Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 18H (Galois Theory)

A primitive root of unity ${\xi }_n$ satisfies ${\xi }_n^n=1$ and ${\xi }_n^j\ne 1$ for $1\le j<n$, equivalently it has multiplicative order $n$. A positive characteristic of a field must be prime: if its least positive value $p$ factored as $ab$ with $1<a,b<p$, the two nonzero field elements $a\cdot 1$ and $b\cdot 1$ would have product zero. If this prime $p$ divides $n$, put $m=n/p$. In characteristic $p$, the binomial coefficients $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ for $0<j<p$ vanish because their integer numerators contain $p$ and their denominators do not. Thus $(X-1{)}^p=X^p-1$. It follows that $({\xi }_n^m-1{)}^p={\xi }_n^n-1=0$. A field has no nonzero nilpotent element, so ${\xi }_n^m=1$, contradicting primitivity. Consequently $\boxed{\mathrm{char}K\nmid n}$.

A primitive tenth root $\xi$ has ${\xi }^5=-1$ and $\xi \ne -1$, so division of $X^5+1$ by $X+1$ givesIt is irreducible over $\mathbb{Q}$: replacing $X$ by $-X$ gives ${\Phi }_5(X)$, and then replacing $X$ by $X+1$ gives $X^4+5X^3+10X^2+10X+5$. Any nontrivial monic integer factorization would reduce modulo five to factors that are powers of $X$, forcing both constant terms divisible by five and their product divisible by $25$, contrary to the constant term five. Rational factorization reduces to integer factorization by Gauss lemma: products of primitive integer polynomials are primitive, since reduction modulo any prime preserves a nonzero product. ThusLet $t=\xi +{\xi }^{-1}$. Dividing ${\Phi }_{10}(\xi )=0$ by ${\xi }^2$ gives $t^2-t-1=0$. Hence $(2t-1{)}^2=5$, so $\boxed{\sqrt{5}\in \mathbb{Q}(\xi )}$. Either sign of this square root suffices; for $\xi =e^{\pi i/5}$ the positive one is $2t-1$.

A finite field ${\mathbb{F}}_q$ contains a primitive tenth root exactly when $10\mid (q-1)$. Its multiplicative group is cyclic: the exponent of a finite abelian subgroup of a field is attained by an element, by combining its prime-power components; since every element is a root of $X^e-1$, the polynomial root bound forces the group order to be at most that exponent, hence equal to it. A cyclic group contains an element of order ten exactly when ten divides its order.

Over ${\mathbb{F}}_q$, the conjugates of $\xi$ are its Frobenius automorphism orbit $\xi ,{\xi }^q,{\xi }^{q^2},\dots$. The extension degree is the least $d>0$ with $q^d\equiv 1(\mathrm{mod}10)$. For $q=3$ this order is four, for $q=11$ it is one, and for $q=19$ it is two. Therefore