In this wikipedia article state the following :"a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field . In other words, a polynomial a primitive polynomial if its degree is m and it has a root $\alpha$ such that $\{0,1,\alpha,\alpha ^{2},\alpha ^{3},\dots ,\alpha ^{p^{m}-2}\} $ is the entire field".
i don't get how the two definitions are equivalent.
My second question: let say I have a polynomial $P$ of degree $m$ over $\mathbb{F}_{q}$ and $\alpha$ is a root of $P$ in an algebraic closure of $\mathbb{F}_{q}$, and if $\alpha$, $\alpha^{2}$, $\alpha ^{3}$, ... ,$\alpha ^{p^{m}-2}$ are all different from $1$, does it imply that $P$ is irreducible?
For the first question, the degree of the minimal polynomial of any element in a field extension is the degree of the smallest sub-extension that contains it; for a primitive element$~\alpha$ of a degree $m$ extension there are no strict such sub-extensions, so this degree is$~m$. Then any monic polynomial of degree$~m$ having$~\alpha$ as root must be that minimal polynomial, hence irreducible.
As for the second question (if we exclude the obvious counterexample $\alpha=0$), if you have a nonzero root$~\alpha$ of a degree$~m$ polynomial over $\Bbb F_q$, you know that its minimal polynomial has degree at most$~m$, so $\alpha$ lives in a multiplicative group of order at most $q^m-1$ (of the field $\Bbb F_q[\alpha]$). If that number is indeed the multiplicative order of$~\alpha$ as your hypothesis implies (so $\alpha$ generates said multiplicative group) then all inequalities above must be equalities, an in particular your original polynomial is the minimal polynomial of$~\alpha$, hence irreducible.