Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30
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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 k be an algebraically closed field, let A=k[x1,…,xn]/I be an affine k-algebra, let J⊆A be a radical ideal, and let f∈A∖J. Then there exists a k-algebra map φ:A→k such that J⊆ker⁡φ and φ(f)≠0.

Facts & Assumptions

Given: The Axiom of Choice, an algebraically closed field k, an affine algebra A=k[x1,…,xn]/I, a radical ideal J⊆A, and an element f∉J.

[L1]

In a polynomial ring over an algebraically closed field, I(V(K))=K for every ideal K (Strong Nullstellensatz: I(V(I)) equals the radical of I).

[L2]

Maximal ideals of an affine k-algebra are kernels of k-points (Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points).

Proof

technique · contradiction
1.1givenchoose

Let π:k[x1,…,xn]→A be the quotient map, let J′=π−1(J), and choose f~∈k[x1,…,xn] with π(f~)=f. Because A/J≅k[x1,…,xn]/J′ and J is radical, J′ is a radical ideal of the polynomial ring.

2.1L1step 1.1assume-contracontradiction

Assume, for contradiction, that every k-algebra map φ:A→k whose kernel contains J also satisfies φ(f)=0. Then every point a∈kn annihilating J′ also annihilates f~, so f~∈I(V(J′)). By [L1], I(V(J′))=J′=J′, whence f=π(f~)∈J, contradiction.

3.1L2step 2.1discharge-contradiction∎

Therefore some k-algebra map φ:A→k annihilating J satisfies φ(f)≠0. Such a map is a point of the affine algebra by [L2].

Depends on

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Dependency tree · two levels

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