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I have attached a snapshot of an example I'm working through in Galois Theory.

enter image description here

I was wondering if anyone could explain to me how the roots $\eta_{2}$ and $\eta_{3}$ are determined? The text mentions that the starting polynomial was $f(x) = x^3 - 3x + 1$ was somehow inspired by the root $\eta_{1} = \zeta + \zeta^8$ where $\zeta = e^{2\pi i/9}$.

I was able to verify that $\eta_{1} = \zeta + \zeta^8$ is indeed a root of $f(x)$, but I don't follow the remaining analysis of the relationship between $\eta_{1}$ and the other two roots.

I get the bigger picture idea of what this example is trying to illuminate: that is, an example of where the splitting field for a cubic polynomial is a degree $3$ extension, as opposed to (the previous example was this case) a degree $6$ extension. My questions summarized:

1) How are $\eta_{2}$ and $\eta_{3}$ obtained?

2) How can I show that $\eta_{1}$ is real (or is there a typo with the $2\cdot\text{cos}(2\pi/9)$ form?

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1 Answer 1

up vote 1 down vote accepted

Imagine the regular 9-gon on the complex plane around $0$, with nodes $\zeta^k$ ($k=0..8$), and let $\phi:=2\pi/9$. Then $$\zeta=\cos\phi+i\sin\phi$$ And $\zeta^9=1=\zeta^0$, so $\zeta^8=\zeta^{-1}=\bar\zeta$, hence $\eta_1=2\cos\phi$, which is the double of the real part of $\zeta$, hence real. Anyway, any polynomial of odd degree has at least one real root.

For $\eta_2$ and $\eta_3$ it may really be a guess, omitting the three nodes $1,\zeta^3,\zeta^6$.

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Your derivation of $\eta_{1}$ makes great sense! Thanks. –  Kyle Schlitt Nov 22 '12 at 2:14
2  
Just a note: A $9$-gon is called a 'nonagon' or 'enneagon'. –  000 Nov 22 '12 at 2:28

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