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 $b_j$.

I am not getting any clue for part 3). Please help.


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.

  • 1
    $\begingroup$ Thanks a lot, for rectifying my error. I'll check back again $\endgroup$ – Shadow May 2 at 10:14

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