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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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naive homogenization adds component

Statement refuted

Raw homogenization of arbitrary generators always gives the projective closure.

Counterexample

Given: I=(x−1,xy)⊆k[x,y], with projective coordinates [Z:X:Y].

1.1

The raw homogenized generators are X−Z and XY, whose common projective zeros include [0:0:1]: both polynomials vanish there.

givenalgebra
2.1

Since xy−y(x−1)=y, one has I=(x−1,y); hence V(I)={(1,0)} and its closure is V+(X−Z,Y), which does not contain [0:0:1].

step 1.1algebra
3.1

Thus raw generators introduce a spurious point at infinity; saturation removes it and is necessary.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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