Affine subspace of finite field intersecting prime subfield Consider a finite field $\mathbb F_{p^n}$ and let $V = \{ y -y^p \mid y \in \mathbb F_{p^n} \} \subset \mathbb F_{p^n}$. It is not hard to see that $V$ is an $(n-1)$-dimensional vector space over $\mathbb F_p$. Now let $x \in \mathbb F_{p^n}$ and consider the affine subspace $x + V$ and its intersection with the prime subfield: $(x+V) \cap \mathbb F_{p}$.
After some computations I came to believe that $(x + V) \cap \mathbb F_{p}$ is non-empty for all $x \in \mathbb F_{p^n}$ if and only if $n$ is coprime to $p$ (and in this case $|(x+V) \cap \mathbb F_p| = 1$). Can you give me a hint what a proof might encompass? Or where I could look this up?
 A: The following is the first way of proving this that occurred to me.
You may be familiar with the trace function
$$
tr:\Bbb{F}_{p^n}\to\Bbb{F}_p, x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}.
$$
Its following properties are relevant here:


*

*We have $tr(x)^p=tr(x)$ for all $x\in\Bbb{F}_{p^n}$. This follows from the fact we can raise to power $p$ term-by-term. This implies that, indeed, $tr(x)\in\Bbb{F}_p$.

*We have $tr(x+y)=tr(x)+tr(y)$ for all $x,y\in\Bbb{F}_{p^n}$. This is because the Frobenius automorphism and its iterates are homomorphisms of additive groups. Consequently $tr$ is linear over $\Bbb{F}_p$.

*We have $tr(x^p)=tr(x)$ (see item !.)

*If $x\in\Bbb{F}_p$ is an element of the prime field, then $x^{p^i}=x$ for all $i$. Consequently $tr(x)=n\cdot x$.


Item 3. implies that your subspace $V$ is contained in the kernel of the trace map. The polynomial formula of $tr$ shows that $\operatorname{ker}(tr)$ cannot have more than $p^{n-1}$ elements. Therefore
$$
V=\operatorname{ker}(tr).
$$
Item 4. implies that $\Bbb{F}_p$ is contained in the kernel of the trace if and only if $p\mid n$. This allows us to conclude:


*

*If $p\nmid n$, then $\Bbb{F}_p\cap V=\{0\}$ and, as the dimensions add up correctly,
$$\Bbb{F}_{p^n}=V\oplus\Bbb{F}_p$$
where the direct sum is that of vector subspaces over $\Bbb{F}_p$. The observation that every coset of $V$ contains a unique element from $\Bbb{F}_p$ follows from this.

*On the other hand, if $p\mid n$, then $\Bbb{F}_p\subset V$.

