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.


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,


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


You do not need to know the roots of the cubic to find its Galois group. You should consult an algebra book about Galois groups and discriminants here. Then the solution is as follows. By the Rational Root Theorem you know that $x^3-3x+1$ is irreducible and its discriminant is $81$, which is a square in $\mathbb{Q}$. Therefore the Galois group of $x^3-3x+1$ is the alternating group $A_3$. If the discriminant of a cubic is not a square, and the polynomial is irreducible, then its Galois group is $S_3$. This is the case, for example, for $x^3+3x+1$.

  • $\begingroup$ This answer is correct if a bit mysterious. $\endgroup$ – Charles Jun 2 '16 at 19:25
  • $\begingroup$ @Charles: basic Galois theory, which is one of the tags in the question. It shouldn't be that misterious and in fact it is precise, short and to the point. +1 $\endgroup$ – DonAntonio Jun 2 '16 at 19:35
  • $\begingroup$ Thanks. So since this is a 'reduced' cubic we can use: $disc=-4(-3)^3-27(1)^2=81=9^2 \implies Gal(L/\mathbb{Q})=A_3$. What can we say about the second case, is it true we must consider: $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3}), \mathbb{Q}]$? $\endgroup$ – thinker Jun 2 '16 at 19:44
  • $\begingroup$ @DietrichBurde All I meant is that a beginner might find it a bit hard to follow. $\endgroup$ – Charles Jun 2 '16 at 20:47

For the second one, you should think about those polynomials separately. In that case, it's pretty easy to see that the splitting field of $(x^3 - 2)(x^2+3)$ is $\mathbb{Q}(\sqrt[3]{2} , \zeta_3, \sqrt{-3})$.

Since $\zeta_3 = \frac{-1 \pm \sqrt{-3}}{2}$, this is the same as $\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3})$. Since $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt{-3}) : \mathbb{Q}] = 2$, and $(2,3) =1$, we get $[\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3}) :\mathbb{Q}] = 6$.

So, $Gal(\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3})/\mathbb{Q})$ is either $\mathbb{Z}/6\mathbb{Z}$ or $S_3$. But, since the subfield $\mathbb{Q}(\sqrt[3]{2})$ is not Galois over $\mathbb{Q}$, the Galois group cannot be abelian, and so it is $S_3$.


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