# If $[F : F_p] = n$, does $F$ have $p^n$ elements?

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!

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Yes it is true. In fact, if $K$ is a field with $q$ elements, and $[F:K] = n$, then $F$ is a field with $q^n$ elements. To see why this true, what does $[F:F_p] = n$, mean exactly? – JavaMan Apr 10 '11 at 23:42
$F$ is an $n$-dimensional vector space over $F_p$...so there are $n$ vectors in $F$ that are not in $p$. So everything in $F$ can be written as $f_1v_1 + f_2v_2 + ... f_nv_n$. ...Okay, now I feel stupid. – badatmath Apr 10 '11 at 23:45
There is NO reason to feel stupid (or to think you are "bad" at math)! Talking things out is helpful... (: – JavaMan Apr 10 '11 at 23:46
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.