## 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.