# Galois group of a reducible polynomial over a arbitrary field

How to proceed to determine the Galois group of a reducible polynomial over a field $F$. As an example I tried to compute the Galois group of $f(X)=X^4+4\in\mathbb{Q}[X]$; one can check that $f(X)=(X^2-2X+2)(X^2+2X+2)$ and the polynomials $h(X)=X^2-2X+2$ and $g(X)=X^2+2X+2$ are irreducible (Eisenstein). Computing the the discriminant $d_g=d_h=-4$, then $\Delta_g,\Delta_h\not\in\mathbb{Q}$. Therefore, $G_h=S_2=G_g$. The roots of $h(X)$ are $\{1+i,1-i\}$ and the roots of $g(X)$ are $\{-1+i,-1-i\}$. As all the roots are in the same field, namely $\mathbb{Q}(i)$, it follows that $G_f=S_2$ and thus a cyclic group of order $2$ in $S_4$. Is that right? If not all the roots were not in the same field, the Galois group of a reducible polynomial $f(X)=h_n(X)\dots h_0(X)$ would be the direct sum of the Galois group of each $h_i(X)$?

In this case, you are correct. In general, Galois groups of different polynomials don't always combine as easily as you have stated. Consider $(x^3-2)(x^3-5)$. The roots of the first factor are $\sqrt{2}, \pm \sqrt{2}z$ (where $z$ is a primitive third root of unity) and those of the second are $\sqrt{5}, \pm \sqrt{5}z$. Now, each has Galois group $S_3$, but the splitting field of either polynomial contains $z$, so that together, the splitting field only has dimension 18 as opposed to 36 if the Galois group were $S_3 \times S_3$.