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Dedekind factorization theorem states that if $K=\mathbb{Q}(\theta )$ with integral $\theta$ and $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, then a factorization $\overline{f}={\prod }_i{\overline{f}}_i^{e_i}$ of its minimal polynomial of an algebraic element into distinct monic irreducible polynomials over ${\mathbb{F}}_p$ givesThe factors are distinct prime ideals; the residue degrees and ramification indices satisfy ${\sum }_i{e}_i\mathrm{deg}{f}_i=[K:\mathbb{Q}]$. Unique prime ideal factorization holds in the ring of integers.

For a square-free integer $d\equiv 1(\mathrm{mod}4)$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, so the index condition is automatic. If $p_i\mid d$, the polynomial $X^2+d$ reduces to $X^2$ modulo $p_i$. HenceTherefore $d{\mathcal{O}}_K=({\prod }_i{P}_i{)}^2=(\sqrt{-d}{)}^2$. By unique prime ideal factorization, ${\prod }_i{P}_i=(\sqrt{-d})$.

Let $S$ be a nonempty proper subset and put $m={\prod }_{i\in S}{p}_i$. For $d>1$, $1<m<d$ and $m$ is a square-free integer. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, its ideal norm givesThe inequality $m<d$ forces $v=0$, leaving $m=u^2$, impossible for a square-free integer greater than one. Thus every such subproduct is nonprincipal.

These are ramified prime classes for discriminant minus four d. The ideal classes $[P_i]$ have order dividing two. Their product is the identity, and the preceding proof rules out every other nonempty subset relation. The homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by these classes consequently has kernel exactly $\{0,(1,\dots ,1)\}$. Its image is thereforeThe boundary case $d=1$ has $k=0$ and makes the printed exponent $k-1$ meaningless; its field is $\mathbb{Q}(i)$ with trivial ideal class group. The subgroup statement is understood for $d>1$, while the prime-factor and proper-subset statements are empty for $d=1$.