I've been studying for my final exam, and came across this homework problem (that has already been solved, and graded.):(adsbygoogle = window.adsbygoogle || []).push({});

"Show that the Galois group of ##f(x)=x^3-1## over ℚ, is cyclic of order 2."

I had a question related to this problem, but not about this problem exactly. What follows is the line of thought that lead me to my question.

The polynomial has roots, 1, ##\lambda = \frac{-1+i\sqrt{3}}{2}##, and ##\bar\lambda = \frac{-1-i\sqrt{3}}{2}## over ℂ.

In this case, ##\lambda^2 = \bar\lambda##. So, the conjugate of a root is the same as the root squared (which isn't always true, weird?). Further, ##\lambda^3 = \lambda\cdot\bar\lambda=\bar\lambda\cdot\lambda = 1##. So, from that the set of roots, has inverses and the identity, is closed under multiplication, and is generated by ##\lambda##. Thus, the set of roots is cyclic of order 3.

I was trying to think of the conditions under which this happens, and I thought maybe if the Galois group is cyclic, but it seems like 1 wouldn't always be a root of a polynomial whose Galois group is cyclic, so then the roots wouldn't even form a group.

Is there something "interesting" going on here? Like, when do the roots form a group? Cyclic group?

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# When do roots of a polynomial form a group?

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