Nonisomorphic number fields with the same Galois group and discriminant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:42:59Z http://mathoverflow.net/feeds/question/107402 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107402/nonisomorphic-number-fields-with-the-same-galois-group-and-discriminant Nonisomorphic number fields with the same Galois group and discriminant Gene Ward Smith 2012-09-17T16:52:12Z 2012-09-18T04:18:02Z <p>It is possible for nonisomorphic fields to have the same discriminant, and even to be arithmetically equivalent, meaning with the same Dedekind zeta function. Previous questions which discussed this are "Number fields with same discriminant and regulator?" and "Are there two non-isomorphic number fields with the same degree, class number and discriminant?". The usual way this comes about involves fields with isomorphic subfields. I would find it useful if the discriminant sufficed to characterize number fields whose splitting field had a simple Galois group, so that there were no subfields. So the question is, can there be two nonisomorphic fields with the same degree, the same discriminant and the same simple Galois group for the splitting field?</p> http://mathoverflow.net/questions/107402/nonisomorphic-number-fields-with-the-same-galois-group-and-discriminant/107419#107419 Answer by Rob Harron for Nonisomorphic number fields with the same Galois group and discriminant Rob Harron 2012-09-17T19:33:50Z 2012-09-18T04:18:02Z <p>My <a href="http://mathoverflow.net/questions/3448/are-there-two-non-isomorphic-number-fields-with-the-same-degree-class-number-and/29467#29467" rel="nofollow">answer</a> to one of those previous questions provides an example of two Galois cubic fields of the same discriminant. The cyclic group of order three is simple.</p> <p><strong>Added:</strong> Just to give you an idea, if you look in Cohen's book <em>A course in computational algebraic number theory</em>, Theorem 6.4.6 says that if \$e\$ is the product of 9 times \$t-1\$ distinct primes that are congruent to 1 mod 3, then there are exactly \$2^{t-1}\$ cyclic cubic fields of discriminant \$e^2\$. There's a similar statement for \$9\nmid e\$.</p> <p><strong>And more:</strong> So using sage/pari, I just found the following example of two non-isomorphic quintic \$A_5\$-extensions with the same discriminant. Take \$f=x^5 - x^2 - 2x - 3\$ and \$g=x^5 - x^4 + 5x^3 - 3x^2 + 4x - 3\$. The associated fields (and the polynomials themselves) have discriminant 243049 (and 1 real place), which is a square. Hence, the Galois group is \$A_5, D_5\$, or \$C_5\$. Their factorization mod 3 are \$x(x + 1)(x^3 + 2x^2 + x + 1)\$ and \$x (x + 2) (x^3 + 2x + 2)\$, respectively. Hence, both of their Galois groups contain a 3-cycle and hence must be \$A_5\$.</p> <p>In fact, these both have class number 1. However, 5 is inert in the second field, but splits into primes of degree 1,1, and 3 in the first field. Their regulators are 10.7855048337065 and 7.68746944439494, respectively.</p>