Integral closure of $\ZZ$ in $\QQ(i)$

The integral closure is \[ \ZZ[i]. \]

Every Gaussian integer \(a+bi\) with \(a,b\in\ZZ\) is integral over \(\ZZ\), since it satisfies \[ x^2-2ax+(a^2+b^2)=0. \] Thus \(\ZZ[i]\) is contained in the integral closure.

Conversely, let \[ \alpha=a+bi\in\QQ(i) \] be integral over \(\ZZ\), with \(a,b\in\QQ\). Its conjugate \(\bar\alpha=a-bi\) is also integral. Hence the trace and norm \[ \alpha+\bar\alpha=2a, \qquad \alpha\bar\alpha=a^2+b^2 \] are rational algebraic integers, hence integers.

Write \[ 2a=m\in\ZZ. \] Then \[ (2b)^2=4(a^2+b^2)-m^2\in\ZZ. \] Since \(2b\in\QQ\) and its square is an integer, \(2b=k\in\ZZ\). Now \[ m^2+k^2=4(a^2+b^2) \] is divisible by \(4\). Squares modulo \(4\) are \(0\) or \(1\), so \(m\) and \(k\) must both be even. Therefore \(a,b\in\ZZ\), and hence \(\alpha\in\ZZ[i]\).

Thus the integral closure of \(\ZZ\) in \(\QQ(i)\) is exactly \(\ZZ[i]\).