Why is $\text{Gal}(K/k) \cong \text{Gal}(K(\omega)/k(\omega))$ with $\omega \notin K$? The situation is as follows:
$K$ is the splitting field of $\; x^3+px+q \in k[x]$ and $ K(\omega)$ is the field over which $\; x^3-1$ splits.
That's probably one of those things you just see, but I as a beginner haven't got the slightest idea how to show $\text{Gal}(K/k) \cong \text{Gal}(K(\omega)/k(\omega))$. $\omega$ is a root of $x^3-1$ and $\omega \notin K$.
Can anyone help?
 A: A somewhat high-brow explanation comes from the theory of linearly disjoint extensions. Surely that theory is overkill for the purposes of your question, but it's past midnight and I need to wake up in six hours, so...
Here $K/k$ and $k(\omega)/k$ are both Galois extensions of $k$ (assuming $6\cdot 1_k\neq 0_k$). The assumption $\omega\notin K$ implies that the two extensions intersect trivially, $K\cap k(\omega)=k$. For two Galois extensions this implies that the two extensions are linearly disjoint. This has many consequences, for example it implies that $Gal(K(\omega),k)$ is a direct product of $Gal(K(\omega)/K)$ and $Gal(K(\omega)/k(\omega))$. Your claim follows from this by applying the parallelogram rule (aka the second isomorphism theorem) and basic facts from Galois theory.
If you don't know about linearly disjoint field extensions, then I recommend that you study Pete L. Clark's lecture notes on Field Theory. I once wrote a quick-and-dirty intro to the topic that also suffices here. Pete has put a couple orders of magnitude more effort into his notes, and it shows.
A: The 3rd roots complex roots of unity are $1$, $j$, $j^2$, so the extension $k(j)$/$k$ is galois of degree $2$ (you must assume that $K$ is not of characteristic $3$) . Let $a$ be a root of the given cubic, supposed to be irreducible (if it were reducible, the problem would not make sense). Then $k(a)$/$k$ is of degree $3$, hence the composite extension $k(a, j)$ has degree  $6$ over $k$ . But the roots of the cubic are given by well known formulae  which show that the splitting field $K$ is $K$ = $k$(a, cubic root of D), where D is the discriminant  $4 p^3 - 27 q^2$ . Hence $K$ contains $k(a, j)$. Since Gal(K/k) is a subgroup of $S_3$, we conclude that $K$ = $k(a, j)$. The rest of the proof is obvious, because $K$ = $k(a)$($j$) = $k(j)$($a$) .
A: To Miriam's intention, allow me to summarize the discussion about the roots of the cubic f($X$) = $X^3$ + p$X$ + q (although it goes beyond her original question). I keep all my previous notations and I assume that k is of characteristic different from 2 and 3. Since $S$ := Gal(K/k) is a subgroup of $S_3$ , its order is 3 or 6. In both cases, it contains a unique subgroup $C$ of order 3, generated by the 3-cycle which permutes the 3 roots of the cubic, say a, b, c. To distinguish between the two cases, consider the "discriminant", which is here D := $((a-b)(a-c)(b-c))^2$, so that k(sqr D) is contained in K, and is actually the fixed field of $C$ . Consequently  $S$ = $S_3$ iff  D is a square in k (not simply in K ). 
Let us now answer the question : when does j belong to K ? A simple calculation using the derivative f '($X$) shows that D = -(4 $p^3$ + 27 $q^2$). Since k(sqr D) is the unique quadratic subextension of K/k (because of the unicity of C), j belongs to K iff -3 is multiplicatively congruent to D modulo a square of k. Another discriminant now comes into play, that is the discriminant of the second degree equation which gives $u^3$ and $v^3$, namely d = (27 $q^2$ + 4 $p^3$)/27   (brute force), or d =(-$3$). $3^2$. D . So j belongs to K iff d itself is a square in k, which means that $u^3$ and $v^3$ belong to k .
Now I have done my home work !
