Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 20G (Number Fields)

The supplied integral basis differs unimodularly from $1,\zeta ,{\zeta }^2,{\zeta }^3$, since $1=-\zeta -{\zeta }^2-{\zeta }^3-{\zeta }^4$. With $f(t)={\Phi }_5(t)=(t^5-1)/(t-1)$, the field discriminant is $\mathrm{Norm}(f^{\prime }(\zeta ))$, the degree-four sign being positive. At a nonidentity fifth root,soThe Minkowski bound for ideal classes gives an integral ideal in every ideal class with norm at most $[3/(2{\pi }^2)]\sqrt{125}<2$. Its positive integer norm must be one, so it is the whole integer ring. Every class is trivial, proving all ideals are principal.

For $1\le n\le 4$, $u_n=(1-{\zeta }^n)/(1-\zeta )=1+\zeta +\cdots +{\zeta }^{n-1}$ is an algebraic integer. The numerator has the same norm five as the denominator, because its conjugates permute the nonidentity fifth roots. Thus $\mathrm{Norm}(u_n)=1$, proving it is a unit. Multiplying the four numerator factors gives five, henceThe ideal $P=(1-\zeta )$ has norm five; explicitly reduction modulo it sets $\zeta =1$ and gives the quotient ${\mathbb{F}}_5$. Thus it is prime and $(5)=P^4$: five is totally ramified, with ramification index four and residue degree one. A rational prime can ramify only if it divides the field field discriminant, so no rational prime other than five ramifies, and there are no other ramified prime ideals.