Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
3  
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

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.