Past exam of the number theory course of the University of Oxford / 2010 / B / 5 / Solution

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.

For the computation put $s=\sqrt{-6}$. The Kummer-Dedekind theorem says that when $p$ does not divide the power-basis index, a factorization $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ gives $(p)={\prod }_i(p,g_i(s){)}^{e_i}$, with each factor having norm $p^{\mathrm{deg}{g}_i}$. The given full power basis has index one. Since $X^2+6\equiv X^2(\mathrm{mod}2)$ and $X^2+6\equiv (X-1)(X+1)(\mathrm{mod}7)$,Here $N(P)=2$ and $N(Q)=N(Q^{\prime })=7$. A prime ideal dividing an ideal of norm $98=2\cdot 7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and norm multiplicativity therefore leave exactlyThey are distinct by their prime-ideal valuations, and each has norm $98$.