 Integral basis of the fourth-root-of-two field

Write $\alpha =\sqrt[4]{2}$ and an algebraic integer as $a+b\alpha +c{\alpha }^2+d{\alpha }^3$ with rational coefficients. Traces to $\mathbb{Q}(\sqrt{2})$ show $2a,2b,2c,4d$ are integers. The norm has coefficients $a^2+2c^2-4bd$ and $2ac-b^2-2d^2$ in the integral basis $1,\sqrt{2}$. Set $A=2a$, $B=2b$, $C=2c$, $D=4d$. Integrality gives $A^2+2C^2-2BD\equiv 0(\mathrm{mod}4)$ and $4AC-2B^2-D^2\equiv 0(\mathrm{mod}8)$. The second forces $D$ even and then $B$ even; the first forces $A,C$ even; returning to the second forces $D$ divisible by four. Thus all original coefficients are integers, proving the integral basis $1,\alpha ,{\alpha }^2,{\alpha }^3$.