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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.

  • 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.2012-08-19
  • 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.2012-08-20
  • 0
    @WillJagy, I would appreciate it very much if you send me the pdf. Regards,2012-08-20
  • 0
    Done. ${}{}{} $2012-08-20
  • 0
    @WillJagy Thank you so much.2012-08-20
  • 0
    @Matt I don't know the standard proof of the case $Q(\sqrt{-23})$. Regards,2012-08-20

2 Answers 2