Why do we have that $\Bbb F_{p^d}$ is a sub-field of $\Bbb F_{p^n}$ iff $d\mid n$?

Question

Why do we have that $\Bbb F_{p^d}$ is a sub-field of $\Bbb F_{p^n}$ iff $d\mid n$? Why is any field of cardinal $p^k$ with $k\leq n$ not a sub-field? I thought what is important is for the cardinal of the sub-field to divide the cardinal of the field.

Answer 1

Suppose $ \mathbb{F}_p\subset\mathbb{F}_{p^k}\subset\mathbb{F}_{p^d} $, Since $ [\mathbb{F}_{p^d}:\mathbb{F}_p]=d $, then $$ d=[\mathbb{F}_{p^d}:\mathbb{F}_{p}]=[\mathbb{F}_{p^d}:\mathbb{F}_{p^k}][\mathbb{F}_{p^k}:\mathbb{F}_p]=[\mathbb{F}_{p^d}:\mathbb{F}_{p^r}]\cdot k ,$$ hence $ k|d $.

To prove the inverse, suppose $ k|d $, since elements of $ \mathbb{F}_{p^d} $ are the roots of $ x^{p^d}-x $ over $ \mathbb{F}_{p} $, and $ (x^{p^k}-x)|(x^{p^d}-x) $, then we claim the roots of $ x^{p^k}-x $ form a subfield $ \mathbb{F}_{p^k} $ of $ \mathbb{F}_{p^d} $. Now prove the claim:

If $ x, y $ are roots of $ x^{p^k}-x $, then $$ (x+y)^{p^k}-(x+y)=x^{p^k}+y^{p^k}-x-y=x+y-x-y=0 ;$$ $$ (xy)^{p^k}-xy=x^{p^k}y^{p^k}-xy=xy-xy=0 ;$$ and $ 0, 1 $ are roots of $ x^{p^k}-x $. Therefore the roots of $ x^{p^k}-x $ form a subfield. Since $ (x^{p^k}-x)'=-1 $ is relatively prime to $ x^{p^k}-x $ which implies there is no multiple root for $ x^{p^k}-x $, and we are done.

Answer 2

The elements of a finite field of size $p^k$ are the zeroes of $x^{p^k }-x$ in some extension of $\mathbb{F}_p$.

So we have that $\mathbb{F}_{p^k} \subseteq \mathbb{F}_{p^n} \iff x^{p^k} - x \mid x^{p^n} - x$. Can you see why $x^{p^k} - x \mid x^{p^n} - x \iff k \mid n$?

To see the latter note that $x^{p^k} - x$, has a root $\alpha$ s.t. $\alpha^{p^k-1} = 1$, and $p^k-1$ is the smallest such integer. Then we have that $\alpha^{p^n-1} = 1$ and so $p^k-1\mid p^n -1$. Now $\gcd(p^k - 1, p^n -1) = p^{\gcd(n,k)}-1$, so we must have $\gcd(n,k) = k$, which implies that $k \mid n$

For the other direction suppose $k \mid n$, then $x^{p^n} - x = x(x^{p^{mk}-1} - 1)$. Now $p^k-1 \mid p^{mk}-1$ and so hence $x^{p^k-1} - 1 \mid x^{p^{km}-1} - 1$, so $x^{p^k} - x \mid x^{p^n} - x$

Answer 3

A totally different argument (the vector space dimension is the real go to argument): If $\Bbb F_{p^d}$ is a subfield of $\Bbb F_{p^n}$, then the multiplicative group $\Bbb F_{p^d}^\times $ is a subgroup of $\Bbb F_{p^n}^\times$. In particular, $$ p^d-1\mid p^n-1.$$ But then also $p^d-1\mid p^n-1-p^{n-d}(p^d-1)=p^{n-d}-1$ and by repeating the process, $p^d-1\mid p^{n\bmod d}-1$. As $n\bmod d<d$, this is only possibile if $n\bmod d=0$.