What is the Galois group over rationals of the splitting field of the reducible polynomial $f(x)=x^8+14x^4+1=(x^4 - 2 x^3 + 2 x^2 + 2 x+1)(x^4 + 2 x^3 + 2 x^2 - 2 x + 1)=(x^4+2i\sqrt3 x^2+1)(x^4-2i\sqrt3 x^2+1)$
Any help would be much appreciated! In my previous question, I got help on determining the Galois group of the first factor (Finding Galois group over Rationals)
The quartic factors can be written as sums of two squares $$ \begin{aligned} x^4-2x^3+2x^2+2x+1&=(x^2-x)^2+(x+1)^2,\\ x^4+2x^3+2x^2-2x+1&=(x^2+x)^2+(x-1)^2. \end{aligned} $$ Therefore over $\Bbb{Q}(i)$ we get the four quadratic factors $$ x^2+x\pm i(x-1),x^2-x\pm i(x+1). $$ The discriminant of those quadratics are $\pm 6i$. By the quadratic formula the roots are in the field $L=\Bbb{Q}(i,\sqrt{6i})$. As $\sqrt{6i}=\pm\sqrt3 (1+i)$ we can identify $L=\Bbb{Q}(i,\sqrt3)$. This is then also the splitting field. For the coefficients of any factor must be in the splitting field, as do the square roots of the discriminants of those factors. The Galois group is thus $C_2\times C_2$. For both quartic factors as well as the octic.
The zeros of your octic are obviously units in the ring of integers $\mathcal{O}=\Bbb{Z}[i,\omega]$ of $L$, where $\omega=(-1+i\sqrt3)/2$. We know that $u=2+\sqrt3$ is a fundamental unit of $\Bbb{Z}(\sqrt3)$. On the other hand, the zeros of $x^2+14x+1=0$ are $-7\pm4\sqrt3=-u^{\pm2}$. Therefore I got interested in writing the zeros of the octic as elements of $\mathcal{O}$. After a bit of experimenting I found the number $$ v=-1-\omega+i\omega $$ with $$ v^2=iu.\qquad(*) $$ From $(*)$ it follows that $v^4=-u^2=-7-4\sqrt3$, so $v$ and its conjugates are four of the zeros of the octic. The other four are the negatives of these.
By Dirichlet's unit theorem we know that the unit group of $\mathcal{O}$ has rank one. Its torsion part consists of the twelfth roots of unity ($L$ is the twelfth cyclotomic field). I suspect that $v$ may well be a generator, i.e. $\mathcal{O}^*=\mu_{12}\times\langle v\rangle$, but I'm too ignorant about the necessary methods of algebraic number theory to verify this.
Another look at this lead to the following. Let $\zeta=e^{\pi i/6}$ be a primitive twelfth root of unity. Its minimal polynomial is $\Phi_{12}(x)=x^4-x^2+1$. Therefore $$ v=-1-\zeta^4+\zeta^7=-\zeta^2+\zeta^7=-\zeta^2(1-\zeta^5). $$ By basic properties of cyclotomic polynomials (or simply using that $\Phi_{12}(x)=x^4-x^2+1$) we have $$1=\Phi_{12}(1)=(1-\zeta)(1-\zeta^5)(1-\zeta^7)(1-\zeta^{11})$$ implying that $1-\zeta^5$ is a unit. Hence so is $v$. WP on cyclotomic units refers to Washington's book for more.