Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 20G (Number Fields)

Let $K$ be a number field. The class group iswith multiplication induced by ideal multiplication. Equivalently, two fractional ideals $I,J$ represent the same class when $I=(\alpha )J$ for some $\alpha \in K^{\times }$.

Let $I$ be a nonzero integral ideal. By the stated estimate, choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the fractional idealis integral. Its class is $[I{]}^{-1}$ andEvery ideal class therefore has, after taking inverses, an integral representative of norm at most $C$.

There are only finitely many integral ideals of bounded norm. Indeed, if $N(J)=m$, then the finite group ${\mathcal{O}}_K/J$ has order $m$, so multiplication by $m$ kills it andFor each of the finitely many integers $m\le C$, such ideals correspond to ideals of the finite ring ${\mathcal{O}}_K/m{\mathcal{O}}_K$, of which there are only finitely many. Hence $\mathrm{Cl}(K)$ is finite.

If $I=0$, take $n=1$. Otherwise, by part (b), the element $[I]$ of the finite group $\mathrm{Cl}(K)$ has finite order, say $n\ge 1$. Thereforewhich says precisely that $I^n$ is principal. Since $I^n$ is integral, if $I^n=(\alpha )$ then necessarily $\alpha \in {\mathcal{O}}_K$.

First let $I={\mathfrak{p}}^e$ for a prime ideal $\mathfrak{p}$ and $e\ge 1$. If $\alpha \beta \in {\mathfrak{p}}^e$ but $\alpha \notin {\mathfrak{p}}^e$, then in the prime-ideal factorization valuationsThus $v_{\mathfrak{p}}(\beta )>0$, so $\beta \in \mathfrak{p}$ and ${\beta }^e\in {\mathfrak{p}}^e=I$. Hence every prime-ideal power is primary.

Conversely, the radical of any primary ideal is prime. To see this, suppose $ab\in \sqrt{I}$ and $a\notin \sqrt{I}$. For some $m$, $a^m{b}^m\in I$, while $a^m\notin I$. Primaryness gives $(b^m{)}^k\in I$ for some $k$, and hence $b\in \sqrt{I}$.

For nonzero $I$, unique factorization of ideals givessoThis radical is prime only when $r=1$. Therefore a nonzero primary ideal is ${\mathfrak{p}}^e$. The zero ideal is itself prime because ${\mathcal{O}}_K$ is a domain, so it is its own first power. Consequently