Why is $\mathbb{Q}(\operatorname{exp}(\frac{2\pi i}{5}))$ a field extension of degree four not five? Let $\zeta = \operatorname{exp}(\frac{2\pi i}{5})$. Now, consider $\mathbb{Q}(\zeta)$. The minimal polynomial is $P=X^4+X^3+X^2+X+1$, therefore the extension is of degree $4$. But why? As $\zeta$ is a root of $P$, $\{1,\zeta,\zeta^2,\zeta^3\}$ must be a basis of $\mathbb{Q}(\zeta)$ as a $\mathbb{Q}$-Vector space. But where are the linear combinations with $\zeta^4$?
 A: The degree of this field is the degree of the minimum polynomial of $\zeta$.
We certainly know that $x^5-1$ is a polynomial that $\zeta$ satisfies, but it is not irreducible, therefore not minimal.
This is so because $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$ (that's where badjone's hint came from - I felt like I needed to explain that).
There's a reasult that says that $x^{p-1}+x^{p-2}+... + x+1$ is irreducible over $\mathbb{Q}$, hence the minimum polynomial is $ x^4+x^3+x^2+x+1$.
A: I think the general question you are asking is, for an algebraic element$~\zeta$ for which a minimal polynomial$~P$ (over a base field$~F$ that is $\Bbb Q$ here) is known, how to write a power $\zeta^k$ with $k\geq d=\deg P$ as $F$-linear combination of $1,\zeta,\ldots,\zeta^{d-1}$. This can be generalised to any polynomial $A[\zeta]$ of $\zeta$ instead of $\zeta^k=X^k[\zeta]$. The answer is simply to perform Euclidian division of $A$ by$~P$, retaining only the remainder$~R$ of the division; then $Q[\zeta]=R[\zeta]$ gives your $F$-linear combination.
In the example of the question $P=X^4+X^3+X^2+X+1$ and $Q=X^4$; the division gives $A=QP+R$ with quotient $Q=1$ and remainder $R=-X^3-X^2-X-1$, so $\zeta^4=-1-\zeta-\zeta^2-\zeta^3$ is the expression you are after. Another example for primitive $12$-th roots of unity and $k=11$ you get for $P$ the cycloctomic polynomial $P=X^4-X^2+1$, and the division gives $X^{11}=QP+R$ with $Q=X^7+X^5-X$ and $R=-X^3+X$, so $\zeta^{11}=\zeta-\zeta^3$ in this case.
