I am currently reading J.P. Escofier's "Galois Theory" and in the text he discusses the galois group of $\mathbb{Q}(\sqrt[3]{2}, j)$ which is isomorphic to $S_3$. I have become lost in his discussion of finding the field of invariants of a given subgroup of the Galois group, particularly in reference to a specific calculation.
Let $\sigma_3$ denote the $\mathbb{Q}$-automorphism of $\mathbb{Q}(\sqrt[3]{2}, j)$ defined by sending $\sqrt[3]{2} \to j\sqrt[3]{2}$ and $j \to j$. We know that each element $x\in \mathbb{Q}(\sqrt[3]{2}, j)$ can be respresented uniquely in the following way:
$x=a+bj+c\sqrt[3]{2}+dj\sqrt[3]{2}+e\sqrt[3]{4}+fj\sqrt[3]{4}$ (*)
To determine whether this element is invariant we apply $\sigma_3$ to $x$ and determine for what values of $a, b, c, d, e, f$ we will have $\sigma_3(x)=x$. After applying $\sigma_3$ if $x$ is an invariant we should have:
$x=a+bj+cj\sqrt[3]{2}+d(-1-j)\sqrt[3]{2}+e(-1-j)\sqrt[3]{4}+f\sqrt[3]{4}$ (**)
Since $j^2+j+1=0$ I can see how he obtained the $(-1-j)$'s in the second equation. But in the next line he immediately states that this implies that $c=-d, d=c-d, e=-e+f, f=-e$ and hence all the coefficients are $0$.
(1) I do not understand how he obtained the relationships between the coefficients. I am assuming he set (*)=(**), but in doing so myself I was not able to deduce the same relationships. Any clarification on how this was done would be greatly appreciated.
Thanks in advance
By collecting coefficients, you have
$$ x = x=a+bj+c\sqrt[3]{2}+dj\sqrt[3]{2}+e\sqrt[3]{4}+fj\sqrt[3]{4} \\ = a+bj+cj\sqrt[3]{2}+d(-1-j)\sqrt[3]{2}+e(-1-j)\sqrt[3]{4}+f\sqrt[3]{4} \\ = a + bj -d\sqrt[3]{2} + (c - d)j\sqrt[3]{2} + (f - e)\sqrt[3]{4} - ej\sqrt[3]{4} $$
so
$$ c = -d, \, d = c-d, \, e = f-e, \, f = -e. $$