= Solution
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 $\operatorname{Norm}(f'(\zeta))$, the degree-four sign being positive. At a nonidentity fifth root,
$$
f'(\zeta)=\frac{5\zeta^4}{\zeta-1},\qquad
\operatorname{Norm}(1-\zeta)=\Phi_5(1)=5,
$$
so
$$
\boxed{\operatorname{disc}K=5^4/5=125}.
$$
The <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 \b[all ideals are principal].
For $1\le n\le4$, $u_n=(1-\zeta^n)/(1-\zeta)=1+\zeta+\cdots+\zeta^{n-1}$ is an <algebraic integer>. The numerator has the same norm five as the denominator, because its conjugates permute the nonidentity fifth roots. Thus $\operatorname{Norm}(u_n)=1$, proving it is a unit. Multiplying the four numerator factors gives five, hence
$$
\boxed{5/(1-\zeta)^4=\prod_{n=1}^4u_n\ \text{is a unit}}.
$$
The 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.
Back to article page
