Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-27
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The radical of a product of ideals

Statement

Let R be a commutative ring and let I,J⊴R be ideals. Then IJ=I∩J=I∩J.

Facts & Assumptions

Given: A commutative ring R and ideals I,J⊴R.

[L1]

The radical of a finite intersection is the intersection of the radicals (The radical of a finite intersection).

[L2]

The product ideal IJ is generated by finite sums of products ij with i∈I and j∈J (The sum I+J and product IJ of two-sided ideals).

Proof

technique · direct
1.1L1L2given

Every generator ij of IJ lies in both I and J, so IJ⊆I∩J. Therefore IJ⊆I∩J=I∩J by [L1].

1.2L2choosealgebra

If x∈I∩J, choose m,n≥1 with xm∈I and xn∈J. Then xm+n=xmxn∈IJ, so x∈IJ.

2.1step 1.1step 1.2L1∎

Step 1.1 gives IJ⊆I∩J, step 1.2 gives the reverse inclusion, and [L1] identifies that common ideal with I∩J.

Depends on

Used by

Dependency tree · two levels

4 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