# Which subfields of the Galois group of $x^4+8x^2+8x+4$ are Galois and find the splitting field polynomial

This question is an exercise in Dummit and Foote:

1)I want to find the Galois group of $$f(x)=x^4+8x^2+8x+4$$.

2)Which subfields of the splitting field of $$x^4+8x^2+8x+4$$ are Galois over $$\Bbb Q$$?

3) For the subfields which are Galois over $$\Bbb Q$$, find the polynomial $$f(x) \in \Bbb Q[x]$$ for which they are the splitting field over $$\Bbb Q$$.

My attempt:

I have calculated that $$f(x)$$ is irreducible moreover, the resolvent cubic $$h(x)=x^3-16x^2+48x+64$$ is irreducible. Again the discriminant $$315392$$ is not a square so the Galois group has to $$S_4$$. So part 1) is done.

Now, the subfields of the Galois group of $$x^4+8x^2+8x+4$$ which are Galois correspond to the fixed field of a normal subgroup of $$S_4$$ which are $$K_4$$ and $$A_4$$. So in some sense part 2) is solved. If I want to answer this question as a field $$Fix(K_4)=\Bbb Q(a_1,\cdots,a_n)$$ and $$Fix(A_4)=\Bbb Q(b_1,\cdots,b_r)$$ then what will be $$a_i$$ and $$b_j$$.

First of all you have a mistake in computing the discriminant, as it is $$200704 = (7 \cdot 12^6)^2$$. This means that the Galois group is $$A_4$$, instead of $$S_4$$.
Now the normal groups of $$A_4$$ are $$\{e\}, K_4$$ and $$A_4$$. So the only proper Galois subfield is the one corresponsing to $$K_4$$. Now as $$[A_4:K_4] = 3$$ we have that the corresponsing field is cubic. Now use the fact that the splitting field of $$f$$ contains the splitting field of its cubic resolvent. Call the latter $$L$$. Then it's not hard to conclude that $$[L:\mathbb{Q}] = 3$$ and this must correspond to $$K_4$$, as $$A_4$$ has a single subgroup of index $$3$$. Hence, $$L$$ is the splitting field of $$x^3 - 16x^2 + 48x + 64$$.
REMARK: To prove that $$[L:\mathbb{Q}] = 3$$, you can use the fact that the $$f$$ and its resolvent have the same discriminant, which we already found to be a square.