Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.