Rational Maps

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In §Quasi-Projective Varieties, ⁋Definition 7 we defined regular maps between quasi-projective varieties as functions that are defined at every point of their domain. Even if they are written in rational form over \(D(f)\) as in §Affine Varieties, ⁋Definition 14, the only possible denominators are powers of \(f\), so they are defined everywhere.

Nevertheless, many natural functions are not regular maps. For instance, \((x, y) \mapsto [x : y]\) is not a regular map because it is undefined at the origin, yet it looks like a perfectly natural function. In this post we study rational maps, which are functions defined on most points.

Rational Functions

Just as we did for regular maps, we first define the notion of a rational function before defining rational maps.

Definition 1 A rational function on a variety \(X\) is a pair \((U,f)\) consisting of a nonempty open subset \(U\) of \(X\) and a regular function \(f:U \rightarrow \mathbb{K}\) defined on it. Two rational functions \((U,f)\) and \((V,g)\) are equivalent if they agree on \(U\cap V\).

The intuition is as follows. In the Zariski topology closed sets are small and open sets are large. Thus a rational function is a function that may fail to be defined on a small set, but is defined at most other points. Essentially, one may think of open sets in the Zariski topology as being of the form \(D(g)\), and the regular functions \(f/g\) defined on them are now regarded as functions. (§Affine Varieties, ⁋Definition 14) Of course such a function is undefined where \(g=0\), but that is exactly why we consider functions defined on open sets \(U\), and in any case the points where \(g=0\) are small relative to the whole space.

We write \(\mathbb{K}(X)\) for the set of equivalence classes of all rational functions on \(X\). The sum and product of two rational functions are defined on the intersection of their domains of definition, and the inverse of a nonzero rational function is defined where the function is nonzero. Hence \(\mathbb{K}(X)\) is a field, which we call the function field.

Proposition 2 For an affine variety \(X\), we have \(\mathbb{K}(X)=\Frac\mathbb{K}[X]\).

The heart of this proposition is representing an arbitrary regular function \(f:U\rightarrow \mathbb{K}\) defined on an arbitrary open set \(U\) as a fraction. Since \(U\) can be written as a union of \(D(g_i)\) (§Affine Varieties, ⁋Proposition 6), and the coordinate ring of \(D(g_i)\cap X\) is \(\mathbb{K}[X]_{g_i}\), the regular functions on it are rational expressions with powers of \(g_i\) in the denominator, so the proof is not difficult.

What is important is that this proposition provides a practical way to compute rational functions. For example, for \(X = V(\y - \x^2)\) the coordinate ring is \(\mathbb{K}[\x, \y]/(\y - \x^2) \cong \mathbb{K}[\x]\), and therefore \(\mathbb{K}(X) = \Frac(\mathbb{K}[\x]) = \mathbb{K}(\x)\).

Proposition 3 For a variety \(X\) and a nonempty open subset \(U\), we have \(\mathbb{K}(U) = \mathbb{K}(X)\).

Proof

First, the inclusion \(\iota: U \hookrightarrow X\) obviously induces an embedding \(\iota^\ast: \mathbb{K}(X)\rightarrow \mathbb{K}(U)\) of function fields. Since any nonzero field homomorphism is injective, it suffices to show that \(\iota^\ast\) is surjective. ([Field Theory] §Fields, ⁋Proposition 2)

Now for any \(f \in \mathbb{K}(U)\), \(f\) is a regular function on some nonempty open subset \(V\) of \(U\). Then \(V\) is also open in \(X\), so the pair \((V,f)\) belongs to \(\mathbb{K}(X)\).

Example 4 Consider the function field \(\mathbb{K}(\mathbb{P}^n)\) of \(\mathbb{P}^n\). By Proposition 3 it suffices to compute the function field on the open set \(U_0\) of \(\mathbb{P}^n\). Since \(U_0\) is an affine variety, by Proposition 2 it equals the fraction field of \(\mathbb{K}[U_0]\), and hence the function field of \(\mathbb{P}^n\) is the field \(\mathbb{K}(\t_1,\ldots, \t_n)\) generated by \(n\) indeterminates.

Concretely, this is obtained by writing elements of \(\mathbb{P}^n\) as \([x_0:\cdots: x_n]\) and setting \(\t_i=\x_i/\x_0\), where \(\x_i\) is the coordinate function reading off the \(i\)-th coordinate. If we had chosen a different open set \(U_j\), similar rational functions would have been defined via \(\t_i=\x_i/\x_j\), and so in general we know that rational functions on \(\mathbb{P}^n\) are represented as ratios \(F/G\) of homogeneous polynomials of the same degree.

Rational Maps

Recalling how we defined regular maps from regular functions, it is obvious how to define rational maps from rational functions.

Definition 5 A rational map between two varieties \(X, Y\) is a pair \((U,\varphi)\) consisting of a nonempty open subset \(U\) of \(X\) and a regular map \(\varphi: U \rightarrow Y\) defined on it.

As before, two rational maps \(\varphi: U \rightarrow Y\) and \(\psi: V \rightarrow Y\) are regarded as the same if they agree on \(U \cap V\). A rational map is usually denoted \(\varphi: X \dashrightarrow Y\), where the dashed arrow indicates that it may not be defined at every point. The points where it is undefined are called base points.

On the other hand, given a rational map \(\varphi:U\rightarrow Y\), we may consider rational maps equivalent to \((U,\varphi)\). Taking the union of the domains of all such rational maps yields the largest open set on which \(\varphi\) can be defined.

Definition 6 For a rational map \(\varphi: X\dashrightarrow Y\), we write \(\dom(\varphi)\) for the open set obtained by the above procedure.

Example 7 One of the typical examples of a rational map is projection from a point. For instance, consider the line \(\{\x_2=0\}\) in \(\mathbb{P}^2\); this can be thought of as a projective line \(\mathbb{P}^1\) inside \(\mathbb{P}^2\). The point \([0:0:1]\) does not lie on this line, and the line through this point and an arbitrary point \([x_0:x_1:x_2]\) has equation

\[x_1\x_0-x_0\x_1=0\]

Then this line meets the above \(\mathbb{P}^1\) exactly at \([x_0:x_1:0]\), and thus we obtain the following projection

\[[x_0:x_1:x_2]\mapsto [x_0:x_1]\]

in this manner.

Birational Equivalence

If an isomorphism of regular maps means that two varieties have exactly the same structure, then birational equivalence means that two varieties have essentially the same structure. Many geometric properties are preserved not only between isomorphic varieties but also between birationally equivalent varieties.

Definition 8 A rational map \(\varphi: X \dashrightarrow Y\) is called dominant if its image is dense in \(Y\). That is, \(\overline{\varphi(\dom(\varphi))} = Y\).

The dominant condition is needed because the composition of rational maps is not generally defined. For two rational maps \(\varphi: X\dashrightarrow Y\) and \(\psi: Y \dashrightarrow Z\), if the image of \(\varphi\) does not meet \(\dom(\psi)\) there is no way to define \(\psi\circ \varphi\). However, if \(\varphi\) is dominant then \(\varphi(\dom(\varphi))\) is dense in \(Y\), so it must meet the nonempty open set \(\dom(\psi)\), and therefore \(W=\varphi^{-1}(\dom(\psi))\) is a nonempty open subset of \(\dom(\varphi)\) on which \(\psi\circ\varphi\) is defined as a regular map. One can also check that \(\psi\circ\varphi\) obtained in this way is again dominant. Since \(X\) is irreducible, \(W\) is dense in \(\dom(\varphi)\), and \(\varphi^{-1}(\overline{\varphi(W)})\) is a closed subset of \(\dom(\varphi)\) containing \(W\), so \(\varphi(\dom(\varphi))\subseteq \overline{\varphi(W)}\), i.e. \(\overline{\varphi(W)}=Y\). Then \(\varphi(W)\) is dense in \(\dom(\psi)\) as well, so applying the same argument once more to \(\psi\) and \(\varphi(W)\) yields \(\overline{\psi(\varphi(W))}=Z\). Henceforth we consider compositions only for dominant rational maps.

Definition 9 A dominant rational map \(\varphi: X \dashrightarrow Y\) is called a birational map if there exists another dominant rational map \(\psi: Y \dashrightarrow X\) such that \(\psi \circ \varphi = \id_X\) and \(\varphi \circ \psi = \id_Y\) (where defined). Two varieties \(X, Y\) are called birationally equivalent if there exists a birational map between them.

Two birationally equivalent varieties are isomorphic “at most points.” Specifically, as shown in the next proposition, there exist isomorphic open subsets of the two varieties. This shows that birational equivalence is weaker than isomorphism but still a strong relationship.

Proposition 10 For two varieties \(X, Y\) the following are equivalent.

1. \(X\) and \(Y\) are birationally equivalent.
2. There is a \(\mathbb{K}\)-algebra isomorphism \(\mathbb{K}(X) \cong \mathbb{K}(Y)\).
3. There exist nonempty isomorphic open subsets of \(X\) and \(Y\).

Proof

First suppose \(X, Y\) are birationally equivalent. Then considering the domain \(\dom(\varphi)\) of the birational map \(\varphi: X\dashrightarrow Y\), there is an induced \(\mathbb{K}\)-algebra homomorphism \(\varphi^\ast: \mathbb{K}(Y)\rightarrow \mathbb{K}(\dom(\varphi))\) of function fields. Similarly the birational inverse \(\psi: Y\dashrightarrow X\) defines \(\psi^\ast: \mathbb{K}(X)\rightarrow \mathbb{K}(\dom(\psi))\). Now by Proposition 3 we have \(\mathbb{K}(\dom(\varphi))=\mathbb{K}(X)\) and \(\mathbb{K}(\dom(\psi))=\mathbb{K}(Y)\), so using this we see from \(\psi\circ\varphi=\id_X\) and \(\varphi\circ\psi=\id_Y\) that \(\varphi^\ast\) and \(\psi^\ast\) are inverses of each other, and hence \(\mathbb{K}(X)\cong \mathbb{K}(Y)\).

Now suppose a \(\mathbb{K}\)-algebra isomorphism \(\Phi: \mathbb{K}(X) \rightarrow \mathbb{K}(Y)\) is given. For any affine open subset \(U \subseteq X\), the coordinate ring \(\mathbb{K}[U]\) is a finitely generated \(\mathbb{K}\)-subalgebra of \(\mathbb{K}(X)\). Now choose an affine open subset \(V\subseteq Y\) such that the images under \(\Phi\) of the generators of \(\mathbb{K}[U]\) are all regular; then \(\Phi(\mathbb{K}[U])\subseteq \mathbb{K}[V]\), and similarly using \(\Phi^{-1}\) we obtain a nonzero \(f\in \mathbb{K}[U]\) such that \(\Phi^{-1}(\mathbb{K}[V])\subseteq \mathbb{K}[U]_f\). Setting \(h=\Phi(f)\), we have \(\Phi(1/f)=1/h\), so from the two inclusions above we get \(\Phi(\mathbb{K}[U]_f)\subseteq \mathbb{K}[V]_h\) and \(\Phi^{-1}(\mathbb{K}[V]_h)\subseteq \mathbb{K}[U]_f\), and therefore \(\Phi\) restricts to an isomorphism between \(\mathbb{K}[U]_f\) and \(\mathbb{K}[V]_h\). But these are the coordinate rings of the affine varieties \(D(f)\cap U\) and \(D(h)\cap V\) respectively (§Affine Varieties, ⁋Proposition 7), so by §Affine Varieties, ⁋Proposition 18 these two open sets are isomorphic.

That the last condition implies the first is obvious by Proposition 3.

This theorem shows that to determine birational equivalence it suffices to look at the function field.

Example 11 Let us compute the function fields of \(\mathbb{P}^1 \times \mathbb{P}^1\) and of the quadric surface \(Q = V(\x\y - \z\w)\) in \(\mathbb{P}^3\).

First, for \(\mathbb{P}^1 \times \mathbb{P}^1\), by Proposition 3 it suffices to compute on the product open set \(U_0 \times U_0\) of each factor. The function field of the first factor \(\mathbb{P}^1\) is \(\mathbb{K}(\t_1)\) as we saw in Example 4, and similarly the second factor is \(\mathbb{K}(\t_2)\). Hence their function field is \(\mathbb{K}(\t_1,\t_2)\).

Now consider the quadric surface \(Q = V(\x\y - \z\w) \subseteq \mathbb{P}^3\). Again by Proposition 3 it suffices to compute on the affine patch \(\{\w \ne 0\}\). On this patch set \(\x' = \x/\w\), \(\y' = \y/\w\), \(\z' = \z/\w\); then the equation \(\x\y - \z\w = 0\) becomes \(\x'\y' - \z' = 0\). Hence \(\z' = \x'\y'\), and the coordinate ring of this patch is \(\mathbb{K}[\x', \y', \z']/(\x'\y' - \z') \cong \mathbb{K}[\x', \y']\). By Proposition 2 we have \(\mathbb{K}(Q) = \Frac(\mathbb{K}[\x', \y']) = \mathbb{K}(\x', \y') \cong \mathbb{K}(\t_1, \t_2)\).

Therefore \(\mathbb{K}(\mathbb{P}^1 \times \mathbb{P}^1) \cong \mathbb{K}(Q) \cong \mathbb{K}(\t_1, \t_2)\), so by Proposition 10 the two varieties are birationally equivalent. In fact, the image of the Segre embedding \(\mathbb{P}^1 \times \mathbb{P}^1 \rightarrow \mathbb{P}^3\), \(([x : y], [u : v]) \mapsto [xu : xv : yu : yv]\) discussed in §Projective Varieties, ⁋Example 16 is exactly \(V(\x\w - \y\z)\), which is a quadric that becomes \(Q\) upon swapping \(\y\) and \(\w\). That is, in this case the birational equivalence is actually an isomorphism. This example shows that birational equivalence is weaker than isomorphism, but includes it.

Blow-up

A rational map has the limitation that it is undefined at base points. A representative tool for resolving this limitation is the blow-up. The motivation comes from the very first function we considered, \((x,y)\mapsto [x:y]\). This function takes a point \((x,y)\) in \(\mathbb{A}^2\) and returns the slope of the line through this point and the origin \((0,0)\in \mathbb{A}^2\); it is undefined at the origin because a line requires two distinct points to be defined. In such a case we would usually fix the origin \((0,0)\) and let the other point \((x,y)\) approach \((0,0)\) to compute the limit, but here there are infinitely many directions toward \((0,0)\) so the limit is not well defined.

The idea of blow-up is simple: record all directions toward \((0,0)\) separately.

Example 12 Consider the variety

\[\Bl_{(0,0)} \mathbb{A}^2 = \{((x, y), [u : v]) \in \mathbb{A}^2 \times \mathbb{P}^1 \mid xv = yu\}\]

This set is a closed subvariety of \(\mathbb{A}^2 \times \mathbb{P}^1\). The condition \(xv = yu\) means that the point \((x, y)\) and the line \([u : v]\) lie in the same direction. That is,

• For a point \((x,y)\) in \(\mathbb{A}^2\) other than the origin, the condition \(xv=yu\) uniquely determines the point \([u:v]\) in \(\mathbb{P}^1\), and through this the point of \(\Bl_{(0,0)}\mathbb{A}^2\) is uniquely determined.
• At the origin \((0,0)\) of \(\mathbb{A}^2\), every point of \(\mathbb{P}^1\) is possible.

[Hart] p.29. Fig. 3.

Concretely, define the projection \(\pi_1: \Bl_{(0,0)} \mathbb{A}^2 \rightarrow \mathbb{A}^2\) by \(\pi_1((x, y), [u : v]) = (x, y)\). Then the preimage of every point other than the origin is a single point, while the preimage of the origin is \(\mathbb{P}^1\). This is called the exceptional divisor.

Hence away from the origin the two varieties \(\mathbb{A}^2\) and \(\Bl_{(0,0)}\mathbb{A}^2\) are isomorphic, so \(\pi_1\) is a birational map.

Now consider the rational map \(\varphi: \mathbb{A}^2 \dashrightarrow \mathbb{P}^1\), \((x, y) \mapsto [x : y]\) mentioned earlier. It is undefined at the origin \((0, 0)\), but from the viewpoint of the blow-up \(\Bl_{(0,0)} \mathbb{A}^2\) it is simply the projection \(\pr_2\) to the \(\mathbb{P}^1\) factor, which in particular is a regular map. In this way we can resolve base points where a birational map is undefined.

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References

[Hart] R. Hartshorne, Algebraic Geometry, Springer, 1977.
[Har] J. Harris, Algebraic Geometry: A First Course, Springer, 1992.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Varieties in Projective Space, Springer, 2013.