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The polynomial $T^4-2$ is irreducible by Eisenstein criterion at two, so $1,\alpha ,{\alpha }^2,{\alpha }^3$ is a rational basis and each of these elements is integral. The nontrivial automorphism of $K/k$ sends $\alpha$ to $-\alpha$. For an algebraic integer $\theta$, its relative trace and the trace of the algebraic integer $\alpha \theta$ areBoth are in ${\mathcal{O}}_k$, proving $2a,2b,2c,4d\in \mathbb{Z}$. Multiplying $\theta$ by its conjugate gives the field normAgain this is in ${\mathcal{O}}_k$, so both displayed coefficients are integers.

To finish the denominator argument, put $A=2a$, $B=2b$, $C=2c$, $D=4d$, all integers. The two norm coefficients implyThe second congruence modulo two first gives $D$ even. Modulo four it then gives $B$ even. Therefore $2BD$ vanishes modulo four, and the first congruence reads $A^2+2C^2\equiv 0(\mathrm{mod}4)$. The possible squares modulo four force $A$ even and then $C$ even. Returning to the second congruence now gives $D^2\equiv 0(\mathrm{mod}8)$, so $D$ is divisible by four. Thus $a,b,c,d$ are all integers.

Conversely, every integer linear combination of these powers is integral. We have proved the integral basis of the fourth-root-of-two field: