What’s the Galois group of \(x^2 + 1\) over \(\mathbb{Q}\)? What’s the integral closure of \(\mathbb{Z}\) in \(\mathbb{Q}(i)\)?
Solution
The polynomial \(x^2+1\) is irreducible over \(\mathbb Q\), and its splitting field is \[ \mathbb Q(i). \] Thus \[ [\mathbb Q(i):\mathbb Q]=2, \qquad \operatorname{Gal}(\mathbb Q(i)/\mathbb Q) =\{1,\text{complex conjugation}\} \cong C_2. \]
Now let \(\alpha=a+bi\in\mathbb Q(i)\) be integral over \(\mathbb Z\). Its trace and norm are integers: \[ 2a=\operatorname{Tr}(\alpha)\in\mathbb Z, \qquad a^2+b^2=N(\alpha)\in\mathbb Z. \] Put \(m=2a\in\mathbb Z\). Then \[ (2b)^2=4(a^2+b^2)-m^2\in\mathbb Z. \] Since \(2b\in\mathbb Q\) and its square is an integer, \(2b\in\mathbb Z\). Moreover \[ m^2+(2b)^2=4(a^2+b^2)\equiv0\pmod4. \] Squares modulo \(4\) are \(0\) or \(1\), so both \(m\) and \(2b\) are even. Hence \(a,b\in\mathbb Z\).
Conversely, \(i\) is integral because it satisfies \(x^2+1=0\), so every element of \(\mathbb Z[i]\) is integral. Therefore the integral closure of \(\mathbb Z\) in \(\mathbb Q(i)\) is \[ \boxed{\mathbb Z[i]}. \]