Elements in extension field

Question

Suppose $\alpha\in\text{GF}(q^n)\setminus\text{GF}(q)$. Then there exists an irreducible polynomial $f(x)\in\text{GF}(q)[x]$ such that $f(\alpha)=0$. My question is that whether this $f(x)$ is unique for each $\alpha$ ? Is there any well-known result on this ?

Answer 1

Up to multiplication by a nonzero element of $GF(q)$, it's unique. Fix some $0\ne f(x)\in GF(q)[x]$ with minimal degree such that $f(\alpha)=0$. Since $f(x)$ has minimal degree, it must be irreducible. Then, given any other polynomial $g(x)\in GF(q)[x]$ with $g(\alpha)=0$, you can divide $g(x)$ by $f(x)$ to get polynomials $q(x)$ and $r(x)$ in $GF(q)[x]$ with $$ g(x)=q(x)f(x)+r(x), \ \ \ \ \deg r(x) < \deg f(x). $$ Then, since $g(\alpha)=f(\alpha)=0$, $r(\alpha)=0$, so by minimality of $\deg f(x)$, $r(x)$ must be the zero polynomial. This proves that $f(x)$ divides $g(x)$, so, if $g(x)$ is also irreducible, $f(x)$ and $g(x)$ must be nonzero constant multiples of each other.

This is a standard result that works for any $\alpha$ algebraic over any field $K$ just as well as for $\alpha$ algebraic over $GF(q)$.