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I read in Jaames E. Humphreys' book "Linear Algebraic Groups", that a rational function need not be regular.

Suppose that $X$ is a varity over an algebraically close field $K$. If I am not mistaken, a rational function over $X$ is a function in $K(X)$. And a function on $X$ is regular at a point $x \in X$ if there exists $g,h \in K[X]$, and an open subset $V \subseteq X$ containing $x$, such that for all $y \in V$, $h(y) \neq 0$ and $f(y) =g(y)/h(y)$. And $f(x)$ is a regular function if it is regular at every point in $X$.

I not understand why rational functions need not be regular, and I am wondering if there is any example of rational function being not regular.

Thank you very much~

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up vote 5 down vote accepted

Here is an example: let $X=\mathbb{A}^1_K$, and let $f(y)=\frac{1}{y}$. This is a rational function $f:X-\,\,-\,\,\rightarrow K$ that is regular at all $y\in X$, $y\neq 0$, but which is not defined at $y=0$. It cannot be extended to a regular function at $y=0$, for suppose $$\frac{f}{g}=\frac{1}{y}$$ in some open neighborhood of $y=0$; this open neighborhood is in particular infinite, because the topology on $\mathbb{A}^1_K$ is the cofinite topology and $K$ (being algebraically closed) is infinite. Then $f(y)y-g(y)=0$ for infinitely many $y\in K$, but because $fy-g$ can only have finitely many roots, we must actually have $fy=g$, hence $\frac{f}{g}=\frac{1}{y}$ everywhere, and therefore it cannot be defined at $y=0$.

One of the things that I found a little counterintuitive about rational functions at first is that they are not functions, per se, because they might only be defined on an open subset of the variety in question.

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Thank you very much~ This is really an illuminating example. In fact, I was confused partly because I thought that regular functions on $X$ were functions defined on $X$. – ShinyaSakai May 10 '11 at 3:19

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