Why is $\mathbb{F}_{p}\subseteq\mathbb{F}_{p^{k}}$ a field extension? In our algebra course our professor said, during a the beginning of
a chapter on field extension, that $\left[\mathbb{F}_{p^{k}}:\mathbb{F}_{p}\right]=k$
(where $p$ is obviously prime). 
My question is: Why can we even assume that $\mathbb{F}_{p}$ is a subfield of $\mathbb{F}_{p^k}$ ?
$\mathbb{F}_{p}$
isn't closed under the restriction of the multiplication in $\mathbb{F}_{p^{k}}$
to $\mathbb{F}_{p}$, so $\mathbb{F}_{p}$ doesn't form a subfield
(for example, for $1,2\in\mathbb{F}_{3}\subseteq\mathbb{F}_{3^{2}}$
we have that $2+1=3\in\mathbb{F}_{3^{2}}$ and not $0$, as we should
obtain in $\mathbb{F}_{3}$), so in my opinion we can't even talk
about $\left[\mathbb{F}_{p^{k}}:\mathbb{F}_{p}\right]$, since we
defined this only for field extension (I realize that we still can
talk about $\left[F:G\right]$, if we define this in a more general
way just for fields $F,G$ such that $F$ is a $G$-vector space;
but in that case I would be annoyed by the slopiness of our course).
 A: Every field has $1$ in it.  In a finite field, $1+1+\cdots+1=n$ must eventually be $0$ for large enough $n$.  The smallest such $n$ cannot be composite or you would have zero-divisors.  If $n$ is a prime other than $p$, then since $\gcd(p^k,n)=1$, we could conclude $1=0$ using $s\ p^k+t\ n=1$.  So $1, 2, 3, \ldots, p-1$ are all inside $\mathbb{F}_{p^k}$ and nonzero, with $p=0$.  These elements add together as you expect.  Further, since multiplication by an integer is just repeated addition, they also multiply like you would expect. And there is your $\mathbb{F}_p\subset\mathbb{F}_{p^k}$.
A: $\mathbb{F}_{p^m} \subseteq \mathbb{F}_{p^n}$ is a field extension $ \iff m|n$
1.) $\mathbb{F}_{p^m} \subseteq \mathbb{F}_{p^n}$ is an extension $ \Rightarrow m|n$
$\mathbb{F}_{p^m}^* = <\alpha>$
$\mathbb{F}_{p^n}^* = <\beta>$
so we have (and this is trivial):
$\mathbb{F}_{p^m} = \mathbb{Z}_p[\alpha]$
$\mathbb{F}_{p^n} = \mathbb{Z}_p[\beta]$
Therefore:
$\mathbb{Z}_p[\alpha] \subseteq \mathbb{Z}_p[\beta]$ is a field extension so 
$\left[ \mathbb{Z}_p[\alpha] :\mathbb{Z}_p \right] | \left[ \mathbb{Z}_p[\beta] :\mathbb{Z}_p \right]$
so: $m | n$
2.) $m|n \Rightarrow \mathbb{F}_{p^m} \subseteq \mathbb{F}_{p^n}$  is an extension
we prove that $\alpha \in \mathbb{Z}_p[\beta]$:
$n = k \times m$ with induction over k we can prove that $\alpha^{p^n} - \alpha = 0$
Infact: if $k = 1$ is trivial because $\alpha^{p^m} - \alpha = 0$ by definition of $\mathbb{F}_{p^m}$
$\alpha^{p^{(k + 1)\times m}}= (\alpha^{p^{k\times m}}) ^ {p^m} = \alpha^{p^m} = \alpha$ 
So we have finished.
