I want to find the Galois groups of the following polynomials over $\mathbb{Q}$. The specific problems I am having is finding the roots of the first polynomial and dealing with a degree $6$ polynomial.
$X^3-3X+1$
Do we first need to find its roots, then construct a splitting field $L$, then calculate $Gal(L/\mathbb{Q})$?
I am having difficulties finding roots. If we let the reduced cubic be: $U^2+qU+\frac{p^3}{27}=U^2+U+\frac{27}{27}=U^2+U+1$. The roots of this are: $x=\frac{-1 \pm \sqrt{-3}}{2}$
How do we use this to find the roots of the cubic?
Once I can decompose the polynomial I know that the Galois group will be $\{e\}, Z_2, A_3$ or $S_3$ depending on the degree of the splitting field and and how many linear factors there are,
$(X^3-2)(X^2+3)$
I have never encountered finding the Galois group of a degree $6$ polynomial but I am guessing that since it is factorised this eases things somewhat.
Let $f(X)=(X^3-2)(X^2+3)=(X-\sqrt[3]{2})(X^2+aX+b)(X-\sqrt{-3})(X+\sqrt{3})$
I am not sure how to find the coefficients of $X^2+aX+b$. Is it irreducible?
Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$ then (assuming $X^2+aX+b$ is irreducible) $L=\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$.
If this is true what would $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3}), \mathbb{Q}]$ be?
I think this degree would be the order of the Galois group, so it could narrow down to one of $S_3, S_4, A_3, A_4...$ etc