# Why is the domain of a rational function necessarily nonempty

Let $V$ be an irreducible affine variety. A rational map $f : V \to \mathbb A^n$ is an $n$-tuple of maps $(f_1, \ldots , f_n)$ where there $f_i$ are rational functions i.e. are in $k(V)$. Th map is called regular at the point $P$ if all the $f_i$ are regular at the point $P$. Hence $\mathrm{dom}(f) = \bigcap_{i=1}^n \mathrm{dom}(f_i)$.

My book then says that $\mathrm{dom}(f)$ is therefore a nonempty open subset of $V$. I can see why it's open, but I can't see why it has to be nonempty. I can see that each of the $\mathrm{dom}(f_i)$ are nonempty, and that it's true in the case where $V$ embeds into $\mathbb A^1$ and the polynomials are in one variable.

So, why is $\mathrm{dom}(f)$ necessarily nonempty?

Thanks

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You've noted that the $\operatorname{dom}(f_i)$ are non-empty open subsets of $V$. We haven't used irreducibility yet, and in fact that adjective seems to be crucial for working with rational functions and function fields because of the following.
Lemma. If $X$ is an irreducible topological space and $U_1, U_2$ are non-empty open subsets of $X$, then $U_1 \cap U_2 \neq \varnothing$.
Thanks, I've yet to go through Bruno's answer closely. Rather than starting a new thread: my book discusses when it's possible to define to composite of rational maps. It says $f : V \to W$ and $g : W \to \mathbb A^1$ can be composed when $f$ is dominant, i.e. when $f(\mathrm{dom}(f))$ is a dense subset of $W$. This is all fine - but why is the codomain of $g$ just $\mathbb A^1$? I don't see how the argument fails for a more general codomain – Matt Mar 6 '12 at 17:58
@Matt I think that you can compose them. Probably the author is headed towards obtaining a map of function fields $K(W) \to K(V)$ obtained by sending such $g$ to $g \circ f$, in order to get the equivalence of categories between varities with dominant rational maps and the category of finitely generated field extensions of the ground field. – Dylan Moreland Mar 6 '12 at 18:19