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This posting aims at creating an “abstract duplicate target” that some other postings can point to, whenever appropriate. The central question here is to ask for an elementary proof of the following:

Proposition. Let $\mathbb F$ be a field, $f\in\mathbb F[x]$ be a non-constant polynomial and $m\in\mathbb F[x]$ be a monic and non-constant polynomial that splits in $\mathbb F$. If $\mathbb F$ has a positive characteristic, we also assume that $\deg m\le \operatorname{char} \mathbb F$. Then there exists a polynomial $g$ such that $h(x):=f(g(x)) - x$ is divisible by $m$ if and only if for each root $\lambda$ of $m(x)=0$, one can find some $\mu\in\mathbb F$ such that $f(\mu)=\lambda$, and that $f'(\mu)\ne0$ if $\lambda$ is a repeated root of $m$.

One major application of this proposition is to determine the existence of a matrix $k$-th root under appropriate conditions.

Corollary. Let $A\in M_n(\mathbb F)$ and $k$ be a positive integer. Suppose $A$ has a full spectrum in $\mathbb F$ and each eigenvalue $\lambda$ of $A$ has a $k$-th root $\mu\in\mathbb F$ such that whenever $\lambda$ is not semi-simple, $k\mu^{k-1}$ is nonzero. Then $A$ has a $k$-th root expressible as $g(A)$ for some $g\in\mathbb F[x]$.

The following questions, for instances, can be answered using the proposition or the corollary above:

user1551
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  • @BenjaminWright I want to weaken the requirement on $\mathbb F$. If $\mathbb F$ is algebraically closed, I can no longer apply the corollary above to deal with $k$-th roots of a matrix over a finite field. The price of this weakening is that one needs to check, case by case, whether or not $f(\mu)=\lambda$ is solvable. I also need to remedy this weakening by requiring $m$ to split in $\mathbb F$, so that I can deal with each root $\lambda$ of $m(x)=0$ directly. – user1551 Jan 06 '25 at 07:24

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Proof of the proposition. I follow Ramen Nii-chan’s answer to follow Maxime Bôcher’s strategy (cf. Introduction to Higher Algebra, 1907, pp 297-299).

For necessity, let $\lambda$ be any repeated root of $m(x)=0$. If $g$ exists, let $\mu=g(\lambda)$. Since $m$ divides $h$, $h(\lambda)$ has to be zero. Therefore $f(\mu)=\lambda$. If $\lambda$ is a repeated root of $m$, it must also be a repeated root of $h$. Therefore $h'(\lambda)=0$, i.e., $f'(\mu)g'(\lambda)-1=0$. Hence $f'(\mu)\ne0$.

For sufficiency, consider the special case where $m(x)=(x-\lambda)^s$ first. If $s=1$, pick any $\mu$ with $f(\mu)=\lambda$, set $g(x)=(x-\lambda)+\mu$ and we are done. Now suppose $s>1$. Then $h$ is divisible by $(x-\lambda)^s$ if and only if $$ h^{(k)}(\lambda)=0 \quad\text{for}\quad k=0,1,\ldots,s-1.\tag{1} $$ By mathematical induction, for each $k\ge1$ we have $$ h^{(k)}(\lambda)=F_k\left(x,g(\lambda),\ldots,g^{(k-1)}(\lambda),f(g(\lambda)),\ldots,f^{(k)}(g(\lambda))\right) \ +\ f'(g(\lambda))\,g^{(k)}(\lambda), $$ for some polynomial $F_k$ that depends on $k$ but not on $f$ and $g$. By assumption, we may pick some $\mu$ such that $f(\mu)=\lambda$ and $f'(\mu)\ne0$. Hence $(1)$ is satisfied if we can pick a polynomial $g$ such that \begin{align*} g(\lambda)=c_0 &:=\mu\quad\text{(which solves $h(\lambda)=0$)},\\ g^{(k)}(\lambda)=c_k &:=-\frac{F_k\left(x,c_0,\ldots,c_{k-1},f(\mu),\ldots,f^{(k)}(\mu)\right)}{f'(\mu)}\quad\text{for}\quad k=1,2,\ldots,s-1. \end{align*} In particular, $(1)$ is always solved by the Taylor polynomial $g(x)=\sum_{k=0}^{s-1}\frac{1}{k!}c_k(x-\lambda)^k$ (note that $k!\ne0$ because $k<s\le \operatorname{char}\mathbb F$).

Now consider the general case. Let $m=\prod_{i=1}^rp_i$ where $p_i(x)=(x-\lambda_i)^{s_i}$ and $\lambda_i$s are distinct. By the result in the special case above, we can find for each $i$ a polynomial $g_i$ such that $f\circ g_i \equiv x \bmod{p_i}$. Since the $p_i$s are pairwise coprime, by Chinese remainder theorem we can find a polynomial $g$ such that $g \equiv g_i \bmod{p_i}$. It follows that $f\circ g \equiv f\circ g_i \equiv x \bmod{p_i}$ for each $i$. Therefore $f\circ g \equiv x \bmod{m}$ because $p_1,p_2,\ldots,p_r$ are pairwise coprime.

user1551
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