Galois groups of truncated $\cosh(x)$ Taylor polynomials and related results?

Question

(Part of this question was written with ChatGPT because english is not my native language). I am currently translating my diploma thesis from 2010 in english and thought to think about the topic again: It is about investigating the Galois groups of Taylor polynomials, particularly focusing on those derived from the hyperbolic cosine function. Inspired by a theorem by Schur on the Galois groups of exponential Taylor polynomials, I have been exploring similar structures in the Taylor polynomials of the hyperbolic cosine function, $ \cosh(x)$.

Background:

Schur's theorem states that the Galois group over $ \mathbb{Q} $ of the polynomial $ E_n(x) := \sum_{k=0}^n \frac{x^k}{k!} $ is the symmetric group $ S_n $ if $ n \not\equiv 0 \mod 4 $ and the alternating group $ A_n $ otherwise. This result has been proven using methods from analytic number theory (Schur-Sylvester's theorem on divisibility) and algebraic number theory (Dedekind's theorems), as well as alternative methods involving Newton polygons as demonstrated by Coleman.

I am attempting to determine the Galois groups of truncated $ \cosh(x) $ Taylor polynomials, defined as $ C_n(x) := \sum_{k=0}^n \frac{x^{2k}}{(2k)!} $. Initial computations for low-degree polynomials suggest that the Galois group of $ C_n(x) $ over $ \mathbb{Q} $ may be $ S_2 \wr S_n $, the wreath product of $ S_2 $ with $ S_n $.

Challenges:

At the time of writing the diploma (2009-2010) proving this conjecture for $ C_n(x) $ involved challenges, particularly in identifying when a subgroup $ G \leq S_2 \wr S_n $ constitutes the entire group $ S_2 \wr S_n $. Tools similar to Jordan's theorem, which provides criteria for a group being the full symmetric or alternating group, would be invaluable. Unfortunately, no such criteria seem to exist for the wreath product $ S_2 \wr S_n $.

Questions for the Community:

1. Has anyone encountered similar structures or results concerning the Galois groups of other special functions or their Taylor expansions?
2. Are there known results or theorems that might help in proving that the Galois group of $ C_n(x) $ over $ \mathbb{Q} $ is indeed $ S_2 \wr S_n $?
3. Any recommendations on methodologies or literature that could be useful in tackling problems involving the Galois groups of functions beyond the exponential function?

Any insights or suggestions would be greatly appreciated as they could significantly impact the direction and feasibility of proving this conjecture.

Answer 1

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

EDIT 2: in fact Assumption 1 below implies Assumption 2 by the question's author's work, so by the arguments below Assumption 1 is equivalent to $\mathrm{Gal}(C_n)=S_2\wr S_n$. It appears likely that it holds for all $n$, but I don't have a proof.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ is odd (and the same for $(2n)!\cdot\mathrm{disc}(F_n)$).

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).