Exercise 15.5.4

Proof. If $L=\mathbb{Q}\left(i,\phi \left(\frac{\varpi }{p}\right)\right)$, then, by Theorem 15.5.1, $\mathbb{Q}(i)\subset L$ is a Galois extension, and $\mathrm{Gal}(L∕\mathbb{Q}(i))$ is isomorphic to a subgroup of ${\left(\mathbb{Z}[i]∕\mathit{p\mathbb{Z}}[i]\right)}^{\text{∗}}$, thus $\left|\mathrm{Gal}(L∕\mathbb{Q}(i))\right|=2^m$ for some integer $m\ge 0$. As in the proof of Theorem 10.1.12, Since $\mathrm{Gal}(L∕\mathbb{Q}(i))$ is a $p$-group for $p=2$, $\mathrm{Gal}(L∕\mathbb{Q}(i))$ is solvable. This means that we have subgroups

such that $G_i$ is normal in $G_{i-1}$ of index $2$. Then the Galois correspondence gives

where $[L_{G_i}:L_{G_{i-1}}]=2$ for all $i$. We can add at the beginning of the chain $\mathbb{Q}\subset \mathbb{Q}(i)$, where $[\mathbb{Q}(i):\mathbb{Q}]=2$.

By Theorem 10.1.6, where $\alpha =\phi \left(\frac{\varpi }{p}\right)\in L$, this proves that $\alpha =\phi \left(\frac{\varpi }{p}\right)$ is constructible. We recall the proof:

$i$ is constructible, and $𝒞$ is a subfield of $ℂ$, therefore $\mathbb{Q}\subset \mathbb{Q}(i)\subset 𝒞$.

Write $L_i=L_{G_i}$. By Exercise 7.1.12, $L_i=L_{i-1}(\sqrt{{\alpha }_i})$ for some ${\alpha }_i\in L_{i-1}$. If $L_{i-1}\subset 𝒞$, then ${\alpha }_i\in L_{i-1}$ is constructible, which implies $\sqrt{{\alpha }_i}\in 𝒞$ by Theorem 10.1.4. Thus $L_i=L_{i-1}(\sqrt{{\alpha }_i})\subset 𝒞$. This shows by induction that $L=L_n\subset 𝒞$, thus $\alpha =\phi \left(\frac{\varpi }{p}\right)\in L$ is constructible. □