Congruence relation on polynomial ring over finite fields

Question

Let $\mathbb{F}$ be a finite field with char $\mathbb{F} =p$, $A =\mathbb{F}[T]$ be its polynomial ring. Let $P$ be an irreducible polynomial in $A$.

Consider the modulo $P$ map between the group of units: $$f:(A/P^{n}A)^* \to (A/PA)^*.$$ Let $d$ be an integer coprime to $p$. I already know that

1. $\ker f$ is an abelian $p$-group.
2. raising to $d$-th power is an automorphism on $\ker f$.

My question is why $a\in A$ is a d-th power mod $P^n$ if and only if $a$ is a d-th power mod $P$? I know this should be true but can't find a rigorous statement.

($a$ is a $d$-th power mod $f$ means the equation $x^d \equiv a$ mod $f$ is solvable.)

Answer 1

I think the conclusion comes from

$$(A/P^nA)^* \cong \ker f \times (A/PA)^*.$$

If $\bar{a}$ is a $d$-th power in the second factor on the right, and $a=k\times\bar{a}$, we can find $k'$ such that $k'^d=k$ from the automorphism condition.