Notation

Let

$p$ be a prime,

$F$ be the field of order

$p$,

$K$ be an extension of

$F$, and

$\varphi :K\to K$ 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

$x\in K$,

$\varphi (f(x))=\varphi (f)(\varphi (x))$.

Let us further assume that the degree

$d$ of

$K/F$ is even, so that

$K^{\ast }$ has a subgroup

$S$ of order

$p+1$. We will often use the property that

$\varphi (x)=x^{-1}$ 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

$\mathring{f}(x)=f(x)+\varphi (f(x))$ has image in

$F$ since

$\varphi (\mathring{f}(x))=\mathring{f}(x)$. 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

$$\mathring{f}(x)=f(x)+\varphi (f)(\varphi (x))=f(x)+\varphi (f)(x^{-1})x^n$$
where we used

$\varphi (x)=x^{-1},x^n=1$ 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

$$P(X)=f(X)+\varphi (f)(X^{-1})X^n$$
for some

$f\in K[X{]}^{<n/2}$ with

$f(0)=0$.

We now consider the polynomial

$P(\tau X)$ for some

$\tau \in S-H$:

$$P(\tau X)=f(\tau X)+\varphi (f)({\tau }^{-1}{X}^{-1}){\tau }^n{X}^n=f(\tau X)+\varphi (f)(\varphi (\tau )X^{-1}){\tau }^n{X}^n,$$
where we used

$\varphi (\tau )={\tau }^{-1}$. We define

$f_{\tau }(X)=f(\tau X){\tau }^{-n/2}$ and observe that

$$P(\tau X){\tau }^{-n/2}=f_{\tau }(X)+\varphi (f_{\tau })(X^{-1})X^n.$$
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

$$F_i=\{x\in K\mid {\varphi }^i(x)=x\}=\{x\in K\mid x^{p^i-1}=1\}\cup 0.$$
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

$$S_i=\{x\in K\mid {\varphi }^i(x)=x^{-1}\}=\{x\in K\mid x^{p^i+1}=1\}$$
of

$K^{\ast }$.

Polynomials

$H\to F$ WORK IN PROGRESS

Let

$f\in K[X]$ be a polynomial of degree

$<n/2$. The map

$\mathring{f}(x)=\sum _{i=0}^{d-1}{\varphi }^i(f(x))$ has image in

$F$ since

$\varphi (\mathring{f}(x))=\mathring{f}(x)$. 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

$$\mathring{f}(x)=\sum _{i=0}^{d-1}{\varphi }^i(f){\varphi }^i(x)=\sum _{i=0}^{d/2-1}{\varphi }^{2i}(f)(x)+(\sum _{i=0}^{d/2-1}{\varphi }^{2i+1}(f)(x^{-1}))x^n$$
where we've used

$\varphi (x)=x^{-1}$,

${\varphi }^2(x)=x$, and

$x^n=1$ on

$H$. This expression is a polynomial

$P(x)$ of degree

$<n$ satisfying

$P(x)=\mathring{f}(x)$ for all