Suppose that \(K\) is a Galois extension of \(\mathbb{Q}\) of dimension \(2^n\) for some positive integer \(n\). Prove that every element of \(K\) is constructible.
Suppose \(f(x) \in \mathbb{Q}[x]\) is an irreducible quartic such that the splitting field of \(f(x)\) over \(\mathbb{Q}\) has dimension 24. Suppose \(\alpha\) is a root of \(f(x)\). Prove that \(\alpha\) is not constructible.
Solution
View the number fields inside \(\mathbb C\). A complex number is constructible when its real and imaginary coordinates are straightedge-and-compass constructible from \(0\) and \(1\).
1. A complex number \(z\) is constructible if and only if it liesin a finite tower\[ \mathbb Q=E_0\subset E_1\subset\cdots\subset E_r\subseteq\mathbb C, \qquad [E_i:E_{i-1}]=2. \]The tower of length zero is allowed.
Use the real quadratic-tower criterion and the fact that the real constructible numbers form a field closed under nonnegative square roots [@DF04]. For constructible real and imaginary coordinates, combine their two real quadratic towers and adjoin \(i\). Each adjunction still has degree at most \(2\) after the earlier fields are enlarged; omit any degree-\(1\) steps. This gives a quadratic tower containing \(z\).
Conversely, complex numbers with constructible coordinates form a field, by the coordinate formulas for addition, multiplication, and division. This field is also closed under taking square roots. Indeed, for \(z=a+ib\) with constructible real \(a,b\), put \[ s=\sqrt{a^2+b^2},\qquad u=\sqrt{\frac{s+a}{2}},\qquad v=\sqrt{\frac{s-a}{2}}. \] All radicands are nonnegative, and all three numbers are constructible. We have \(u^2-v^2=a\) and \(2uv=|b|\). Choose \(\varepsilon\in\{1,-1\}\) with \(\varepsilon|b|=b\); either sign is allowed when \(b=0\). Then \((u+i\varepsilon v)^2=z\). The quadratic formula now shows that every quadratic extension of a field of constructible complex numbers still consists of constructible numbers. Induction up the tower proves the converse.
2. A group of order \(2^n\) admits a chain from the whole groupto the identity subgroup with every successive index equal to \(2\).
We induct on \(n\), with the trivial group as the case \(n=0\). For a nontrivial \(2\)-group \(G\), the conjugacy-class equation shows that \(|Z(G)|\) is even: every noncentral conjugacy class has size \([G:C_G(g)]\), a power of \(2\) greater than \(1\). Since the center contains the identity, it therefore contains some \(z\ne1\). By Lagrange’s theorem \(z\) has order \(2^e\) for some \(e\geq1\). The element \(z^{2^{e-1}}\) is central of order \(2\); let \(C\) be its generated subgroup. Then \(C\lhd G\) and \(|G/C|=2^{n-1}\).
Apply induction to \(G/C\). Taking inverse images of its subgroup chain gives a chain from \(G\) to \(C\) with all successive indices \(2\), because the quotient map identifies the relevant coset spaces. Append \(C\supset\{1\}\) to obtain the required chain.
3. In part (a), every element of \(K\) is constructible.
The group \(G=\operatorname{Gal}(K/\mathbb Q)\) has order \(2^n\). Choose the chain \[ G=G_0\supset G_1\supset\cdots\supset G_n=\{1\} \] from step 2, and set \(F_i=K^{G_i}\). The Galois correspondence and its degree formula [@DF04] give \[ \mathbb Q=F_0\subset F_1\subset\cdots\subset F_n=K, \qquad [F_{i+1}:F_i]=[G_i:G_{i+1}]=2. \] Thus every element of \(K\) lies in a quadratic tower and is constructible by step 1.
4. Every finite quadratic tower over \(\mathbb Q\) is containedin a finite Galois extension of \(\mathbb Q\) of power-of-two degree.
Let \(\mathbb Q=E_0\subset\cdots\subset E_r\) be such a tower. In characteristic zero, completing the square in a quadratic minimal polynomial gives \[ E_i=E_{i-1}(\sqrt{a_i}),\qquad a_i\in E_{i-1}^{\times}. \] We construct finite Galois fields \(N_i\supseteq E_i\) of power-of-two degree over \(\mathbb Q\), together with polynomials \(g_i\in\mathbb Q[T]\) having splitting field \(N_i\). Start with \(N_0=\mathbb Q\) and \(g_0(T)=T\).
Suppose \(N_{i-1}\) and \(g_{i-1}\) have been constructed, and put \(H=\operatorname{Gal}(N_{i-1}/\mathbb Q)\). Define \[ q_i(T)=\prod_{\sigma\in H}(T^2-\sigma(a_i)),\qquad N_i=N_{i-1}\bigl(\sqrt{\sigma(a_i)}:\sigma\in H\bigr). \] The coefficients of \(q_i\) belong to \(N_{i-1}\) and are fixed by \(H\), because each group element permutes the factors. Since the fixed field of \(H\) is \(\mathbb Q\), we have \(q_i\in\mathbb Q[T]\). The field \(N_i\) is precisely the splitting field of \(g_i=g_{i-1}q_i\): the roots of \(g_{i-1}\) generate \(N_{i-1}\), and the additional roots are the two signs of each displayed square root. Therefore \(N_i/\mathbb Q\) is finite Galois.
There are finitely many square roots to adjoin. Each adjunction has degree \(1\) or \(2\), so \([N_i:N_{i-1}]\) is a power of \(2\). The tower law makes \([N_i:\mathbb Q]\) a power of \(2\) as well. The identity element of \(H\) ensures that \(\sqrt{a_i}\in N_i\), and hence \(E_i\subseteq N_i\). This completes the induction; \(N_r\) is the required extension.
5. The root in part (b) is not constructible.
Suppose \(\alpha\) were constructible. Steps 1 and 4 put it in a finite Galois extension \(N/\mathbb Q\) with \([N:\mathbb Q]=2^m\) for some \(m\geq0\). The irreducible polynomial \(f\) has a root in the normal extension \(N/\mathbb Q\), so it splits in \(N\). Consequently its splitting field \(L\) is contained in \(N\). The tower law would give \[ 24=[L:\mathbb Q]\mid[N:\mathbb Q]=2^m, \] which is impossible since \(3\mid24\) but \(3\nmid2^m\). Thus \(\alpha\) is not constructible.