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