If $[F : F_p] = n$, does $F$ have $p^n$ elements?
My book seems to be implying that this is true but I'm not sure why.
Thanks!
If $[F : F_p] = n$, does $F$ have $p^n$ elements?
My book seems to be implying that this is true but I'm not sure why.
Thanks!
Comment converted to answer:
Yes it is true:
If $K$ is a finite field with $q$ elements, and $F$ is an extension field of degree $[F:K] = n$, then $F$ is a finite field with $q^n$ elements.
To see this, note that $[F:K] = n$ means that $F$ is an $n$-dimensional vector space over $K$, and hence, $F$ exhibits an $n$-dimensional basis over $K$. Calling that basis $x_1, \dots , x_n$, then everything in $F$ can be written uniquely as $a_1 x_1 + \dots + a_n x_n$ (where $a_i \in K$). Since there are $q$ choices for each $a_i$, it follows $F$ must have $q^n$ elements.