Show that if the field of $p^a$ elements is a subfield of the field of $p^b$ elements if and only if $a\vert b$.

Question: Show that if the field of $p^a$ elements is a subfield of the field of $p^b$ elements if and only if $a\vert b$.

If $|F|=p^a$, then any nonzero element of $F$ is a root of $x^{p^a}-x$. Similarly, any nonzero element of a field of $p^b$ elements is a root of $x^{p^b}-x$. So, if a field of $p^a$ elements is a subfield of a field of $p^b$ elements, then I suppose we can say that $b>a$, and any nonzero element in a field of $p^a$ elements is also a root of $x^{p^b}-x$... but I don't believe this would imply that $a\vert b$. Moreover, if $a|b$, then I was trying to just prove definition of a subfield, but that wasn't getting me anywhere. Any help is greatly appreciated! Thak you.

First, you are aware that an irreducible polynomial $f(x)$ of degree $a$ divides $x^{p^b}-x$ in $\mathbb{F}_p[x]$ iff $a \ | \ b$, right:

A question based on a n irreducible polynomial dividing $x^{q^n}-x$ in finite fields

So let $K$ be a field of cardinality $p^b$. Then for some irreducible polynomial $h(x) \in \mathbb{F}_p[x]$ of degree $b$, the field $K$ can be written as $K = \mathbb{F}_p(\alpha)$ where $\alpha$ is a root of $h(x)$, and every element $y \in K$ can be written $y=y(\alpha)$, where $y(x)$ is a polynomial in $\mathbb{F}_p[x]$ of degree $<b$. Now, note that all $p^b$ elements $y(\alpha)$ of $K$ satisfy $y^{p^b}(\alpha) -y(\alpha) = 0$, or equivalently, $$h(\alpha) \ | \ (y^{p^b}(\alpha) -y(\alpha)).$$ Thus for every $y(\alpha)\in K$ there is an irreducible polynomial $f(x) \in \mathbb{F}_p[x]$ that divides the polynomial $x^{p^b}-x$, that has $y(\alpha)$ as a root i.e., $h(\alpha) \ | \ f(y(\alpha))$. Note also that $p^b$ is the degree of the polynomial $x^{p^b}-x$, so the polynomial $x^{p^b}-x$ must factor completely in $K$. So conversely, for every irreducible polynomial $f(x) \in \mathbb{F}_p[x]$ that divides the polynomial $x^{p^b}-x$, there is a $y(\alpha) \in K$ such that $y(\alpha)$ is a root of $f$, i.e., $h(\alpha) | f(y(\alpha))$.

Now, let $f$ be an irreducible polynomial of degree $a$. Then as already noted near the top, $f(x)$ divides the polynomial $x^{p^b}-x$ in $\mathbb{F}_p[x]$. So then there is a $y_f(\alpha) \in K$ such that $y_f(\alpha)$ is a root of $f$. Note however as $y_f(\alpha)$ is a root of an irreducible polynomial, namely, $f$ in $\mathbb{F}_p[x]$ of degree $a$. So $y_f(\alpha)$ generates a field $F$ of precisely $p^a$ elements. As $y(\alpha)$ is in $K$, it follows that $F$ is a subfield in $K$. So if $a \ | \ b$, there is a subfield $F$ of $p^a$ elements inside a subfield $K$ of $p^b$ elements.

Note the converse by noting that every field $F$ of $p^a$ elements can be written $F=\mathbb{F}_p(\gamma)$ for some $\gamma$ that is the root of an irreducible polynomial $f(x) \in \mathbb{F}_p[x]$ of degree $a$. As also noted above then, if $\gamma$ is in $K$, then $f(x)$ must divide $x^{p^b}-x$, but as noted in the top paragraph, this does not happen unless the degree $a$ of $f(x)$, divides $b$. So here, only if $a \ | \ b$ is there a subfield of $p^a$ elements inside a subfield $K$ of $p^b$ elements.

***For the only if direction, one could also note that a field $F$ of $p^a$ elements satisfies $F^×=F \setminus \{0\}$ and is a group w exactly $p^a-1$ elements, while a field $K$ of $p^b$ elements satisfies $K^×=K \setminus \{0\}$ and is a group with $p^b-1$ elements. So $p^a-1$ must divide $p^b-1$, and this happens only if $a$ divides $b$.

For one direction, let $F_a$ be a subfield of $F_b$ with $|F_a|=p^a$ and $|F_b|=p^b.$ Let $G_a=F_a\setminus \{0\}$ and $G_b=F_b$ \ $\{0\}.$ Then $G_a$ and $G_b$ are groups under the fields' multiplication, and $G_a$ is a subgroup of $G_b.$ Therefore $$p^a-1=|G_a|\text { divides }|G_b|=p^b-1.$$ Therefore $\sum_{i=0}^{a-1}p^i=(p^a-1)/(p-1)$ divides $(p^b-1)/(p-1)=\sum_{j=0}^{b-1}p^j.$

Now let $b=an+c$ where $n\in \Bbb N$ and $0\le c<a.$

If $c=0$ then $a|b.$

But if $c\ne 0$ then the positive integer $$\sum_{j=0}^{b-1}p^j\,-(\sum_{i=0}^{a-1}p^i)(\sum_{k=0}^{n-1}p^{ak})=\sum_{\ell=0}^{c-1} p^{an+\ell}=p^{an}\sum_{\ell=0}^{c-1}p^{\ell}$$ would be divisible by $\sum_{i=0}^{a-1}p^i.$ But this is impossible because $\sum_{i=0}^{a-1}p^i$ is co-prime to $p^{an}$ and $0<\sum_{\ell=0}^{c-1}p^{\ell}<\sum_{i=0}^{a-1}p^i.$