For a field extension $E/F$ and $|E| = p^r, |F| = p,$ why is $E$ a degree r extension over $F?$

Question

I am currently wondering the above question. So intuitively $E$ should be a dimension $r$ vector space over $F,$ but why exactly? My book seems to just assume this. I feel like it should be obvious but it isn't to me.

Answer 1

Since $E$ is an extension of $F$ and is a finite field, it follows that $E$ is an $F$-vector space and that the degree $[E:F]$ is finite.

If $d=[E:F]$, then $E$ has an $F$-basis $\{x_1,\dots,x_d\}$, hence $E$ consists of all linear combinations of the form $$ c_1x_1+\dots+c_dx_d $$ where $c_j\in F$. Since there are $p$ choices for each $c_j$, it follows that $E$ has $p^d$ elements. Since $|E|=p^r$, we get $d=r$.