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A radical ideal omitting a function admits a point that kills the ideal but not the function
Statement
Assume the Axiom of Choice.
Let be an algebraically closed field, let be an affine -algebra, let be a radical ideal, and let . Then there exists a -algebra map such that and .
Facts & Assumptions
Given: The Axiom of Choice, an algebraically closed field , an affine algebra , a radical ideal , and an element .
In a polynomial ring over an algebraically closed field, for every ideal (Strong Nullstellensatz: I(V(I)) equals the radical of I).
Maximal ideals of an affine -algebra are kernels of -points (Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points).
Proof
Let be the quotient map, let , and choose with . Because and is radical, is a radical ideal of the polynomial ring.
Assume, for contradiction, that every -algebra map whose kernel contains also satisfies . Then every point annihilating also annihilates , so . By [L1], , whence , contradiction.
Therefore some -algebra map annihilating satisfies . Such a map is a point of the affine algebra by [L2].
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
9 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
- Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (15.7) (standard reference, not scraped)