# Rational mapping

In

mathematics , in particular the subfield ofalgebraic geometry , a**rational map**is a kind ofpartial function betweenalgebraic varieties . In this article we use the convention that varieties are irreducible.**Definition****A first attempt**Suppose we take the prescription literally that a rational map is to be a

partial function between two varieties. Then we must make the (ultimately incorrect) definition:* A rational map from $V$ to $W$ is a pair $F\; =\; (U,\; f)$, where $U\; subset\; V$ is an open set and $f\; colon\; U\; o\; W$ is a morphism of varieties.

To see how far we can get with this definition, we will try to define the composition of two rational maps $F\; =\; (U,f),\; G\; =\; (T,\; g)$ with $U\; subset\; V,\; T\; subset\; W$. This should just be the composition $g\; circ\; f$, defined on at least a subset of $U$, but this is meaningless unless the range of $f$ is inside $T$. Nonetheless, we have at least a partially defined composition law, whenever $f(U)\; subset\; T$. We might try to extend the definition by saying that the composition $G\; circ\; F$ is obtained by extending $g$ onto a larger open set containing $f(U)$ (if possible) and then composing with $f$. This is easily seen not to depend on the extension we take.

With this definition, when $g$ is the identity map on $T$ then $g\; circ\; f\; =\; f$. Likewise, if $f$ is the identity map on $U$, then $g\; circ\; f\; =\; g$. This means that we have a different rational map which is the identity for every open subset of every variety, which is at least philosophically offensive. However, we have a sort of built-in correction if we take the extended definition of composition, in that if we have identities defined on nested open sets $T\; subset\; T\text{'}$, they compose equally with every $(U,\; f)$ with $f(U)\; subset\; T$. Since we would like, if possible, to have only one identity map, we can try to identify $(T,\; \{\; m\; id\}\_T),\; (T\text{'},\; \{\; m\; id\}\_\{T\text{'}\})$ in this case. Since of course the identity map makes sense on "any" domain, and since every open set is contained in the entire variety $W$, this means that we have at least succeeded in identifying all prospective identity maps. The result is the following:

* The identity rational map on a variety $W$ is the collection of all pairs $(T,\; \{\; m\; id\}\_T)$ with $T\; subset\; W$ an open subset.Considering that this was motivated by the extended definition of composition, we are led to make a similar definition for all rational maps.**Formal definition**Formally, a

**rational map**$f\; colon\; V\; o\; W$ between two varieties is an equivalence class of pairs $(f\_U,\; U)$ in which $f\_U$ is a morphism of varieties from $U$ to $W$, and two such pairs $(f\_U,\; U)$ and $(f\_V,\; V)$ are considered equivalent if $f\_U$ and $f\_V$ coincide on the intersection $U\; cap\; V$ (this is, in particular, vacuously true if the intersection is empty). The proof that this defines an equivalence relation relies on the following lemma:* If two morphisms of varieties are equal on any dense open set, then they are equal.

Given such an equivalence class, $f$ itself is supposed to be interpreted as the function obtained by gluing together the partial functions $f\_U$ to obtain a function on the union of all $U$ which appear. This itself may be a partial function if the $U$ do not form an

open cover of $V$.We return to the problem of composition. Two arbitrary rational maps $f\; colon\; V\; o\; W,\; g\; colon\; W\; o\; X$ can not necessarily be composed, but if $f$ has dense image then they can: pick any representative pairs $(f\_U,\; U),\; (g\_V,\; V)$ and note that since $f\_U(U)$ is dense in $W$ it must intersect $V$, and so $f\_U^\{-1\}(V)$ is a nonempty open subset of $V$ (as a morphism is necessarily continuous). Then $g\_V\; circ\; f\_U$ is defined on this set, so we take $g\; circ\; f$ to be the equivalence class containing the pair $(g\_V\; circ\; f\_U,\; f\_U^\{-1\}(V))$.

**Dominant and birational maps**We encountered the following condition in considering composition: $f$ is said to be

**dominant**if the range of any of the $f\_U$ is dense in $W$ (this then implies that the range of any of them is dense, since open sets are dense in a variety). Since dominant rational maps can be composed, we can go further and talk about their inverses: a dominant rational map $f$ is said to be**birational**if there exists a rational map $g\; colon\; W\; o\; V$ which is its inverse, where the composition is taken in the above sense.The importance of rational maps to algebraic geometry is in the connection between such maps and maps between the function fields of $V$ and $W$. Even a cursory examination of the definitions reveals a similarity between that of rational map and that of rational function; in fact, a rational function is just a rational map whose range (or codomain, more categorically precisely) is the projective line. Composition of functions then allows us to "pull back" rational functions along a rational map, so that a single rational map $f\; colon\; V\; o\; W$ induces a homomorphism of fields $K(W)\; o\; K(V)$. In particular, the following theorem is central: the

functor from the category of projective varieties with dominant rational maps (over a fixed base field, for example $mathbb\{C\}$) to the category of field extensions of the base field with reverse inclusion of extensions as morphisms, which associates each variety to its function field and each map to the associated map of function fields, is anequivalence of categories .**An example of birational equivalence**Two varieties are said to be

**birationally equivalent**if there exists a birational map between them; this theorem states that birational equivalence of varieties is identical to isomorphism of their function fields as extensions of the base field. This is somewhat more liberal than the notion of isomorphism of varieties (which requires a globally defined morphism to witness the isomorphism, not merely a rational map), in that there exist varieties which are birational but not isomorphic.The usual example is that $mathbb\{P\}^2\_k$ is biratonal to the variety $X$ contained in $mathbb\{P\}^3\_k$ consisting of the set of projective points $[w\; :\; x\; :\; y\; :\; z]$ such that $xy\; -\; wz\; =\; 0$, but not isomorphic. Indeed, any two lines in $mathbb\{P\}^2\_k$ intersect, but the lines in $X$ defined by $w\; =\; x\; =\; 0$ and $y\; =\; z\; =\; 0$ cannot intersect since their intersection would have all coordinates zero. To compute the function field of $X$ we pass to an affine subset (which does not change the field, a manifestation of the fact that a rational map depends only on its behavior in any open subset of its domain) in which $w\; eq\; 0$; in projective space this means we may take $w\; =\; 1$ and therefore identify this subset with the affine $xyz$-plane. There, the coordinate ring of $X$ is:$A(X)\; =\; k\; [x,y,z]\; /(xy\; -\; z)\; cong\; k\; [x,y]$via the map $p(x,y,z)\; mapsto\; p(x,y,xy)$. And the

field of fractions of the latter is just $k(x,y)$, isomorphic to that of $mathbb\{P\}^2\_k$. Note that at no time did we actually produce a rational map, though tracing through the proof of the theorem it is possible to do so.**ee also***

Birational geometry **References*** | year=1977, section I.4.

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