I'm trying to do I solve this exercise:
Let $K$ be a field contained in the algebraic closure of $Q$, and $a ∈ K$. Describe the splitting field and the Galois group on $K$ of the polynomial $f(X)=X^3-a$ in the following situations:
a) $K$ contains a third primitive root of unity and $a$ is a cube in $K$.
b) $K$ does not contain a third primitive root of unity and $a$ is a cube in $K$.
c) $K$ contains a third primitive root of unity and $a$ is not a cube in $K$.
d) $K$ does not contain a third primitive root and $a$ is not a cube in $K$.
The roots of $f(X)$ are $\sqrt[3]{a},\omega\sqrt[3]{a},\omega^2\sqrt[3]{a}$ with $\omega$ third primitive root of unity.
- Now, in the first point we have $\omega \in K$ and $a=b^3$ where $b\in K$ so all roots of $f(X)$ are in $K$ and so the splitting field is $K$. Assuming that my reasoning is correct, since the elements of the Galois group cause the permutation of roots of $f(x)$ and the roots contained in $K$ are three, I would say that the Galois group is $S_3$.
- In the second point with a similar reasoning, since $\omega$ is not in $K$, I found that the breaking field is $K(\omega)$ and in this case I don't know if the Galois group is trivial because permute $\omega$ with itself or if it is again $S_3$ because the splitting field still contains all the roots of $f(x)$.
Could someone help me understand?