What is the Fixed Subfield of this Group?

Question

Suppose $\mathbb{K}$ is a field of characteristic zero. Let $G$ be the subgroup of the Galois group of $\mathbb{K} \subset \mathbb{K}(x)$ (the field of rational functions over $\mathbb{K}$ with indeterminate $x$) generated by the automorphism $x \rightarrow x+1$. Find the fixed subfield of $\mathbb{K}(x)$ corresponding to $G$.

My attempt: During the test, I went on a tangent trying to show that the fixed subfield is $\mathbb{K}$ because a rational function being periodic seemed weird to me; however I could not prove it and I'm sure the Galois group isn't generated by $x \rightarrow x+1$. Alternatively, I tried to find a rational function that is fixed by this automorphism to no avail.

Context: This kind of question was on a HW assignment and a test in my Algebra course (that's an indicator it might be on the Qualifying Exam). Any help would be appreciated, Thank you.

$\underline{Edit:}$ I'm seeing answers that result in the fixed field being $\mathbb{K}$ (as I had guessed). Does this imply that the $Gal(\mathbb{K}(x) / \mathbb{K}) \cong <\beta >$ by Gaois Correspondence? Where $\beta$ takes $x \rightarrow x+1$.

Answer 1

Suppose that $f(x) = \frac{p(x)}{q(x)}$ is fixed, with $p,q$ nonzero relatively prime polynomials in $K[x]$. Then $f(x) = f(x+1)$ implies $p(x) q(x+1) = p(x+1) q(x)$. By the assumption that $p(x)$ and $q(x)$ are relatively prime, we see that $p(x) \mid p(x) q(x+1) = p(x+1) q(x)$ implies $p(x) \mid p(x+1)$. Similarly, since $p(x+1)$ and $q(x+1)$ are also relatively prime, we also have $p(x+1) \mid p(x)$.

Therefore, $p(x+1)$ must be a unit of $K[x]$ times $p(x)$. Comparing leading coefficients of both sides, that unit must be 1. Now, $\Delta p(x) = p(x+1) - p(x) = 0$; but also, if $p(x) = a_n x^n + \cdots + a_0$ with $a_n \ne 0$, then the leading term of $\Delta p(x)$ is $n a_n x^{n-1}$. This implies that we must have $n = 0$, i.e. $p(x)$ is a constant polynomial.

A similar argument shows $q(x)$ must also be a constant polynomial, so $f(x) = \frac{p(x)}{q(x)}$ is a constant. (The only case this argument does not cover is $f(x) = 0$, which is also obviously constant.)

Answer 2

Let $\Phi(x)\in \mathbb K(x)$ denote a rational function invariant under $x \mapsto x+1$. Choose $a\in \mathbb K$ for which $\Phi(a)$ exists (always possible since $\mathbb K$ is infinite) and let $k=\Phi(a)$ and let $\Psi(x)=\Phi(x)-k$.

Then $\Psi(x)\in \mathbb K(x)$ has infinitely many zeroes, as $\Psi(a+n)=0\;\forall n\in \mathbb N$. It follows that the numerator of $\Psi(x)$ is identically $0$ which implies that $\Phi(x)=k \;\forall x\in \mathbb K$.

(Note: characteristic $0$ was needed to get infinitely many zeroes. If the characteristic were $p$ then this argument would only yield $p$ zeroes.)