Let $w$ be a primitive 10th root of unity. Find the irreducible polynomial of $w+w^{-1}$.
I know that the cyclotomic polynomial of $g_{10}(x)=x^4-x^3+x^2-x+1$ but I can't apply the same techniques used in this question (Cyclotomic polynomials and Galois groups ) where $w$ is a 7th root of unity instead. Any help?
2 Answers
$$(w+w^{-1})^2 = w^2 + 2 + w^{-2}$$ but $$ w^2 - w + 1 - w^{-1} + w^{-2} = w^{-2} (w^4 - w^3 + w^2 - w + 1) = 0 $$ so $$ (w+w^{-1})^2 = w + 1 + w^{-1}$$ i.e. $w+w^{-1}$ is a root of $x^2 - x - 1$.
First off, note that the $10$th roots of unity and $5$th roots of unity both define the $5$th cyclotomic field, $Q(ζ_5)$. If $w$ is a $10$th root of unity, then the minimal polynomial of $w$ is:
$x^4-x^3+x^2-x+1$
Note that this is true for $w^5$ as well.
Now $w^{-1}+w$ is $1/w$ + $w$, or using the equivalent definition of $w^5=w$, $w^{-1}+w$ = $w^4+w$. This element generates the minimal polynomial $x^2-x-1$, which defines the quadratic subfield of the 5$th$ cyclotomic field, namely $K = Q(√5)$.
