# 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).

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$\mathbb{F}_{p^k} \neq \mathbb{Z}/p^k\mathbb{Z}$, so your example is not correct. $1+2 = 0$ in $\mathbb{F}_9$. $\mathbb{F}_{p^k}$ is defined as the unique (up to isomorphism) field extension of degree $k$ of $\mathbb{F}_p$ (which can be proven to exist using irreducible polynomials of degree $k$). –  KevinDL Jan 8 '12 at 9:54
The thing that you need to remember about field extensions like $L:K$ is that when you say that $K$ is a subfield of $L$, you are really saying that $L$ contains a field isomorphic to $K$. In other words, $K$ embeds in $L$ with some injective homomorphism. –  Aleks Vlasev Jan 8 '12 at 10:02
Dear @KevinDL: I think you're answering the question precisely as it is asked, and suggest you upgrade your comment to an answer. –  Pierre-Yves Gaillard Jan 8 '12 at 10:05
$\mathbb{F}_9$ does not have the same addition and multiplication as $\mathbb{Z}_9$. As a set, $\mathbb{F}_9$ might be explicitly described like this: $\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}$ subject to the reduction $x^2=2$. Surely you have learned that fields do not have zero-divisors. In $\mathbb{Z}_9$, $3\cdot3=0$, so that should make it clear that $\mathbb{F}_9$ is not $\mathbb{Z}_9$. –  alex.jordan Jan 8 '12 at 10:06
OP is clearly confused about the definition of $\mathbb F_{p^k}$ so I would suggest she study that first (with comments by KevinDL and alex.jordan as hints). Then the question will disappear. –  Marc van Leeuwen Jan 8 '12 at 10:26

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 $\mathbb{F}_p\subset\mathbb{F}_{p^k}$.

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$\mathbb{F}_{p^m} \subseteq \mathbb{F}_{p^n}$ 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.

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I think 1) can be justified without using primitive elements: if you know that $[ \mathbb{F_{p^k}}: \mathbb{Z_p}] \forall k \in \mathbb{N}$ then the formula for degrees yields: $[ \mathbb{F_{p^n}}: \mathbb{Z_p}]=[ \mathbb{F_{p^n}}: \mathbb{F_{p^m}}][ \mathbb{F_{p^m}}: \mathbb{Z_p}]$, therefore $m=[ \mathbb{F_{p^m}}: \mathbb{Z_p}]$ divides $n=[ \mathbb{F_{p^n}}: \mathbb{Z_p}]$. –  Emilio Ferrucci Jan 8 '12 at 12:58
That's right, it's obvious. –  Kraw Jan 8 '12 at 13:26