# Notation
Let $p$ be a prime, $F$ be the field of order $p$, $K$ be an extension of $F$, and $\phi\colon K \to K$ be the Frobenius automorphism. The Frobenius automorphism can be extended to an automorphism of $K[X]$ by coefficient-wise application. Then for a polynomial $f\in K[X]$ and $x\in K$, $\phi(f(x)) = \phi(f)(\phi(x))$.

Let us further assume that the degree $d$ of $K/F$ is even, so that $K^*$ has a subgroup $S$ of order $p+1$. We will often use the property that $\phi(x)=x^{-1}$ for all $x\in S$.

Finally, let $H$ be a subgroup of $S$ of order $n$, which is also assumed to be even.

# Case $d=2$
For simplicity we first start with the case $d=2$, i.e., the extension $K/F$ is quadratic.

Let $f\in K[X]$ be a polynomial of degree $<n/2$ such that $f(0)=0$. The map $\mathring{f}(x)= f(x) + \phi(f(x))$ has image in $F$ since $\phi(\mathring{f}(x)) = \mathring{f}(x)$. In general $\mathring{f}$ has high degree, but over $H$ we can transform it into a low-degree polynomial. For $x\in H$, it can be rewritten as
$$
\mathring{f}(x) = f(x) + \phi(f)(\phi(x)) = f(x) + \phi(f)(x^{-1})x^n
$$
where we used $\phi(x)=x^{-1}, x^n=1$ on $H$. This expression is a polynomial $P(X)$ of degree $<n$ satisfying $P(x)=\mathring{f}(x)\in H$ for all $x \in H$ and $P(0)=0$.

This transformation gives a linear map
$$
\{f\in K[X]^{<n/2} \mid f(0)=0 \} \to \{P\in K[X]^{<n}\mid P(H)\subseteq F, P(0)=0\}.
$$
We observe that the map is injective and that both vector spaces have the same dimension over $F$. Therefore the map is an isomorphism. This means that every polynomial $P\in K[X]^{<n}$ with $P(H)\subseteq F, P(0)=0$ has the form
$$
P(X) = f(X) + \phi(f)(X^{-1})X^n
$$
for some $f\in K[X]^{<n/2}$ with $f(0)=0$.

We now consider the polynomial $P(\tau X)$ for some $\tau \in S-H$:
$$
P(\tau X) = f(\tau X) + \phi(f)(\tau^{-1}X^{-1})\tau^n X^n = f(\tau X) + \phi(f)(\phi(\tau)X^{-1})\tau^n X^n,
$$
where we used $\phi(\tau)=\tau^{-1}$. We define $f_\tau(X) = f(\tau X)\tau^{-n/2}$ and observe that
$$
P(\tau X)\tau^{-n/2} = f_\tau(X) + \phi(f_\tau)(X^{-1})X^n.
$$
Since $f_\tau(X)$ is a polynomial of degree $<n/2$ with $f_\tau(0)=0$, we conclude that $P(\tau X)\tau^{-n/2}$ is in $\{P\in K[X]^{<n}\mid P(H)\subseteq F, P(0)=0\}$. We thus recover the theorem of [HNL23] showing that $P(\tau H) \subseteq \tau^{n/2} F$.

# Case $d$ even WORK IN PROGRESS
We now consider the general case where $d$ is even. For $i\mid d$ we define
$$F_i=\{x\in K \mid \phi^i(x)=x\}=\{x\in K \mid x^{p^i-1}=1\} \cup 0.$$
In particular, $F_1=F$, $F_d=K$, and $F_i$ is the extension of $F$ of degree $i$, i.e., we have a tower of extensions
$$
F - F_i - K.
$$
For $i\mid \frac d 2$, we have $p^i + 1 \mid p^d -1$ and thus there exists a subgroup
$$
S_i = \{x\in K \mid \phi^i(x)=x^{-1}\}=\{x\in K \mid x^{p^i+1}=1\}
$$
of $K^*$.

# Polynomials $H \to F$ WORK IN PROGRESS
Let $f\in K[X]$ be a polynomial of degree $<n/2$. The map $\mathring{f}(x)=\sum_{i=0}^{d-1} \phi^i(f(x))$ has image in $F$ since $\phi(\mathring{f}(x)) = \mathring{f}(x)$. In general $\mathring{f}$ has high degree, but over $H$ we can transform it into a low-degree polynomial. For $x\in H$, it can be rewritten as
$$
\mathring{f}(x) = \sum_{i=0}^{d-1} \phi^i(f)\phi^i(x) = \sum_{i=0}^{d/2-1} \phi^{2i}(f)(x) + \left(\sum_{i=0}^{d/2-1} \phi^{2i+1}(f)(x^{-1})\right)x^n
$$
where we've used $\phi(x)=x^{-1}$, $\phi^2(x)=x$, and $x^n=1$ on $H$. This expression is a polynomial $P(x)$ of degree $<n$ satisfying $P(x)=\mathring{f}(x)$ for all