On the Galois group of the compositions of polynomials

Question

We reprint an old math SE question here (see https://math.stackexchange.com/questions/1241224/composition-of-polynomials-and-galois-theory):

" Let $f(x)$ be a polynomial of degree $n$ over $\mathbb{Q}$, with Galois group isomorphic to the symmetric group $S_n$. How do I show that $f$ cannot be expressed as a composition $g(h(x))$ of two polynomials $g$ and $h$ of degrees $> 1$. "

This old question does not have an answer, but one comment refers to the article

http://www.ccms.or.kr/data/pdfpaper/jcms22_3/22_3_497.pdf

of Choi. Therein, in the paragraph after Lemma 3.2, it is written:

"One of the important results about Gal$(f(g(x))/K)$ is that the Galois group is a wreath product of certain groups ([6])."

Here, $K$ denotes any field and the reference $[6]$ points to the article

"The Galois theory of iterates and composites of polynomials" by Odoni.

Alas, Choi does not give a particular Lemma or Theorem of $[6]$ as a reference.

The closest we could find is Lemma $4.1$ in $[6]$:

$K$ is an arbitrary field.

LEMMA $4.1.$ Let $f(g(X))$ be separable over $K$, and let deg$(f)= k$, deg$(g)=l$, with $k,l\geq 1$. Then $f(X)$ is also separable over $K$. Let $\mathcal{F}$ be Gal $f(X)/K$, identified with a subgroup of the permutations of its zeros in the usual way. Then there is an injective homomorphism of Gal $f(g(X))/K$ into the wreath product of $\mathcal{F}$ with the symmetric group $S_l$.

The question is now:

How to derive the statement in Choi's article from Lemma $4.1$ of Odoni's article?

Or is there another result of Odoni's article needed?

Any additional references are very welcome.

EDIT:

I am interested in the question, if the Galois group of two such polynomials is a wreath product in a non-trivial way.

Thank you very much for the help.

Answer 1

Since $\mathcal{F} \le S_k$, the wreath product of $\mathcal{F}$ with $S_l$ is no larger than the wreath product of $S_k$ with $S_l$. This has cardinality $$ (l!)^k k! < (k\cdot l)! = |S_{k\cdot l}|.$$ (Perhaps the wreath product we need is the opposite one, with cardinality $(k!)^l l!$. The result is the same.)

To see the inequality, write $$\frac{(k\cdot l)!}{(l!)^k} = \frac{(1\cdot\ldots\cdot l)}{(1\cdot\ldots\cdot l)}\cdot\frac{ ((l+1)\cdot\ldots\cdot (2l))}{(1\cdot\ldots\cdot l)}\cdot\ldots\cdot \frac{((kl-l+1)\ldots\cdot(kl))}{(1\cdot\ldots\cdot l)}$$ $$> \frac{l}{l}\cdot\frac{2l}{l}\cdot\ldots\cdot\frac{kl}{l} =k!,$$ where to pass from the first line to the second we take the last factor from the numerator and denominator of each fraction (it is $>$ and not just $\ge$ if $k>1$).

Therefore $\mathrm{Gal}(f\circ g(X)/K)$ has cardinality less than $S_{k\cdot l}$.

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Edit: Regarding whether the Galois group is the wreath product, the answer is no. Consider the Galois group of $f\circ f(x)$ over $\mathbb{Q}$, where $f(x)=x^2 - 2$. Its order is $4$, while the wreath product of $S_2$ with itself has order $8$.

Answer 2

You might find Satz 6.1 from the my diploma thesis from 2010 relevant:

Let $k$ be a field of characteristic $0$, $f(x)$ an irreducible polynomial from $k[x]$ with roots set $\Omega_f$ in an algebraic closure. Furthermore let $K = k(\Omega_f)$ be the splitting field of $f(x)$ and $G=Gal(K/k)$ the Galoisgroup. The following two statements $(1) \iff (2.a) \text{ and } (2.b)$ are equivalent:

(1) There exist polynomials $g(x)$ from $K[x]$ and $h(x)$ from $k[x]$, both of degree $\ge 2$ such that $f(x) = g(h(x))$.

(2.a) $G$ is imprimitive with a block $\Delta$,

(2.b) If $H = \{ \sigma \in G| \sigma(\Delta) = \Delta \}$, $F$ is the fixed field of $H$ and $b$ from $\Delta$, so we have: $\operatorname{Irr}(b,F,x) = h(x)-a$ for a polynomial $h(x)$ in $k[x]$ and $a$ in $F$.