Decide whether or not each of the following fields is a splitting field over \(\mathbb{Q}\).
\(\mathbb{Q}(\sqrt[3]{2})\).
\(\mathbb{Q}(\sqrt{2} + \sqrt{5})\).
Solution
1. The field \(\mathbb Q(\sqrt[3]{2})\) is not a splittingfield over \(\mathbb Q\).
Put \(a=\sqrt[3]{2}>0\). The polynomial \(T^3-2\) is Eisenstein at two, hence irreducible over \(\mathbb Q\) [@DF04]. Its roots are \(a,\zeta a,\zeta^2 a\), where \(\zeta=e^{2\pi i/3}\) is a nonreal primitive cube root of unity. The field \(\mathbb Q(a)\) lies in \(\mathbb R\) and therefore does not contain the latter two roots. An irreducible polynomial over \(\mathbb Q\) has a root in this field without splitting there, so the extension is not normal. Every splitting field over a field is a normal extension [@DF04]. Consequently \(\mathbb Q(a)\) cannot be the splitting field of any polynomial over \(\mathbb Q\).
2. The field \(\mathbb Q(\sqrt2+\sqrt5)\) is the splittingfield of \((T^2-2)(T^2-5)\) over \(\mathbb Q\).
Let \(s=\sqrt2+\sqrt5>0\). Since \[ (\sqrt5-\sqrt2)s=5-2=3, \] one has \(\sqrt5-\sqrt2=3/s\in\mathbb Q(s)\). Adding and subtracting this identity from the definition of \(s\) gives \[ \sqrt5=\frac{s+3/s}{2},\qquad \sqrt2=\frac{s-3/s}{2}. \] Thus both radicals lie in \(\mathbb Q(s)\), and the reverse containment is immediate from \(s=\sqrt2+\sqrt5\). Therefore \(\mathbb Q(s)=\mathbb Q(\sqrt2,\sqrt5)\).
The roots of \((T^2-2)(T^2-5)\) are exactly \(\sqrt2,-\sqrt2,\sqrt5,-\sqrt5\). They all belong to this field, and the field they generate is exactly \(\mathbb Q(\sqrt2,\sqrt5)=\mathbb Q(s)\). This proves that it is the asserted splitting field.