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Every Zariski-closed subset has a unique radical defining ideal

Statement

Assume the Axiom of Choice.

Let R be a commutative ring and let ZSpec(R) be Zariski-closed. Then I(Z):=pZp is a radical ideal, Z=V(I(Z)), and if Z=V(J) for an ideal J, then I(Z)=J. In particular, Z has a unique radical defining ideal.

Facts & Assumptions

Given: A commutative ring R, a Zariski-closed subset ZSpec(R), and the Axiom of Choice.

[L1]

For every ideal J, its radical is the intersection of the prime ideals containing J (The radical of an ideal is the intersection of the prime ideals containing it).

[L2]

Two vanishing sets are equal exactly when the radicals of their defining ideals are equal (Vanishing sets detect radicals).

[A1]

Because Z is Zariski-closed, there exists an ideal JR with Z=V(J).

Proof

technique · direct
1.1

Choose J with Z=V(J) as in [A1]. Then I(Z)=pV(J)p=J by [L1]. In particular, I(Z) is radical.

L1A1
2.1

Since step 1.1 gives I(Z)=J, fact [L2] yields V(I(Z))=V(J)=V(J)=Z.

L2step 1.1A1
2.2

If also Z=V(K) for an ideal K, then V(K)=V(J), so [L2] gives K=J=I(Z). Thus any radical ideal defining Z equals I(Z).

L2step 1.1A1
3.1

Therefore every Zariski-closed subset has the unique radical defining ideal I(Z).

step 2.1step 2.2

Depends on

Used by

Dependency tree · two levels

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Sources