Which numbers are constructible? Give an example of a non-constructible number whose degree is nevertheless a power of 2.
Solution
A real algebraic number \(\alpha\) is constructible by straightedge and compass iff there is a tower of fields \[ \QQ=K_0\subset K_1\subset\cdots\subset K_r \] with \[ [K_i:K_{i-1}]=2 \] for every \(i\), such that \(\alpha\in K_r\subseteq\RR\).
Equivalently, \(\alpha\) lies in the smallest subfield of \(\RR\) obtained from \(\QQ\) by repeatedly adjoining square roots of positive elements.
1. Every constructible algebraic number has degree a power of \(2\) over \(\QQ\).
If \(\alpha\in K_r\), then \[ [\QQ(\alpha):\QQ]\mid[K_r:\QQ]=2^r \] by the tower law. Hence its degree is a power of \(2\).
2. The converse is false.
Let \(\alpha\) be a real root of \[ f(x)=x^4-x-1. \] The polynomial is irreducible over \(\QQ\): modulo \(2\) it becomes \[ x^4+x+1, \] which has no root in \(\FF_2\) and is not divisible by the only irreducible quadratic \(x^2+x+1\), so it is irreducible over \(\FF_2\), hence over \(\QQ\). Thus \[ [\QQ(\alpha):\QQ]=4. \]
The discriminant of \(f\) is \[ -283, \] which is not a square. Its cubic resolvent is \[ y^3+4y-1, \] which is irreducible over \(\QQ\) by the rational-root test. For an irreducible quartic, irreducibility of the cubic resolvent implies that the Galois group is \(A_4\) or \(S_4\); the nonsquare discriminant excludes \(A_4\). Hence the Galois closure has Galois group \(S_4\).
If \(\alpha\) were constructible, it would lie in an iterated quadratic extension; the normal closure of such an extension has degree a power of \(2\), so the Galois group of the normal closure of \(\QQ(\alpha)\) would be a \(2\)-group. But \(S_4\) has order \(24\), not a power of \(2\). Contradiction.
Thus degree a power of \(2\) is necessary, but not sufficient, for constructibility.