We are given that $E/\Bbb{Q}$ is a splitting field extension for some quartic polynomial $f(x)\in \Bbb{Q}[x]$ such that $G = \operatorname{Gal}(E/\Bbb{Q}) = S_4$. We are told $f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)\in E[x]$, and we must determine the degree of $[B:\Bbb{Q}]$ where $B = \Bbb{Q}(\alpha_1+\alpha_2)$, and we must determine the degree of the subfields of $B$.

Progress Thus Far

I believe I have determined the degree $[B:\Bbb{Q}]$. We need all the permutations that will fix $\alpha_1+\alpha_2$, which are the following: $$ H = \{(1), (12), (34),(12)(34)\}.$$ This means that $[B:\Bbb{Q}] = |G:H| = 24/4 = 6.$

Now to determine the subfields, I want to find the subgroups of $S_4$ that contain $H$. I believe the only possibilities are $$H_1 = \{(1),(12),(34),(12)(34), (1324),(1423),(14)(23),(13)(24)\} \simeq D_4,$$ and $S_4$ itself. Clearly, the fixed field of $S_4$ is $\Bbb{Q}$ and $[\Bbb{Q}:\Bbb{Q}] = 1$. I know that the degree over $\Bbb{Q}$ of the fixed field of $H_1$ would be $24/8 = 3$, but I am having a hard time figuring out the fixed field of $H_1$.

It seems like the field $B_1 = \Bbb{Q}((\alpha_1+\alpha_2)(\alpha_3+\alpha_4))$ would be fixed by $H_1$, but how would I know that this is THE fixed field? Also, how do I know that this is in fact a subfield of $B$, because it is not clear to me that it is? Is there some other way that we could write $B$ that would make it more clear that $B_1$ is a subfield of $B$?

I have some insight into why $B_1 = \Bbb{Q}((\alpha_1+\alpha_2)(\alpha_3+\alpha_4))$ is a subfield of $B$. First, if we expand out $f$ we can find that $\alpha_1+\alpha_2+\alpha_3+\alpha_4$ is one of the coefficients. This means $\alpha_1+\alpha_2+\alpha_3+\alpha_4 \in \Bbb{Q}$, which also implies that $(\alpha_1+\alpha_2+\alpha_3+\alpha_4) - (\alpha_1+\alpha_2) = \alpha_3 + \alpha_4$ must be in $B$ as well. So in fact $(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)\in B$ since $B$ is a field.


1 Answer 1


If I understand your question, you basically want to show that $B_1$ is a proper subset of $B$. Alternatively, you want to show that $\operatorname{Gal}(E/B)$ is strictly smaller than $\operatorname{Gal}(E/B_1)$ or equivalently, you want to find a permutation that fixes $B_1$ but moves $B$.

Well, how about $\alpha_1 \to \alpha_3, \alpha_2 \to \alpha_4$ and vice versa. This swaps $\alpha_1+\alpha_2 \to \alpha_3 + \alpha_4$ and so fixes $B_1$ but certainly not $\alpha_1 + \alpha_2$ unless the two are equal. In that case, define $B_2$ the same as $B_1$ except use $\alpha_k^2$ instead of $\alpha_k$. If even this doesn't work, try $B_3$ with $\alpha_k^3$ and so on...

One of these is guaranteed to work since otherwise you would have $\alpha_i = \alpha_j$ for $i\neq j$. The reason $B_n$ is contained in $B$ is because $\sum_k \alpha_k^n \in \Bbb Q$ because every permutation fixes it and then the proof proceeds as in your edit.

EDIT: Actually, you don't need to go above $B_2$. This is because $\alpha_1^2 + \alpha_2^2 = (\alpha_1+\alpha_2)^2 - 2\alpha_1\alpha_2$ and if $B_2$ is fixed by that permutation, then $\alpha_1\alpha_2 = \alpha_3\alpha_4$.

In particular, $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ are all roots of the same degree $2$ equation $$x^2 - (\alpha_1+\alpha_2)x + \alpha_1\alpha_2 = x^2 - (\alpha_3+\alpha_4)x + \alpha_3\alpha_4$$


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