What is the Galois group of \(x^2 - 2\)? Why is \(x^2 - 2\) irreducible?
Solution
1. The polynomial \(x^2-2\) is irreducible over \(\QQ\).
A reducible quadratic over \(\QQ\) has a rational root. If \(a/b\) in lowest terms satisfied \((a/b)^2=2\), then \(a^2=2b^2\). Thus \(a\) is even, say \(a=2c\), giving \(b^2=2c^2\), so \(b\) is even, contradicting coprimality.
2. Its splitting field is \(\QQ(\sqrt2)\).
The roots are \(\pm\sqrt2\), both contained in \(\QQ(\sqrt2)\), and irreducibility gives degree \(2\).
3. Therefore\[ \operatorname{Gal}(x^2-2/\QQ)\cong C_2. \]
There are exactly two \(\QQ\)-automorphisms of the quadratic splitting field: the identity and conjugation \(\sqrt2\mapsto-\sqrt2\).