CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For flat $M$ , one has $I M \cap J M = (I \cap J) M$

Statement

Let $R$ be a commutative ring, let $I, J \subseteq R$ be ideals, and let $M$ be a flat $R$ -module. Then

$I M \cap J M = (I \cap J) M .$

Facts & Assumptions

Given: Ideals $I, J$ of a commutative ring $R$ and a flat $R$ -module $M$ .

[L1]

Tensoring an exact sequence with a flat module preserves exactness (Flat and faithfully flat modules and ring homomorphisms).

[L2]

There is a natural isomorphism $M \otimes_{R} (R / I) ≅ M / I M$ , and similarly for $J$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L3]

The submodule $I M$ consists of finite sums of products $i m$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

[L4]

Exactness at a module is equality of the incoming image and outgoing kernel (Exact sequences and short exact sequences of modules).

[L5]

Tensor products commute with direct sums (Tensor products commute with arbitrary direct sums).

[L6]

Over a commutative ring the natural symmetry $σ_{A, B} : A \otimes_{R} B \to B \otimes_{R} A$ , $a \otimes b \mapsto b \otimes a$ , is an isomorphism (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

Proof

technique · direct

1.1givenL4

The sequence $0 \to I \cap J \to R \to R / I \oplus R / J$ , whose last displayed map sends $r$ to $(r + I, r + J)$ , is exact because its kernel is exactly $I \cap J$ .

2.1step 1.1L1L5

Tensor step 1.1 with the flat module $M$ . By [L1], the resulting sequence is exact and begins $0 \to (I \cap J) \otimes_{R} M \to M \to (R / I \otimes_{R} M) \oplus (R / J \otimes_{R} M)$ , using [L5].

3.1step 2.1L2L6

The symmetry of [L6] identifies $R / I \otimes_{R} M$ with $M \otimes_{R} (R / I)$ and likewise for $J$ , so [L2] applies and the last map in step 2.1 is $m \mapsto (m + I M, m + J M)$ , whose kernel is $I M \cap J M$ .

3.2step 2.1L3

The image of $(I \cap J) \otimes_{R} M \to M$ is the set of finite sums of products $a m$ with $a \in I \cap J$ , namely $(I \cap J) M$ by [L3].

4.1step 3.1step 3.2L4∎

Exactness in step 2.1 identifies the image in step 3.2 with the kernel in step 3.1, proving $(I \cap J) M = I M \cap J M$ . The calculation also covers $I = 0$ , $J = 0$ , $I = R$ , or $J = R$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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

Stacks Project, Lemma 10.39.2 (standard reference, not scraped)

For flat MM, one has IM∩JM=(I∩J)MIM∩JM=(I∩J)M