If A² is principal over an algebraic number field, then A² is principal over any extension

Let K_1 and K_2 be algebraic number fields with K_1 \subseteq K_2, and having integer rings \mathcal{O}_1 and \mathcal{O}_2, respectively. Suppose (A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1 is an ideal such that (A)^2 is principal. Show that (A)_{\mathcal{O}_2}^2 is principal in \mathcal{O}_2.

We will let G_1 and G_2 denote the ideal class groups of K_1 and K_2, respectively. For brevity, if S \subseteq \mathcal{O}_1 is a subset, then we will denote by (S)_1 the ideal generated by S in \mathcal{O}_1 and by (S)_2 the ideal generated by S in \mathcal{O}_2.

We claim that if (A)_1 \sim (B)_1, then (A)_2 \sim (B)_2. To see this, suppose we have \alpha and \beta such that (\alpha)_1(A)_1 = (\beta)_1(B)_1. Then (\alpha A)_1 = (\beta B)_1. Now if x = \sum r_i \alpha a_i \in (\alpha A)_2, then each \alpha a_i is in (\beta B)_1, and so has the form \sum s_{i,j} \beta b_j. So x \in (\beta B)_2. Conversely, (\beta B)_2 \subseteq (\alpha A)_2, and so (A)_2 \sim (B)_2.

The mapping A \mapsto (A)_2 then induces a well-defined mapping \Psi : G_1 \rightarrow G_2 given by [A] \mapsto [(A)_2]. Moreover, since \Psi([A][B]) = \Psi([AB]) = [(AB)_2] = [(A)_2][(B)_2] = \Psi(A) \Psi(B), \Psi is a group homomorphism.

In particular, if A^2 \sim (1) in \mathcal{O}_1, then \Psi(A) \sim (1) in \mathcal{O}_2, as desired.

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