Now suppose that a finite prime $\mathfrak{p}$ of $\mathbb{Q}(\sqrt{\Delta_K})$ ramifies in $N$. As $N/\mathbb{Q}(\sqrt{\Delta_K})$ is Galois of prime degree, we must have that $\mathfrak{p}$ is in fact totally ramified in $N$. The ram-rel identity $efg=6$ for the extension $N/\mathbb{Q}$ yields that $3|e$ and by multiplicativity of the ramification index it follows that the rational prime $p$ lying under $\mathfrak{p}$ ramifies totally in $K$, say $p\mathcal{O}_K=\mathfrak{q}^3$. But this implies that $\mathfrak{q}^2|\mathfrak{D}_K$, and hence upon taking norms that $p^2|\Delta_K$, which is a contradiction and hence $N/\mathbb{Q}(\sqrt{\Delta_K})$ is unramified at the finite primes.