Roots of a cubic equation When will the roots of a cubic polynomial $ax^3+bx^2+cx+d$ generate a cyclic group of order 3?
I literally mean: the cubic equation $ax^3+bx^2+cx+d$ will have three roots (possibly all three real, possibly one real and two complex), say $\zeta_1$, $\zeta_2$ and $\zeta_3$. Under what conditions is $\{\zeta_1, \zeta_2, \zeta_3\}$ isomorphic to $C_3$?
 A: If a complex number has multiplicative order 3, then it is a primitive cubic root of unit. Hence, two of the roots are primitive cubic roots of unit and the other one is 1. In other words, the polynomial is $a(x^3-1)$.
A: I put up an answer several weeks ago, but I think I have a better answer now.
The irreducible cubic has roots $a$, $b$, and $c$. In general, each of these roots generates an extension field which is a vector space over the rationals of dimension three. For the Galois group to be cyclic of order three, each of these vector spaces must be the same vector space. In other words, considering the extension field,
$A + Ba + Ca^2$ (with $A$, $B$, and $C$ rational)
we must find that $b$ and $c$ are also elements of this field. Typically then $b$ must be expressible as something like
$b = 3a^2 -2a + 5$
But if we can transform $a$ into $b$ by means of this operation, then surely we must be able to transform $b$ into $c$ by the same operation; and again, $c$ into $a$. If we apply this function recursively three times, we get an eight degree polynomial for $a$ in terms of $a$.
Interestingly, this polynomial is not irreducible. Why not? Because we haven't excluded the possibility that the operation returns the same argument it started with: namely,
$a = 3a^2 - 2a + 5$
So it will turn out that the eighth-degree polynomial is divisible by $(3a^2 - 3a + 5)$. (Note this expression is slightly different from the formula for the operator!). You end up with a sixth degree equation whose Galois group is $S_3$. This isn't exactly what we were trying for...we wanted a third degree equation with just three roots. We almost got it..in fact, we got three roots $a_1$, $b_1$, and $c_1$ who can be permuted cyclically with each other...but we also got their complex conjugates, $a_2$, $b_2$ and $c_2$, with obviously the same properties. Neither of these triplets are the roots of a cubic...at least, not a cubic with rational coefficients. We can fix this up by adding together complex conjugates:
$a = a_1 + a_2$
$b = b_1 + b_2$
$c = c_1 + c_2$
And with this definition, $a$ $b$ and $c$ are the roots of a cubic with Galois group $C_3$. I have to guess that all such Galois groups can be generated by this method. It also appears to me that the differences $a_1-a_2$ etc should also generate a cubic equation, but I haven't checked this out.
A: One obvious condition to get C3 as the Galois group is that there are no simple permutations of the roots. So a pair of complex roots is out. You must have three real roots. Then there must be some algebraic condition between the roots which does not allow you to permutate them willy-nilly. An example of three roots with built-in structure would be cos(pi/7), cos(2pi/7), and cos(3pi/7). They are the roots of the cubic equation which comes from adding together conjucate roots of x^7-1=0. If they don't have a Galois group of C3 then I'll be very surprised.
