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Dominant rational maps to an affine variety correspond to injective homomorphisms of function fields

Statement

Let $X$ and $Y$ be classical affine varieties. Sending a dominant rational map $η : X ⇢ Y$ to its pullback on function fields gives a canonical bijection ${dominant rational maps X ⇢ Y} ≅ {Hom}_{k -fld}^{i n j} (k (Y), k (X)) .$

Facts & Assumptions

Given: Classical affine varieties $X \subseteq A_{k}^{n}$ and $Y \subseteq A_{k}^{m}$ over an algebraically closed field $k$ .

[L1]

A dominant rational map $X ⇢ Y$ induces an injective pullback homomorphism $k (Y) ↪ k (X)$ , functorially (Dominant maps pull back function fields functorially).

[L2]

The function field of a classical affine variety is the fraction field of its coordinate ring (The function field of an irreducible classical affine variety).

[L3]

If $Y \subseteq A_{k}^{m}$ , then $k [Y] = k [y_{1}, \dots, y_{m}] / I (Y)$ (The coordinate ring of an affine algebraic set).

[L4]

The ideal $I (Y)$ consists exactly of the polynomials vanishing on $Y$ (The vanishing ideal of a subset of affine space).

[L5]

Every affine open of $X$ is a principal open on this page, and for $U = D_{X} (d)$ one has $k [U] = k [X]_{d}$ (Affine open subsets of a classical affine variety, Regular functions on a principal open are the principal localization of the coordinate ring).

[L6]

Morphisms from an affine variety to $Y$ correspond to $k$ -algebra homomorphisms out of $k [Y]$ (Affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms).

Proof

technique · direct

1.1L1given

By [L1], every dominant rational map $η : X ⇢ Y$ determines an injective $k$ -homomorphism $η^{*} : k (Y) ↪ k (X)$ .

1.2L2L3L5givenconstruct

Conversely, let $σ : k (Y) ↪ k (X)$ be an injective $k$ -homomorphism. For the coordinate classes ${\overline{y}}_{1}, \dots, {\overline{y}}_{m} \in k [Y]$ from [L3], write $σ ({\overline{y}}_{i}) = \frac{a_{i}}{d_{i}}$ with $a_{i}, d_{i} \in k [X]$ and $d_{i} \neq 0$ in the domain $k [X]$ . Put $d = d_{1} \dots d_{m}$ and $U = D_{X} (d)$ . By [L5], each $a_{i} / d_{i}$ is regular on $U$ , so each $σ ({\overline{y}}_{i})$ lies in $k [U]$ .

2.1L2L3L4step 1.2algebra

Define a $k$ -algebra homomorphism $α : k [Y] \to k [U]$ by sending ${\overline{y}}_{i}$ to $σ ({\overline{y}}_{i}) ∣_{U}$ . This is well defined because [L4] identifies $I (Y)$ as the defining ideal of $Y$ , and if $g \in I (Y)$ , then its class is $0$ in $k [Y]$ by [L3], hence $g (σ ({\overline{y}}_{1}), \dots, σ ({\overline{y}}_{m})) = σ (g) = 0$ in $k (X)$ . Therefore every relation of $I (Y)$ is respected.

3.1L6step 2.1

By [L6], the homomorphism $α$ corresponds to a morphism $φ : U \to Y$ . Its rational-map class gives a rational map $η_{σ} : X ⇢ Y$ .

4.1step 2.1step 3.1

The rational map $η_{σ}$ is dominant. If its image were not dense in $Y$ , then some nonzero $h \in k [Y]$ would vanish on $φ (U)$ . Then $α (h) = 0$ in $k [U]$ , so $σ (h) = 0$ in $k (X)$ by step 2.1, contradicting injectivity of $σ$ .

5.1L2L3step 1.1step 4.1

The pullback $(η_{σ})^{*}$ agrees with $σ$ on the coordinate classes ${\overline{y}}_{i}$ , hence on the whole coordinate ring $k [Y]$ , and therefore on its fraction field $k (Y)$ by [L2]. Thus the construction of steps 1.2-4.1 is inverse to the construction in step 1.1.

6.1step 1.1step 5.1∎

Steps 1.1 and 5.1 give the stated bijection between dominant rational maps $X ⇢ Y$ and injective $k$ -homomorphisms $k (Y) ↪ k (X)$ .

Depends on

Used by

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Sources

J. S. Milne, Algebraic Geometry, Proposition 5.38 (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, Theorem 3.2.1 together with §3.1-§3.2 (standard reference, not scraped)