For a non-normal extension, is the Galois action still transitive on roots that *are* in the extension?

Question

Let $L/F$ be a field extension where $L$ is the splitting field of the irreducible $f \in F[X]$. Then $\text{Gal}(L/F)$ acts on the roots of $f$ transitively - this follows from how every isomorphism of fields in $L/F$ can be extended to an automorphism of $L$.

Now let $K$ be an intermediate field of $L/F$, where $K$ contains only some of the roots of $f$. Is it true that the automorphisms of $K$ fixing $F$ act transitively on the roots of $f$ in $K$? I know it holds if each of the the roots of $f$ in $K$ generate $K/F$, but I'm not seeing a counterexample where $K$ is not simply generated by some of the roots, or some inductive proof applying this fact to simple subextensions.

Answer 1

The following is a specific example showing that it is not always possible to find such automorphisms.

Let $F=\Bbb{Q}$. Let $g(x)=x^4+x+1$. It is not difficult to show that the Galois group of $g(x)$ is $G\simeq S_4$. The complex roots of $g(x)$ come in conjugate pairs $x_1,x_2=\overline{x_1},x_3,x_4=\overline{x_3}$. Furthermore, by the Vieta relations $x_1+x_2+x_3+x_4=0$, implying that the real parts of $x_1$ and $x_3$ are negatives of each other.

Let $L=\Bbb{Q}(x_1,x_2,x_3,x_4)$ be the splitting field of $g(x)$. As we know that the Galois group is the full symmetric group $S_4$, all the permutations of the roots of $g(x)$ are actually automorphisms of $L$.

It is time to define the polynomial $f(x)$ $$ f(x)=\prod_{1\le i<j\le 4}(x-x_i-x_j) $$ to be the degree six polynomial with the sums $x_i+x_j, i<j$, as its zeros. The splitting field of $f(x)$ over $\Bbb{Q}$ is also $L$. This is because the splitting field, temporarily call it $M$, contains the sum $x_1+x_2$ as a root as well as the difference $x_1-x_2=(x_1+x_3)-(x_2+x_3)$ as a difference between two roots of $f(x)$. Hence $x_{1,2}=\frac12[(x_1+x_2)\pm(x_1-x_2)]$ are in $M$. It follows immediately that also $x_3,x_4\in M$ proving the claim.

A calculation involving the symmetric polynomials (or a friendly computer algebra system) shows that $$ f(x)=x^6-4x^2-1. $$ It has two real roots, namely $z=x_1+x_2$ and $w=x_3+x_4=-z$. The other four roots are purely imaginary as we observed earlier that $\mathrm{Re}(x_1)=\mathrm{Re}(x_2)=-\mathrm{Re}(x_3)=-\mathrm{Re}(x_4)$. If so desired, it is possible to write down formulas for the roots of $f(x)$ because they are the square roots of the roots of the cubic $y^3-4y-1$. This fits together with the fact that $S_4$ is solvable.

At long last let's define $K=\Bbb{Q}(x_1,x_2)$. As $[L:\Bbb{Q}]=24$ it follows that $[K:\Bbb{Q}]=4\cdot3=12$. After all, we are "in the maximal degree case" of the splitting field of a quartic. So we go from $\Bbb{Q}$ to $L$ via the sequence of fields $\Bbb{Q}(x_1)$, $\Bbb{Q}(x_1,x_2)=K$, $\Bbb{Q}(x_1,x_2,x_3)=L$ with respective relative extension degrees $4,3,2$. Alternatively, we see that $K$ is the fixed field of $H=\langle\tau\rangle\le G$, where $\tau=(34)$ interchanges $x_3$ and $x_4$ and fixes both $x_1$ and $x_2$. So basic Galois theory also tells us that $[K:\Bbb{Q}]=|G|/|H|=12$.

Any automorphism of $K$ must permute the generators $x_1$ and $x_2$ and is uniquely determined by how it acts on them. As all permutations of the $x_i$s come from $G$, the automorphism $\sigma=(12)\in G$ maps $K$ to itself, and hence its restriction to $K$ is a non-identity automorphism of. We can thus conclude that $\mathrm{Aut}(K)=\langle \sigma_{\vert_K}\rangle\simeq C_2$.

Next we see that $K$ contains both $z$ and $w$. However, $K$ does not contain any other zeros of $f(x)$. For the other zeros all have the form $x_i+x_j$ with $i\in\{1,2\}$ and $j\in\{3,4\}$, and those are not fixed points of $\tau$.

But $z$ and $w$ are both fixed points of $\sigma$. Hence no automorphism of the intermediate field $K$ maps $z$ to $w$. We have our (counter)example.

A curiosity. Obviously there is an automorphism $\delta$ of the smaller intermediate field $E:=\Bbb{Q}(z)\subset K$ that maps $z$ to $-z=w$. However, as we saw, $\delta$ cannot be extended to an automorphism of the bigger field $K$.

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A few words about the translation of this question into the language of group theory.

Let's look at a general (finite) Galois extension $L/F$ with Galois group $G$. Let $K$ be an intermediate field. Galois correspondence says that $K$ is the fixed field of a subgroup $H\le G$. In this set up we can determine the automorphism group $S=\mathrm{Aut}(K/F)$ as follows.

• Because $L/F$ is Galois, any $F$-automorphism $K$ can be extended to an $F$-automorphism of $L$. That is, any element of $S$ is a restriction of an element of $G$ to the intermediate field.
• An automorphism $\sigma\in G$ maps the fixed field of $H$ to the fixed field of the conjugate subgroup $\sigma H\sigma^{-1}$.
• Therefore $\sigma(K)=K$ if and only if $\sigma H\sigma^{-1}=H$ if and only if $\sigma\in N_G(H)$.
• The restriction $\sigma_{\vert_K}$ is the identity mapping if and only if $\sigma\in H$.
• So we have a natural surjective homomorphism $$N_G(H)\to \mathrm{Aut}(K/F), \sigma\mapsto \sigma_{\vert_K}$$ with kernel $H$. The first isomorphism theorem tells us that $$\mathrm{Aut}(K/F)\simeq N_G(H)/H.$$

So in this question we are looking for a situation where a finite group $G$ acts on $X$, the set of zeros of $f(x)$, in such a way that

1. a subgroup $H$ has at least two fixed points in $X$, and
2. the normalizer $N_G(H)$ does not act transitively on the set of fixed points of $H$.

Derek Holt gave a small example of an action of $G=S_4$ meeting the requirements 1 and 2. Porting that into an example of a field extension is then straightforward. Basically we only need to realize the group $G$ as a Galois group of some extension and connect the dots.

Answer 2

Consider a finite Galois extension $K\subset L$ with Galois group $G$. Let $\alpha \in L$ with $\operatorname{Fix}(\alpha) = H$. For $g\in G$ we have $\operatorname{Fix}(g \alpha) = gH g^{-1}$. Consider the subfield $F = K(\alpha, \beta)$ with fixator subgroup $H \cap gHg^{-1}$.

Question: Does there exists an element in $G$ that takes $\alpha$ to $g\alpha$ and invariates $K(\alpha, g \alpha)$?

Translation: Does there exist an element in $gH$ that invariates(normalizes) $H \cap gHg^{-1}$?

We'll produce a counterexample to the above group theoretic fact.

Consider the $G = (\mathbb{Z}/2)^3 \rtimes \mathbb{Z}/3$ inside the wreath product $(\mathbb{Z}/2)^3 \rtimes S_3$, $H = \mathbb{Z}/2^2 \times 0$, and $g = ((0,0,0), (1,2,3) )$. We have $H \cap gHg^{-1} = 0 \times \mathbb{Z}/2 \times 0$, but every element in $gH$ conjugates $0 \times \mathbb{Z}/2 \times 0$ to $0 \times 0 \times \mathbb{Z}/2$.