Explain with field theory why straightedge-and-compass constructions cannot in general trisect an angle, double a cube, or square a circle.
Solution
A real number constructible from \(\QQ\) lies in a tower \[ \QQ=K_0\subset K_1\subset\cdots\subset K_r \] with each extension degree at most \(2\). Hence every constructible algebraic number has degree over \(\QQ\) equal to a power of \(2\).
1. Doubling the cube is impossible.
A cube of side \(1\) has volume \(1\). Doubling its volume requires side length \(\sqrt[3]{2}\). Its minimal polynomial is \[ x^3-2, \] which is irreducible over \(\QQ\) by Eisenstein at \(2\). Thus \[ [\QQ(\sqrt[3]{2}):\QQ]=3, \] not a power of \(2\). Hence \(\sqrt[3]{2}\) is not constructible.
2. A \(60^\circ\) angle cannot be trisected by straightedge and compass.
Trisecting \(60^\circ\) would construct \(20^\circ\), hence \(c=\cos20^\circ\). The triple-angle identity gives \[ 4c^3-3c=\cos60^\circ=\frac12, \] so \(c\) is a root of \[ 8x^3-6x-1. \] This cubic has no rational root, hence is irreducible over \(\QQ\). Therefore \([\QQ(c):\QQ]=3\), so \(c\) is not constructible.
3. Squaring the circle is impossible.
A circle of radius \(1\) has area \(\pi\). A square of the same area would need side length \(\sqrt\pi\). Lindemann’s theorem implies that \(\pi\) is transcendental, hence so is \(\sqrt\pi\). Every constructible number is algebraic over \(\QQ\), so \(\sqrt\pi\) is not constructible.
Thus each classical construction fails because the required length cannot lie in a tower of quadratic extensions of \(\QQ\).