Why the fixed elements under translation transformation $f(x) \mapsto f(x+1)$ of a field of rational functions must be constants?

Question

Let $F$ be a field, and let $E=F(x)$ be the field of rational functions over $F$. We can see that the mapping $f(x) \mapsto f(x+1)$ is an automorphism of the field $E$. My question is, how to prove that the fixed elements under this automorphism are constant? I tried to prove it by comparing coefficients, but then I got stuck in complicate calculations and didn't get any results.

Answer 1

Thanks for the hints in the comments. I think I've come up with a proof, but I'm not sure if it's correct.

I will only consider the case when $\text{char} F = 0$, as the proposition seems not to hold when $\text{char} F > 0$. For instance, when $\text{char} F = 2$, $x^2 - x + 1$ appears to be a fixed element under the automorphism.

Let $f(x) = \frac{p(x)}{q(x)}$ be a fixed element under the isomorphism, where $p(x), q(x)\in F[x]$, and $q(x) \neq 0$. Then $f(x+1)=f(x)$, or equivalently, $p(x+1)q(x) = p(x)q(x+1)$. We can also assume $p(x)≠0$, for otherwise we can directly conclude that $f(x) = 0$ is a constant.

Since the polynomial ring over any field is a unique factorization domain, we can always cancel out the common factors of $p(x)$ and $q(x)$. Hence, we can always assume that $p(x)$ and $q(x)$ are relatively prime.

Write $p(x) = c_p p_1(x) p_2(x) \ldots p_r(x)$, $q(x) = c_{q}q_{1}(x)q_{2}(x) \ldots q_{s}(x)$, where each $p_i(x)$ and $q_{j}(x)$ is a monic irreducible polynomial. From $p(x+1)q(x) = p(x)q(x+1)$ we have $$\begin{aligned} &p_1(x+1) p_2(x+1) \ldots p_r(x+1) q_{1}(x)q_{2}(x) \ldots q_{s}(x)\\&= p_1(x) p_2(x) \ldots p_r(x) q_{1}(x+1)q_{2}(x+1) \ldots q_{s}(x+1),\end{aligned}$$and both the left-hand side and the right-hand side are factorizations of this product into irreducible factors. Since $p_i(x)$ and $q_{j}(x)$ are all monic, these two factorizations are strictly equal. Pick any $p_{i}(x)$ from the right-hand side, it either equals some $p_{j}(x+1)$ or some $q_{k}(x)$. But the latter situation is not possible, as it would contradict the fact that $p(x)$ and $q(x)$ are relatively prime. Therefore, the factors $p_1(x) \ldots p_r(x)$ can only correspond one-to-one with the factors $p_1(x+1) \ldots p_r(x+1)$. Consequently, their products are equal, implying that $p(x+1) = p(x)$, and thus $q(x+1)=q(x)$.

Let $p(x) = a_m x^m + a_{m-1} x^{m-1} + \ldots + a_0$, where $a_m \neq 0$. Suppose $\text{deg} p(x) = m \geq 1$, we have $$\begin{aligned} p(x+1) &= a_m (x+1)^m + a_{m-1} (x+1)^{m-1} + \ldots + a_0 \\&= a_m x^m + (ma_m + a_{m-1}) x^{m-1} + \ldots + a_0 \\=p(x)& = a_m x^m + a_{m-1} x^{m-1} + \ldots + a_0.\end{aligned}$$Since $ma_m + a_{m-1} \neq a_{m-1}$, it follows that $p(x+1) \neq p(x)$, contradicting $p(x+1) = p(x)$. Therefore, $\text{deg} p(x) = 0$ and thus $p(x)$ is a constant. Similarly, $q(x)$ is also a constant, hence $f(x) = \frac{p(x)}{q(x)}$ is a constant.

In conclusion, $f(x)$ must be a constant.

Answer 2

Let $f/g \in F(X)$ be fixed by $X \mapsto X+1$. We may assume that $f,g$ are coprime in $F[X]$. We have $f(X)/g(X) = f(X+1)/g(X+1)$ in $F(X)$, hence $g(X+1) f(X) = g(X) f(X+1)$ in $F[X]$. Since $f,g$ are coprime, it follows that $f(X)$ divides $f(X+1)$ in $F[X]$. Since both polynomials have the same degree and the same lead coefficient, it follows $f(X) = f(X+1)$. The same argument shows $g(X) = g(X+1)$.

So, if $R := \{f \in K[X] : f(X+1)=f(X)\}$, then $R$ is a subalgebra of $F[X]$ and the fixed field is $Q(R)$. We only need to compute $R$.

Claim. When $\mathrm{char}(F)=0$, then $R = F$. If $\mathrm{char}(R) = p > 0$, then $R = F[X^p - X]$.

Proof: Assume $\mathrm{char}(F)=0$ and $f \in R$, we want to prove $f \in F$. If $f \notin F$, then $f$ has a root $\alpha \in \overline{F}$. Since $f(X)=f(X+1)$, then $\alpha+1$ is a root as well. It follows hat $\alpha+n$ is a root for all $n \in \mathbb{N}$, and these are pairwise distinct. The only polnomial with infinitely many roots is $0$, so $f = 0$ and we are done.

Assume $\mathrm{char}(F)=p> 0$. Since $(X+1)^p = X^p+1$, we see that $X^p-X \in R$. Conversely, let $f \in R$, we need to show that $f$ is a polynomial in $X^p - X$. We make an induction on $\deg(f)$, the cases $\deg(f) \leq 0$ being clear. So let $\deg(f) \geq 1$. Write $f = g \cdot (X^p - X) + h$ with $\deg(h) < p$. Then $h \in R$, since $f(X)=f(X+1)=g(X+1) \cdot (X^p - X) + h(X+1)$ and the remainder is unique. If $h \notin F$, the proof in the first case shows that $h$ has at least $p$ roots in $\overline{F}$, hence is of degree $\geq p$, a contradiction. Hence, $h \in F$, and thus $\deg(g) = \deg(f) - p$. We also have $g \in R$ because $g=(f-h)/(X^p-X) \in F(X)$ is fixed by $X \mapsto X+1$. By induction hypothesis, $g \in F[X^p-X]$. Hence, also $f \in F[X^p-X]$. $~\square$