User:Ryan Reich/Rational mapping

In mathematics, in particular the subfield of algebraic geometry, a rational map is a kind of partial function between algebraic varieties. It is defined in a seemingly unnecessarily complicated way, but in fact to be useful the details are important. Formally, a rational map $$f \colon V \to W$$ 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 requires the auxiliary lemma that two distinct morphisms do not coincide on any open set.

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 $$V$$. $$f$$ 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). Two arbitrary rational maps $$f \colon V \to W, g \colon W \to X$$ can not necessarily be composed, but if $$f$$ is dominant 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 $$(g_V \circ f_U, f_U^{-1}(V)$$. Finally, a dominant rational map $$f$$ is said to be birational if there exists a rational map $$g \colon W \to V$$ which is its inverse, where the composition is taken in the above sense.

The complications of the definition are required to make any of the axioms for a category hold when we take the morphisms to be rational maps. For consider what would happen if we were to make the simpler definition that a rational map is simply a morphism defined on an open set, and look at a pair $$f, g$$ of inverse rational maps from a variety to itself. Since they are inverses, $$g \circ f$$ should be the identity map, but on the other hand, the above discussion shows that in defining the composition it is in general necessary to shrink the domain of definition from that of $$f$$, so that this &ldquo;identity&rdquo; map is only defined on a proper open subset. Since it is the identity, composition with $$f$$ should yield $$f$$ again, but since the domain has now changed this equality only holds on a proper subset of the domain of $$f$$, and two functions with different domains cannot be equal. Observe, however, the earlier remark that two distinct morphisms never coincide on an open set, which leads us perhaps to relax the distinction when only the domains differ. But this is just the definition given above. Thus inverses will never exist in the category of varieties with rational maps without the given definition. The discussion also showed that a unique identity map is impossible otherwise. Composition is possible with the naive definition, since the recipe given above works fine, but without the relaxed rules of the correct definition it is ultimately crippled.

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 &ldquo;pull back&rdquo; rational functions along a rational map, so that a single rational map $$f \colon V \to W$$ induces a homomorphism of fields $$K(W) \to K(V)$$. 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 an equivalence of categories.

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 $$w \neq 0$$; 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.