Minimal polynomials
Let $F_q$ be a subfield of $F_r$ . For an element $α$ of $F_r$ , we are interested in nonzero polynomials $f (x) \in Fq [x]$ of the least degree such that $f (α) = 0$.
Definition 3.4.1:
A minimal polynomial of an element $α ∈ F_{q^m}$ with respect to $F_q$ is a nonzero monic polynomial $f (x)$ of the least degree in $F_q [x]$ such that $f (α) = 0.$
Example $3.4.2$: Let $α$ be a root of the polynomial $1 + x + x^2 ∈ F2[x]$. It is clear that the two linear polynomials $x$ and $1+ x$ are not minimal polynomials of $α$. Therefore, $1 + x + x^2$ is a minimal polynomial of $α$. Since $1+(1+α)+(1+α)2 = 1+1+α +1+α2 = 1+α +α2 = 0$ and $1 + α$ is not a root of $x$ or $1 + x, 1 + x + x^2$ is also a minimal polynomial of $1 + α.$
Hi :)
I have an exposition in a few days on this tema, I have already seen this tema in ring theory, but now I can not use some properties ..
I could explain a little this, I know that $F_q$ is field of gallois but that means $F_{q^m}$
$\mathbf F_{q^m}$ is the unique (within a given algebraic closure of the prime subfield $\mathbf F_p$ of $\mathbf F_q$) extension field of degree $m$ of $\mathbf F_q$.
This field is isomorphic to $\mathbf F_q[X]/(f(X))$, where $f(X)$ is any irreducible polynomial in $\mathbf F_q[X]$.
Further, on can prove $\mathbf F_{q^m}\subset \mathbf F_{q^n}$ if and only if $m\mid n$.