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Tensoring is right exact

Statement

Let $R$ be a commutative ring, let

$A \overset{f}{\to} B \overset{g}{\to} C ⟶ 0$

be an exact sequence of $R$ -modules, and let $N$ be an $R$ -module. Then

$A \otimes_{R} N \overset{f \otimes 1}{\to} B \otimes_{R} N \overset{g \otimes 1}{\to} C \otimes_{R} N ⟶ 0$

is exact. Thus tensoring preserves cokernels and surjections, but no injectivity at the left is asserted.

Facts & Assumptions

Given: An exact sequence $A \overset{f}{\to} B \overset{g}{\to} C \to 0$ and an $R$ -module $N$ over a commutative ring $R$ .

[L1]

Exactness means that $g$ is surjective and $im f = \ker g$ (Exact sequences and short exact sequences of modules).

[L2]

Module homomorphisms induce tensor homomorphisms with $(f \otimes 1) (a \otimes n) = f (a) \otimes n$ and functorial composition (Module homomorphisms induce tensor-product homomorphisms functorially).

[L3]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L4]

An elementary-tensor formula descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

[L5]

For a submodule $K \leq X$ , the quotient module $X / K$ has cosets $x + K$ and its induced scalar action (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L6]

A module homomorphism that kills $K$ factors uniquely through $X / K$ (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

[L7]

Tensor products over $R$ carry the scalar action $r (x \otimes n) = (r x) \otimes n$ (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

[L8]

Every tensor is a finite sum of elementary tensors (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums).

Proof

technique · direct

1.1

The map $g \otimes 1$ is surjective: by [L8], every tensor is a finite sum of elementary tensors $c \otimes n$ , and [L1] supplies $b \in B$ with $g (b) = c$ , so $c \otimes n = (g \otimes 1) (b \otimes n)$ .

givenL1L2L8choose

1.2

Functoriality gives $(g \otimes 1) (f \otimes 1) = (g f) \otimes 1 = 0$ , so $im (f \otimes 1) \subseteq \ker (g \otimes 1)$ .

L1L2

2.1

Let $Q = (B \otimes_{R} N) / im (f \otimes 1)$ . The image is a submodule because [L2] and [L7] make $f \otimes 1$ $R$ -linear. Since $g \otimes 1$ kills it, [L6] gives $\overline{g} : Q \to C \otimes_{R} N$ .

step 1.2L2L5L6L7

3.1

For $c \in C$ and $n \in N$ , choose any $b \in B$ with $g (b) = c$ and set $q (c, n) = [b \otimes n] \in Q$ . If $b^{'}$ is another lift, then $b - b^{'} \in \ker g = im f$ by [L1], so $[b \otimes n] = [b^{'} \otimes n]$ ; hence $q$ is well defined.

L1step 2.1choose

4.1

The function $q : C \times N \to Q$ is bilinear: lifts of sums may be taken as sums of lifts, scalar multiples as scalar multiples, and the tensor relations give the required equalities. By [L3] and [L4] it induces $h : C \otimes_{R} N \to Q$ .

step 3.1L3L4L7

5.1

For $b \otimes n$ , one has $h \overline{g} ([b \otimes n]) = h (g (b) \otimes n) = [b \otimes n]$ , while for $c \otimes n$ and a lift $b$ one has $\overline{g} h (c \otimes n) = \overline{g} ([b \otimes n]) = c \otimes n$ ; generators and uniqueness make $h$ and $\overline{g}$ inverse.

step 2.1step 3.1step 4.1L3

6.1

Since $\overline{g}$ is an isomorphism, the kernel of $g \otimes 1$ is exactly the submodule quotiented out in step 2.1, namely $im (f \otimes 1)$ . Together with step 1.1, this is right exactness.

step 1.1step 2.1step 5.1∎

Depends on

Used by

M⊗_RR/I≅ M/IM naturally Corollary
Flat and faithfully flat modules and ring homomorphisms Definition
False: tensoring preserves injections False statement
A short exact sequence with flat quotient remains short exact after tensoring Theorem
Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)
C. Dennis, Week 4 on tensor products and flatness (standard reference, not scraped)