Degree of finite field extension by adjoining root of minimal polynomial

Question

Edited: Let $E$ be a finite degree extension field of $F_q$ where $q = p^n$, $p$ is some prime and $n$ is some integer, and $r \in E$. Let $g(x)$ be the minimal polynomial of $r$ over the prime field $F_p$ of degree $d$. What is the number of elements of $F_p(r)$? The book I am reading claims it is $p^d$, but gives no justification to this and I am having trouble seeing this. Thanks for the help!

Answer 1

$F_p(r)=\{\,a_0+a_1r+\cdots+a_{d-1}r^{d-1}:a_i\in F_p\,\}$, and therefore the number of elements of $F_p(r)$ is $p^d$.

First, since $F_p(r)$ is a field containing $F_p$ and $r$, it must contain every polynomial in $r$ with coefficients in $F_p$, so it must contain all the elements I've listed. Second, the elements I've listed must all be distinct, since if any two were equal there would be a nonzero polynomial of degree at most $d-1$ vaninshing at $r$, contrary to the assumption that the minimal polynomial for $r$ has degree $d$. Third, no polynomial of higher degree is needed; since $r^d$ can be written as a polynomial in $r$ of degree less than $d$, so can any higher power of $r$, and thus any polynomial in $r$ of higher degree.

It remains to show that the set I've given is closed under taking multiplicative inverses. Certainly the set is a vector space over $F_p$ of degree $d$, so every $\beta$ in it satisfies an $F_p$-polynomial of degree at most $d$, and if $c_0+c_1\beta+\cdots+c_d\beta^d=0$, then $\beta\gamma=1$, where $\gamma=-c_0^{-1}(c_1+\cdots+c_d\beta^{d-1})$, which is in the set I've given. .