Intermediate fields of Galois group of $X^4+8T+12$ and minimal polynomials of their generators

Question

I've worked out that $X^4+8T+12$ is irreducible over $\mathbb{Q}$. I also worked out that it's Galois group has to be (isomorphic to) $A_4$. Now I want to make a diagram showing the Galois correspondence. Actually, I also managed to do that but mostly by guessing. Let's say that $$ X^4+8X+12=(X-r_1)(X-r_2)(X-r_3)(X-r_4) \ \ \ \ \ (i) $$ with the $r_i$ all different. $\left \langle (234) \right \rangle$ fixes $r_1$ and has index $4$ in $A_4$, so since $\mathbb{Q}(r_1)$ has degree $4$ over $\mathbb{Q}$, $\mathbb{Q}(r_1)$ corresponds to $\left \langle (234) \right \rangle$ (same for other index $4$ subgroups). Here, I'm not guessing. My question: is there a sure method of finding the other correspondances without having to resort to guessing? Other question. I guessed that $\mathbb{Q}(r_1+r_4)$ corresponds to $\left \langle (14)(23) \right \rangle$ as follows: The orbit of $r_1 + r_4$ under the Galois group action has six elements, $$r_1+r_4, r_2+r_4, r_3+r_4, r_1 + r_3, r_2+r_3, r_1+r_2 $$ therefore the minimal polynomial of $r_1 + r _4$ over $\mathbb{Q}$ has degree $6$. $\left \langle (14)(23) \right \rangle$ fixes $r_1 + r_4$ and has index $6$ in $A_4$, so we are done. What I now would like to do is calculate that minimal polynomial. I have a feeling that I'm missing a trick here. I know that the minimal polynomial is $$(X - (r_1+r_4))(X - (r_2+r_4))(X - (r_3+r_4))( X- (r_1 + r_3))(X - (r_2+r_3))(X - (r_1+r_2)) \ \ \ \ \ \ \ (ii)$$ I also know from $(i)$ that $r_1 + \ldots + r_4 =0$, $r_1r_2 + r_1r_3 + \ldots = 0$ etc... I can find the coefficient of $X^5$ in the minimal polynomial easily from that, but I really don't have the courage to completely work out $(ii)$, expressing the coefficients as polynomials in $r_1 + \ldots + r_4, r_1r_2 + r_1r_3 + \ldots$, etc. etc.. Surely there must be a shorter way?

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In the same way as above, one can show that $\mathbb{Q}(r_1r_2+r_3r_4)$ corresponds to the normal subgroup of $A_4$. Now, here we have $s_1= r_1+r_2+r_3+r_4=0$ and $s_2= r_1r_2+ r_1r_3 + \ldots + r_3r_4=0$. This means that $(r_1+r_2)^2=(r_3+r_4)^2$. So, $$2(r_1+r_2)^2-2r_1r_2-2r_3r_4 = r_1^2+ \ldots + r_4^2= s_1^2-2s_2 = 0$$ hence $(r_1+r_2)^2 =r_1r_2+r_3r_4$. The minimal polynomial of $r_1r_2+r_3r_4$ over $\mathbb{Q}$ is $$(X - (r_1r_2+r_3r_4))(X - (r_1r_3+r_2r_4))(X - (r_2r_3+r_1r_4))$$ Working the coefficients of that out as elementary symmetric polynomials of the $r_i$ takes much less work and we get $X^3-48X-64$. Thus $r_1+r_2$ is a root of $X^6 -48 X^2-64$, which must be its minimal polynomial since it has degree $6$. $r_1 + r_4$ has the same minimal polynomial.

Answer 1

This has to do with the theory of Galois resolvents. Check for example https://en.wikipedia.org/wiki/Resolvent_(Galois_theory) and the references thereby. It might be that in your case you have some smarter tricks, also because when you deal with abelian subextensions you also have CFT available. But I don't think that the general algorithm is much smarter than what you did. Essentially the idea is: you see your subgroup $G\leq A_4$ as a group of permutations on $4$ elements, you compute a polynomial $P\in\mathbb Z[x_1,\ldots,x_4]$ that is invariant under the action of $G$ but not under any bigger group, and you choose a system of representatives $g_1,\ldots,g_m$ for $G$ in $S_4$. Then the polynomial $g(x)=\prod_{i=1}^m(x-g_i(P(r_1,\ldots,r_4)))$ should do the job. You have to be a little careful because you have to make sure that $g$ is squarefree, but there is always a way to choose $P$ that makes $g(x)$ squarefree.