# On the Galois group of the compositions of polynomials

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.

• It is completely unclear what you are asking. The answer to the question in the title of your post is yes, and the result of Odoni that you cite proves that immediately. But there is a disparity between the claim of Choi that the Galois group in question is equal to a wreath product (presumably the wreath product of the Galois groups of the two polynomials), and the result of Odoni, which says only that the Galois group embeds into a wreath product. Was it that disparity that you are really asking about? If so, then you should change the title of your post. Jun 5, 2021 at 12:28
• @DerekHolt Thank you very much for the comment. Yes, I agree, the title in the question was misleading. Yes, I am interested in the question, if the Galois group of two such polynomials is a wreath product in a non-trivial way. Jun 5, 2021 at 12:54
• The edit to my answer answers the edited question as well; a group of order $4$ is not a wreath product of any two nontrivial groups. In the case of $x^2 -2$ composed with itself, the Galois group is in fact cyclic of order $4$, so it is not even a product. Jun 5, 2021 at 12:58
• Ah ok, I did not yet see your edit. Thank you very much ! Jun 5, 2021 at 13:02

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 $$S_{k\cdot l}$$.

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 $$8$$.

• This answers the question in the title of the post, but I am not sure if that was really the intended question! Jun 5, 2021 at 12:29
• @DerekHolt The question was how does the Lemma 4.1 imply the question in the title. This answer correctly explains this by showing that the wreath product is strictly smaller than the full symmetric group (cardinality-wise). Jun 5, 2021 at 12:34
• Yes I agree with that, but it is not clear that Choi's claim that the Galois group is equal to a wreath product follows from Odoni's result that it embeds into a wreath product, and I thought that might be the real point of the question. Jun 5, 2021 at 12:38
• @DerekHolt You may be right, it wasn't clear to me from the question. I have edited the answer. Jun 5, 2021 at 12:50
• Sorry, the title is really misleading. I will make it more explicit. Jun 5, 2021 at 12:55

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 $$F$$.