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Let $K$ be a field of characteristic zero. Can $f/g\in K(X)$ considered as a function $K\to K$ take the same fixed value $\alpha \in K$ infinite times?

By elementary calculus this is not the case if $K$ is a subfield of $\mathbb{R}$, but I don't know how to argue in general (or if the result is true).

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3 Answers 3

up vote 5 down vote accepted

Suppose $a$ is a fixed element of $K$, and $f(x)/g(x)=a$ for infinitely many $x$. Then for such $x$, $f(x)=ag(x)$, so $f(x)-ag(x)=0$. Thus the polynomial $f-ag$ has infinitely many roots, and so must be the zero polynomial. So for any $y$, $f(y)-ag(y)=0$, so $f(y)/g(y)=a$. Thus we have shown that any rational function that takes a given value infinitely many times must in fact be a constant rational function. Note that the assumption that $K$ has characteristic zero turned out to be irrelevant.

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Define $h(x) = f(x) - \alpha g(x).$ As a polynomial, $h$ has some degree $n.$

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Hint $\rm\,\ \dfrac{f}g = \dfrac{h}k\,\ on\,\ S\:\Rightarrow\: fk\!-\!gh=0\,\ on\,\ S,\:$ so $\rm\:|S| > deg(fk\!-\!gh)\:\Rightarrow\:fk=gh\:\Rightarrow\:\dfrac{f}g = \dfrac{h}k$

since a polynomial over a domain with more roots than its degree is the zero polynomial.

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