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 JA be a radical ideal, and let fAJ. Then there exists a k-algebra map φ:Ak such that Jkerφ 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 JA, and an element fJ.

[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
1.1

Let π:k[x1,,xn]A be the quotient map, let J=π1(J), and choose f~k[x1,,xn] with π(f~)=f. Because A/Jk[x1,,xn]/J and J is radical, J is a radical ideal of the polynomial ring.

givenchoose
2.1

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

L1step 1.1assume-contracontradiction
3.1

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

L2step 2.1discharge-contradiction

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