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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.

llf
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    note that this question is equivalent to the updated version of this question: https://math.stackexchange.com/questions/2184351 which is still unanswered – Lukas Heger Aug 30 '21 at 17:48
  • @LukasHeger Thanks, I guess that cooking up a counterexample (if it exists) is trickier when we restrict $f$ in this way. – llf Aug 30 '21 at 19:05
  • This argument covers the case where $K$ is the fixed field of a Sylow subgroup of $Gal(L/F)$. – Jyrki Lahtonen Sep 03 '21 at 16:24
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    I don't see it. Asking for help from group theorists. – Jyrki Lahtonen Sep 03 '21 at 16:58
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    The answer by Derek Holt suggests counterexamples of the following type. Let $f(x)$ be of degree six such that its Galois group $G$ is generated by $\alpha=(132)(456)$ and $\beta=(24)(35)$ as a group of permutations of the six roots. Let $H=\langle\beta\rangle$, and let $K$ be the fixed field of $H$. Then $K$ contains the roots $x_1$ and $x_6$. The normalizer $N=N_G(H)$ of $H$ is generated by $\beta$ and $\gamma=(23)(45)$, so all the elements of $N$ also keep $x_1$ and $x_6$ fixed. But, any automorphism of $K$ is the restriction of an element of $N$, so none of those map $x_1$ to $x_6$. – Jyrki Lahtonen Sep 03 '21 at 21:01
  • @JyrkiLahtonen Thanks. I'll try to realize such an $f \in \mathbb{Q}[X]$ in the coming days, if no one else gets to it first. – llf Sep 03 '21 at 21:32
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    The following should work. Let $g(x)\in\Bbb{Q}[x]$ be an irreducible quartic with roots $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ and Galois group $S_4$. The conjugates of $z=\alpha_1+\alpha_2$ are then the numbers $\alpha_i+\alpha_j$, $1\le i<j\le4$. Let $K=\Bbb{Q}(\alpha_3,\alpha_4)$ be the fixed field of $\alpha_1\mapsto\alpha_2\mapsto\alpha_1$, $\alpha_3\mapsto\alpha_3$, $\alpha_4\mapsto\alpha_4$. Then $K$ contains exactly two conjugates of $z$, namely $z$ itself and, of course, $w:=\alpha_3+\alpha_4$. – Jyrki Lahtonen Sep 04 '21 at 03:46
  • (cont'd) The only non-trivial automorphism $\sigma$ of $K$ interchanges $\alpha_3$ and $\alpha_4$, but $\sigma(z)=z$, $\sigma(w)=w$. Therefore $Aut(K)$ acts trivially on the set of roots $S={z,w}$ of the minimal polynomial of $z$ over $\Bbb{Q}$. This minimal polynomial can thus handle the duties of the polynomial $f(x)$. – Jyrki Lahtonen Sep 04 '21 at 03:56
  • @JyrkiLahtonen: It would be great if you could convert your comments into an answer with some more details (and probably include an explicit polynomial as an example). – Paramanand Singh Sep 08 '21 at 06:28
  • Done, @ParamanandSingh. I have mixed feelings about this contribution. Given the group action example by Derek Holt, this is just its remapping, and not worth much as a contribution. However, the way this played out, I understand why no one else will touch this question with a bargepole. So, like it or lump it, I must step up to the plate :-/ – Jyrki Lahtonen Sep 09 '21 at 18:21

2 Answers2

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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 $K$.

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.

Jyrki Lahtonen
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    Very nice explanation! The Galois group of $x^4+x+1$ is treated here. I am not comfortable with group theory and did not understand the related group theory question or its answer by Derek. – Paramanand Singh Sep 10 '21 at 01:57
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    Fixed a minor typo. – Paramanand Singh Sep 10 '21 at 02:08
  • Thanks, @ParamanandSingh :-) – Jyrki Lahtonen Sep 10 '21 at 03:58
  • @ParamanandSingh I tried to explain the connection to the group theory question in an addendum. Don't know if it works as intended :-) – Jyrki Lahtonen Sep 10 '21 at 05:00
  • I need to revisit normalizer, centralizer, stabilizer (or any other lizers :)) from Dummit and Foote. You are making use of the result $\text{Gal} (L/\sigma(K)) =\sigma\text{Gal} (L/K) \sigma^{-1},,\forall \sigma\in \text{Gal} (L/F) $. I can now make some sense of the related group theoretic aspect. – Paramanand Singh Sep 10 '21 at 06:25
  • @ParamanandSingh Yes. A degree of familiarity with group actions helps immensely when advancing a bit from the basic Galois theory. I will be, again, banging my head against that wall next Spring when I try to introduce a bunch of undergrads to Galois theory in the latter half of general course. Well, it's my job :-) – Jyrki Lahtonen Sep 10 '21 at 06:56
  • @JyrkiLahtonen Thank you for the extensive writeup, I'm busy with exams right now but I'd like to digest all of this in due time. For now I just have another point of curiousity: In the OP I noted that the claim holds when we insist that each of the roots of $f$ in $K$ generates $K/F$. What does this mean in the translation to group theory? My guess is that it involves $S = \operatorname{Aut}(K/F)$ acting on $X$ faithfully, since in this case, the elements of $S$ can be recovered by their action on any particular root. – llf Sep 12 '21 at 22:10
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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 $0 \times 0 \times \mathbb{Z}/2$.

orangeskid
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