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Morphisms from an irreducible affine variety to an affine variety are determined by a dense open subset

Statement

Let $X$ be a classical affine variety, let $Y \subseteq A_{k}^{m}$ be a classical affine variety, and let $U \subseteq X$ be a nonempty open subset. If two morphisms $φ, ψ : X \to Y$ agree on $U$ , then $φ = ψ$ on all of $X$ .

Facts & Assumptions

Given: Classical affine varieties $X \subseteq A_{k}^{n}$ and $Y \subseteq A_{k}^{m}$ , a nonempty open subset $U \subseteq X$ , and morphisms $φ, ψ : X \to Y$ with $φ ∣_{U} = ψ ∣_{U}$ .

[L1]

A morphism pulls every global regular function on the target back to a global regular function on the source (Morphisms of classical affine varieties).

[L2]

Every nonempty open subset of a classical affine variety is dense (Every nonempty open subset of an affine variety is dense).

[L3]

Global regular functions on an affine variety are exactly its coordinate-ring elements (Global regular functions on a classical affine variety are its coordinate ring).

[L4]

For $h \in k [X]$ , the principal open $D_{X} (h)$ is the set of points where $h$ is nonzero (A principal open subset of a classical affine variety).

Proof

technique · direct

1.1

Let ${\overline{y}}_{1}, \dots, {\overline{y}}_{m} \in k [Y]$ be the coordinate classes. By [L1] and [L3], the pullbacks $φ^{*} ({\overline{y}}_{i})$ and $ψ^{*} ({\overline{y}}_{i})$ are elements of $k [X]$ . Since $φ = ψ$ on $U$ , these two functions agree on $U$ for every $i$ .

L1L3given

2.1

Fix $i$ , and put $h_{i} = φ^{*} ({\overline{y}}_{i}) - ψ^{*} ({\overline{y}}_{i}) \in k [X]$ . The function $h_{i}$ vanishes on $U$ . If $D_{X} (h_{i})$ were nonempty, then [L2] would make both $U$ and $D_{X} (h_{i})$ dense open subsets of $X$ , so they would meet. That contradicts [L4], because $h_{i}$ is zero on $U$ and nonzero on $D_{X} (h_{i})$ . Hence $D_{X} (h_{i}) = \emptyset$ , so $h_{i} = 0$ on all of $X$ .

L2L4step 1.1algebra

3.1

Step 2.1 shows that every coordinate function of $Y$ has the same pullback under $φ$ and $ψ$ . Therefore the two maps have the same $m$ coordinate functions on $X$ , so $φ (x) = ψ (x)$ for every $x \in X$ . Thus $φ = ψ$ .

step 2.1algebra∎

Depends on

Used by

The maximal domain of definition of a rational map to an affine variety Definition

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry, Proposition 5.8 (standard reference, not scraped)
Michael Artin, Notes for a Course in Algebraic Geometry, morphism discussion in §2.5 (standard reference, not scraped)