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

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.

• There is a paper maybe you need.
– mayi
Jan 17 at 8:16

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.