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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 $x \rightarrow x+1$.

Sean Nemetz
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  • I think it's true only the constant functions are fixed. My proof, though, is a bit involved - it involves embedding $K(x)$ in the field of formal "inverse Laurent series" $K((x^{-1}))$ and then showing $\Delta (x^n)$ has highest order term $n x^{n-1}$ for $n \in \mathbb{Z}$. So then, going from top coefficient down by induction, you get the non-constant coefficients all being 0. – Daniel Schepler Jul 13 '18 at 21:58
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    In the special case $K = \mathbb{R}$ you could use the mean value theorem to show if $f$ is a fixed point, then either $f'$ has infinitely many roots or $f$ has infinitely many singularities (but since the denominator is a polynomial, the second is impossible), then conclude $f' \equiv 0$, so $f$ is a constant function. – Daniel Schepler Jul 13 '18 at 23:01
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    Somewhat related. – Jyrki Lahtonen Jul 14 '18 at 17:16
  • Thanks Jykri! That answer is great for seeing the non-zero characteristic case – Sean Nemetz Jul 14 '18 at 17:19
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    To address the last question. The extension $\Bbb{K}(x)/\Bbb{K}$ is infinite, so the usual Galois theory does not apply to it. There is a theory of infinite Galois extensions, but that doesn't apply either because this extension is transcendental. – Jyrki Lahtonen Jul 14 '18 at 17:20
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    What little I know about infinite Galois theory I picked up from Jacobson, Basic Algebra I-II. Sounds like you are in grad school, so you can probably use whatever textbook you had in class. It may depend on how much ivy your school has whether you are expected to know infinite Galois theory. I didn't have to, but I was kinda specializing in algebra shortly thereafter :-) Good luck! – Jyrki Lahtonen Jul 14 '18 at 17:24

2 Answers2

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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.)

Daniel Schepler
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    It would be interesting to look through and see where the argument uses the assumption that $K$ has characteristic 0. (Counterexample to show the assumption is necessary: if $K$ has characteristic $p$ and $f(x) = x^p - x$ then $f(x+1) = f(x)$.) – Daniel Schepler Jul 13 '18 at 22:31
  • That's a good point – Sean Nemetz Jul 14 '18 at 17:03
  • Perhaps when you stated " $\dots$ we must have that $n=0$ $\dots$". – Sean Nemetz Jul 14 '18 at 17:08
  • The infiniteness of the ground field gets used in the next to last paragraph, where $m\ne0$ in the ground field. – Lubin Jul 14 '18 at 20:17
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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.)

lulu
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