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$M \otimes_{R} R / I ≅ M / I M$ naturally

Statement

Let $R$ be a commutative ring, $I ⊴ R$ an ideal, and $M$ an $R$ -module. There is a natural $R$ -module isomorphism

$M \otimes_{R} (R / I) ≅ M / I M, m \otimes (r + I) ⟼ r m + I M .$

Both sides also carry the induced $R / I$ -module structure, and the isomorphism is $R / I$ -linear. For $I = 0$ it is the tensor-unit isomorphism, while for $I = R$ both sides are zero.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I$ , and an $R$ -module $M$ .

[L1]

Tensoring an exact sequence ending in zero preserves exactness at the two rightmost terms (Tensoring is right exact).

[L2]

The tensor-unit isomorphism sends $m \otimes r$ to $r m$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[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]

The quotient module $M / I M$ consists of cosets with the induced scalar action (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L5]

A homomorphism that kills a submodule factors uniquely through the quotient module (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

[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

1.1givenL1L6

The sequence $I \to R \to R / I \to 0$ is exact. Tensoring on the right by $M$ and applying [L1] gives the exact sequence $I \otimes_{R} M \to R \otimes_{R} M \to (R / I) \otimes_{R} M \to 0$ . The symmetry isomorphisms of [L6] carry it termwise to $M \otimes_{R} I \to M \otimes_{R} R \to M \otimes_{R} (R / I) \to 0$ , and since $σ$ commutes with the induced maps on elementary tensors, that sequence is exact too.

2.1step 1.1L2L3

Under [L2], the image of $M \otimes_{R} I \to M \otimes_{R} R ≅ M$ consists exactly of finite sums $i m$ , hence is $I M$ by [L3].

3.1step 1.1step 2.1L4L5

Exactness in step 1.1 identifies $M \otimes_{R} (R / I)$ with the cokernel of the first map, which by step 2.1 is $M / I M$ ; [L5] gives the resulting isomorphism.

4.1step 3.1L3L4algebra

Tracing $m \otimes (r + I)$ through the quotient gives $r m + I M$ . Multiplication by an element of $I$ acts as zero on both sides, so the map and its inverse are $R / I$ -linear.

5.1step 4.1L2L3L4

If $I = 0$ , step 4.1 is [L2]. If $I = R$ , then [L3] gives $I M = M$ and $R / I = 0$ , so both sides are zero.

6.1step 3.1step 4.1step 5.1∎

This proves the natural $R$ -linear and $R / I$ -linear isomorphism in every boundary case.

$M \otimes_{R} R / I ≅ M / I M$ naturally

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

M⊗RR/I≅M/IMM⊗R​R/I≅M/IM naturally

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$M \otimes_{R} R / I ≅ M / I M$ naturally