Question:
Let $f,g$ be two polynomials over Finite field $F$ such that $g\mid f$ and $ord(f)=ord(g)$. If $f$ is square-free and $f(0)\ne 0$, can we say that the splitting field of $f$ equals that of $g$ ? If it is true, show it.
Let $f$ be a polynomial over Finite field $F$. The order of $f$, denoted by $ord(f)$, is defined by the least positive integer $e$ such that $f \mid x^e-1$.
Lemma 1:Let $f=\prod_{i=1}^{k}{f_i^{e_i}}$ be a polynomial over Finite field $F$, where $f_i$ are irreducible and $e_i$ are positive integers. Let $p$ be the characteristic of $F$. Then $ord(f)=p^t > lcm(r_1,\ldots,r_k)$, where $r_i=ord(f_i)$ and $t$ is the least positive integer such that $p^t\ge \max(e_1,\ldots,e_k)$.
Thanks for the answer posted by Lord Shark the Unknown. The conclusion is as follows.
Let $f, g$ be two square-free polynomials over finite field $\mathbb{F}_q$. Then the splitting field of $f$ is that of $g$, up to isomorphism, if and only if $\text{ord}(f)=\text{ord}(g)$.
Let $\text{ord}(f)=d$. Then every root of $f(x)=0$ is a $d$-th root of unity and so the splitting field of $f$ is contained in the cyclotomic extension $F(\mu_d)$. Here $\mu_d$ is the group of $d$-th roots of unity in an algebraic closure of $f$. But also $d$ is the least common multiple of the orders of the zeros of $f$. Thus $d$ is the order of the multiplicative group generated by the zeros of $f$. Thus $\mu_d$ is contained in the splitting field of $f$. We conclude that $F(\mu_d)$ is the splitting field of $f$, that is $\text{ord}(f)$ determines the splitting field.