3 add example with A_5 Galois group

My answer 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.

Added: Just to give you an idea, if you look in Cohen's book A course in computational algebraic number theory, 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 $9\nmid e$.

And more: 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 $A_5$.

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.

2 add way to generate infinitely many examples

My answer 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.

Added: Just to give you an idea, if you look in Cohen's book A course in computational algebraic number theory, 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 $9\nmid e$.

1

My answer 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.