Let \[ \zeta=e^{2\pi i/7} \] and set \[ \alpha=\zeta+\zeta^{-1}. \] Find a cubic polynomial with integer coefficients having \(\alpha\) as a root.
Solution
1. Since \(\zeta\) is a primitive seventh root of unity,\[ 1+\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^5+\zeta^6=0. \]
One has \(\zeta^7=1\) and \(\zeta\neq1\). Hence \[ 0 = \frac{\zeta^7-1}{\zeta-1} = 1+\zeta+\cdots+\zeta^6. \]
2. Dividing the relation in step 1 by \(\zeta^3\) gives\[ (\zeta^3+\zeta^{-3}) + (\zeta^2+\zeta^{-2}) + (\zeta+\zeta^{-1}) +1 =0. \]
Since \(\zeta^7=1\), \[ \zeta^{-1}=\zeta^6, \qquad \zeta^{-2}=\zeta^5, \qquad \zeta^{-3}=\zeta^4. \] The displayed equation is exactly the divided cyclotomic relation.
3. In terms of \(\alpha=\zeta+\zeta^{-1}\),\[ \zeta^2+\zeta^{-2} = \alpha^2-2 \]and\[ \zeta^3+\zeta^{-3} = \alpha^3-3\alpha. \]
Squaring \(\alpha\) gives \[ \alpha^2 = \zeta^2+2+\zeta^{-2}. \] Similarly, \[ \alpha^3 = \zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3} = \zeta^3+\zeta^{-3}+3\alpha. \] Rearranging gives both identities.
4. The number \(\alpha\) is a root of\[ \boxed{p(x)=x^3+x^2-2x-1\in\ZZ[x]}. \]
Substitute the identities from step 3 into step 2: \[ (\alpha^3-3\alpha) + (\alpha^2-2) + \alpha +1 =0. \] Collecting terms yields \[ \alpha^3+\alpha^2-2\alpha-1=0. \]
5. Q.E.D.
Step 4 gives the requested cubic polynomial with integer coefficients.