# Different definitions of rational mappings.

In the book "Ideals, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra" by David A. Cox, John Little, Donal O'Shea, the rational mapping is defined as follows.

In the book "Algebraic Geometry: A First Course" by Joe Harris, the rational mapping is defined as follows.

What are differences between these two definitions? What are the regular maps in Definition 4? Thank you very much.

Definition 4 only make sense for affine varieties, but if $X$ and $Y$ are affine, then both definitions are (basically) the same.
Starting with $\phi= (\tfrac{f_1}{g_1},\dots,\tfrac{f_n}{g_n})$ in definition 4, then the associated pair $(U,\varphi)$ of definition 7.3 is the following : $U$ is the non-empty open set of $X$ where $g_i \neq 0$ ($\forall i$), and $\varphi$ is the obvious map.
In the other directtion, let $(U, \varphi : U \rightarrow Y)$ as in definition 7.3. Denote $(\varphi_1,\dots,\varphi_n)$ the components of $\varphi$. The functions $\phi_i : U \rightarrow Y$ are regular functions of an open subset of an affine variety. This implies that $\varphi_i$ can be written $\varphi_i=\tfrac{f_i}{g_i}$ (see lemma 2.1 of Harris' book).
• To give an illustration, $1/x$ is not regular over $k^1$ (since the denominator vanishes at $x=0$). It is rational over $k^1$ though, because it is regular over $k^1 \setminus \{ 0 \}$, which is dense. So, basically, rational functions are quotients of polynomials which makes them only partially defined in case the polynomial in the denominator is nontrivial. Rational functions are more useful for projective varieties because there are no non-constant regular functions there. On affine varieties, regular functions are just polynomials. – Marek Jul 15 '13 at 12:43
• @Marek, thank you very much. How to show that $k^1\backslash \{0\}$ is dense in $k^1$? – LJR Jul 15 '13 at 13:55
• @LJR: well, the only closed sets in $k^1$ (besides $k^1$ itself) are finite collections of points (corresponding to roots of polynomials). Therefore the closure of $k^1 \setminus \{0\}$ can't be anything else than $k^1$. I suggest you review the basics of Zariski topology and its relation to solutions of polynomials. – Marek Jul 15 '13 at 15:53