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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 ax3+bx2+cx+d will have three roots (possibly all three real, possibly one real and two complex), say ζ1,ζ2 and ζ3. Under what conditions is {ζ1,ζ2,ζ3} isomorphic to C3?

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if you dont want an order 2 element of the galois group, you dont want complex conjugation, so the roots must be real –  yoyo May 23 '11 at 21:49
    
@yoyo, except for the tag, it is not clear whether the question is about the Galois group. But +1 for the answer. –  lhf May 23 '11 at 21:57
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What group structure are you considering on $\{ \zeta_1, \zeta_2, \zeta_3 \}$? Any three element set is isomorphic to a cyclic group if I am allowed to choose the group structure. –  David Speyer May 24 '11 at 0:50

3 Answers 3

up vote 4 down vote accepted

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 $a(x^3-1)$.

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I took the question as asked. But did you mean to ask when the Galois group of a cubic is cyclic? –  lhf May 23 '11 at 21:24
    
No I literally meant: 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 $C_3$? –  Clinton Boys May 23 '11 at 22:22

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 a1, b1, and c1 who can be permuted cyclically with each other...but we also got their complex conjugates, a2, b2 and c2, 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 = a1 + a2

b = b1 + b2

c = c1 + c2

And with this definition, a b and c are the roots of a cubic with Galois group C3. I have to guess that all such Galois groups can be generated by this method. It also appears to me that the differences a1-a2 etc should also generate a cubic equation, but I haven't checked this out.

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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.

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I'm surprised this question hasn't been properly answered yet by one of the many people with more expertise than me in Galois Theory. However, since it went unanswered, I thought I made a reasonable stab at it. I wonder why people are downvoting it without explaining what is wrong with it? –  Marty Green May 24 '11 at 2:07

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