Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 74 / 2

For pairwise comaximal ideals $I_1,\dots ,I_r$ of $R$, reduction gives the Chinese remainder theorem for ideals:Its kernel is ${\bigcap }_i{I}_i$. For two comaximal ideals, write $1=u+v$ with $u\in I$, $v\in J$. If $x\in I\cap J$, then $x=xu+xv\in IJ$, showing $I\cap J=IJ$. Induction gives the product formula: a product of ideals individually comaximal with $J$ is still comaximal with $J$, by multiplying identities modulo $J$.

Put $J_i={\prod }_{j\ne i}{I}_j$. Since $I_i+J_i=R$, choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$. The element ${\sum }_i{a}_i{e}_i$ has any prescribed residues $a_i(\mathrm{mod}{I}_i)$, proving surjectivity. Its residue class modulo the product is unique because we already identified the kernel. In a Dedekind domain, powers of distinct nonzero prime ideals are pairwise comaximal, giving the prime-power form. The theorem requires comaximality; it is not a statement about arbitrary lists of ideals.

Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.