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.

The current question is motivated by this question. It is known that the number of imaginary quadratic fields of class number 3 is finite. Assuming the answer to this question is affirmative, I came up with the following question.

Let $f(X) = X^3 + aX + b$ be an irreducible polynomial in $\mathbb{Z}[X]$. Let $p = -(4a^3 + 27b^2)$ be the discriminant of $f(X)$. We consider the following conditions.

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

(2) The class number of $\mathbb{Q}(\sqrt{p})$ is 3.

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

Question Are there infinitely many primes $p$ satisfying (1), (2), (3)?

If this is too difficult, is there any such $p$?

I hope that someone would search for such primes using a computer.

share|improve this question
1  
I believe it is unknown whether there are infnitely many primes satisfying your condition (2). –  Gerry Myerson Aug 22 '12 at 1:42
    
@GerryMyerson I guess there are many(if not infinitely many) primes satisfying the condition (2). I'd like to know, at least, if there exists any prime number satisfying (1), (2), (3). –  Makoto Kato Aug 22 '12 at 7:24
    
So, how many primes have you looked at? –  Gerry Myerson Aug 22 '12 at 12:58
    
@GerryMyerson I looked at the following link. Hilbert class fields of all real quadratic fields of discriminants up to 30000 are given. math.univ-lyon1.fr/~roblot/tables.html I found (p, a, b) = (229, -4, -1), (1373, -8, -5), (2713, -13, -15). Each of these satisfies (1) and (2), but I don't know if they satisy (3). –  Makoto Kato Aug 22 '12 at 17:35
    
How many elliptic curves $y^2 = x^3 + ax + b$ of discriminant $2^4*p$ are there with multiplicative reduction at p? The quadratic twist of this elliptic curve by its discriminant might also be relevant. –  Thom Tyrrell Oct 5 '12 at 14:25

2 Answers 2

up vote 3 down vote accepted
+100

For (229, -4,-1) the polynomial factors as $(x-200)^2(x-58)$

For (1373, -8,-5) the polynomial factors as $(x-860)(x-943)^2$

For (2713, -13,-15) the polynomial factors as $(x-520)^2(x-1673)$

share|improve this answer
    
I take it $(229,-4,-1)$ means $p=229$, $a=-4$, $b=-1$, and I take it you are claiming the class number of ${\bf Q}(\sqrt{229})$ is 3. –  Gerry Myerson Oct 5 '12 at 0:34
    
@GerryMyerson See Kato's remarks above. –  i. m. soloveichik Oct 5 '12 at 13:10
    
Ah.${}{}{}{}{}$ –  Gerry Myerson Oct 5 '12 at 13:24
    
Thanks. May I ask how you found the factorizations? –  Makoto Kato Oct 5 '12 at 22:23
    
I wrote a Maple program. –  i. m. soloveichik Oct 5 '12 at 22:34

Here's some code I wrote using Sage:

def quadClassNumber(p):
    K.<alpha> = NumberField([x^2-p]);
    return K.class_number();  

U=1000;
for p in Primes():
    if p < U and quadClassNumber(p)==3:
        J = EllipticCurve([0,-2^4*3^3*p]);
        L = J.integral_points();
        if len(L) > 0:
            for (a,b,1) in L:
                A = a/(-2^2*3);
                B = b/(2^2*3^3);
                if A.is_integral() and B.is_integral():
                    E = EllipticCurve([A,B]);
                    if E.has_multiplicative_reduction(p):
                        print "p=",p;
                        print E;
                        print E.discriminant().factor();
                        R.<t> = PolynomialRing(IntegerModRing(p),1);
                        print R(E(0,1,0).division_points(2,true)/4).factor();
                        print "    ";
                        break;
    elif p > U:
        break;

The integral points of the elliptic curve J correspond (roughly) to pairs of integers $(A,B)$ satisfying $p = -(4A^3 + 27B^2)$. J is an integral model of the elliptic curve associated with this equation, but the use of J and Sage's integral_points() function comes at the cost of some powers of 2 and 3 that appear in the definitions of $A$ and $B$.

I searched for primes up to 10,000. For instance, with p=8581, a = -16 and b=17 work, and the associated polynomial factors mod 8581 as $(x + 6166)^2(x + 4830)$.

share|improve this answer

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.