Tensor Products and Exact Sequences

Two constructions dominate the homological side of module theory. The tensor product $M \otimes_{R} N$ manufactures a module in which one can form products $m \otimes n$ of elements from two modules, and it is the right tool for extending scalars — enlarging the ring a module is defined over. Exact sequences are the language for the reverse question: how a module $B$ is stitched together from a submodule $A$ and the quotient $C = B / A$ , and how much of $B$ that data determines. Both are governed by universal properties, and the interplay between them — which modules make $Hom$ and $\otimes$ preserve exact sequences — produces the classes of projective, injective, and flat modules.¹

Extension of scalars

Suppose $R$ is a subring of $S$ (with $1_{R} = 1_{S}$ ). Any $S$ -module $N$ is automatically an $R$ -module: the smaller ring already acts. This is restriction of scalars, and it is free. The reverse — taking an $R$ -module and enlarging its action to $S$ — usually fails outright.

For example, $Z$ is a $Z$ -module but cannot be made a $Q$ -module: if it could, $\frac{1}{2} \cdot 1 = z$ would be an integer with $z + z = 1$ . Yet $Z$ embeds in the $Q$ -module $Q$ . By contrast $Z /2 Z$ admits no nonzero map to any $Q$ -module at all: every nonzero element of a $Q$ -vector space has infinite order, so a finite-order element must map to $0$ . The tensor product $S \otimes_{R} N$ is the best possible $S$ -module receiving a map from $N$ ; applied to $R = Z$ , $S = Q$ , it returns $Q$ from $Z$ and $0$ from $Z /2 Z$ , reproducing both behaviors.²

The tensor product

Let $M$ be a right $R$ -module and $N$ a left $R$ -module. Start with the free abelian group on the set $M \times N$ — all formal finite sums of pairs $(m, n)$ , with no relations. To make $\otimes$ additive in each slot and balanced across the ring, quotient by the subgroup generated by

(m_{1} + m_{2}, n) - (m_{1}, n) - (m_{2}, n), (m, n_{1} + n_{2}) - (m, n_{1}) - (m, n_{2}), (m r, n) - (m, r n) .

Care is required: $m \otimes n$ is a coset, so distinct pairs can give equal tensors, and a tensor's expression as a sum of simple tensors is not unique. The third relation is the balancing law that lets a scalar slide across the $\otimes$ sign.

The map $t : M \times N \to M \otimes_{R} N$ , $t (m, n) = m \otimes n$ , is the universal balanced map.

This theorem is the practical tool. To define a map out of $M \otimes_{R} N$ , one need not check that a rule on simple tensors is well defined on cosets; it suffices to check that the corresponding map on ordered pairs is balanced, and the universal property supplies $Φ$ .³

Every balanced map out of

M \times N

factors uniquely through the tensor product; the tensor product is the universal target that converts balancedness into a single linear map.

Module structure and the commutative case

The abelian group $M \otimes_{R} N$ becomes a module when one side carries a second ring action compatibly. If $M$ is an $(S, R)$ -bimodule — a left $S$ -module and right $R$ -module with $s (m r) = (s m) r$ — then $M \otimes_{R} N$ is a left $S$ -module via $s (\sum m_{i} \otimes n_{i}) = \sum (s m_{i}) \otimes n_{i}$ . When $R$ is commutative, every $R$ -module carries its standard $(R, R)$ -bimodule structure ( $m r = r m$ ), so $M \otimes_{R} N$ is always an $R$ -module, and $t$ becomes $R$ -bilinear:

Computing tensor products

The relations force collapses that are invisible until computed.

Nothing changes over $R$ itself. $R \otimes_{R} N ≅ N$ ; in particular $Z \otimes_{Z} A = A$ for any abelian group.
Cyclic groups. $Z / m Z \otimes_{Z} Z / n Z ≅ Z / d Z$ , where $d = gcd (m, n)$ . For coprime $m, n$ the product is $0$ — a tensor-level shadow of the Chinese remainder theorem.
Torsion meets divisible. $Q \otimes_{Z} A = 0$ for any finite abelian $A$ : writing $q = (q / n) n$ and using $na = 0$ gives $q \otimes a = 0$ . This is the vanishing that made extension of scalars from $Z$ to $Q$ kill finite groups.
Free modules. $S \otimes_{R} R^{n} ≅ S^{n}$ ; scalars extend on free modules by extending on each coordinate. Over fields, $K \otimes_{F} F^{n} ≅ K^{n}$ , so a real vector space complexifies to a complex one of the same dimension.

Worked example. Compute $Z /8 Z \otimes_{Z} Z /12 Z$ .

Every simple tensor collapses onto the single generator $1 \otimes 1$ : $a \otimes b = ab (1 \otimes 1)$ by pulling the integers $a$ and $b$ across the tensor sign. The generator is killed by both moduli,

8 (1 \otimes 1) = 8 \otimes 1 = 0 \otimes 1 = 0, 12 (1 \otimes 1) = 1 \otimes 12 = 1 \otimes 0 = 0,

so its order divides $gcd (8, 12) = 4$ . To rule out further collapse, exhibit a balanced map onto $Z /4$ : the rule $φ (a, b) = ab mod 4$ is well defined ( $a \mapsto a + 8$ changes $ab$ by $8 b \equiv 0 mod 4$ , and $b \mapsto b + 12$ changes it by $12 a \equiv 0$ ), additive in each slot, and balanced. The universal property gives a homomorphism $Φ : Z /8 \otimes_{Z} Z /12 \to Z /4$ with $Φ (1 \otimes 1) = 1$ , an element of order $4$ . Hence $1 \otimes 1$ has order exactly $4$ and

Z /8 Z \otimes_{Z} Z /12 Z ≅ Z /4 Z .

The same two-step argument — collapse to $1 \otimes 1$ , then a balanced map to certify the order — proves the general case $Z / m \otimes_{Z} Z / n ≅ Z / gcd (m, n)$ .

The companion computation for $Hom$ lands on the same answer: homomorphism groups between cyclic modules are also cyclic of gcd order.

Operations on tensors

Tensor products interact with maps and with direct sums as expected.

Tensor of homomorphisms. For $φ : M \to M^{'}$ and $ψ : N \to N^{'}$ there is a unique $φ \otimes ψ : M \otimes_{R} N \to M^{'} \otimes_{R} N^{'}$ with $(φ \otimes ψ) (m \otimes n) = φ (m) \otimes ψ (n)$ .
Associativity. $(M \otimes_{R} N) \otimes_{T} L ≅ M \otimes_{R} (N \otimes_{T} L)$ , so an iterated tensor product $M_{1} \otimes \dots \otimes M_{n}$ is unambiguous and is the universal object for multilinear maps.
Distributivity. $(M \oplus M^{'}) \otimes_{R} N ≅ (M \otimes_{R} N) \oplus (M^{'} \otimes_{R} N)$ , and likewise in the second variable — tensoring commutes with direct sums.

Exact sequences

The extension problem runs opposite to the tensor construction: given modules $A$ and $C$ , build the modules $B$ that contain a copy of $A$ with $B / A ≅ C$ . The bookkeeping device is exactness.

Unwinding the endpoints: exactness at $A$ says $ψ$ is injective; exactness at $C$ says $φ$ is surjective; exactness at $B$ says $im ψ = ker φ$ . So a short exact sequence encodes " $A$ is a submodule of $B$ (via $ψ$ ) and $C$ is the quotient" — $B$ is an extension of $C$ by $A$ .⁴

A short exact sequence:

A

injects into

B

as a submodule,

B

surjects onto the quotient

C

, and the image of the first map coincides with the kernel of the second.

Worked example. Check that

0 ⟶ Z /2 ψ Z /4 φ Z /2 ⟶ 0, ψ (x) = 2 x, φ (y) = y mod 2,

is a short exact sequence.

Homomorphisms. $ψ$ is well defined: $x$ and $x + 2$ have images $2 x$ and $2 x + 4 = 2 x$ in $Z /4$ . $φ$ is reduction mod $2$ , well defined since $4$ is a multiple of $2$ .
Exact at $Z /2$ (left). $ker ψ = 0$ : $ψ (1) = 2 \neq = 0$ in $Z /4$ .
Exact at $Z /4$ . $im ψ = {0, 2}$ , and $ker φ = {y : y even} = {0, 2}$ . Equal.
Exact at $Z /2$ (right). $φ (1) = 1$ , so $φ$ is onto.

So $Z /4$ is an extension of $Z /2$ by $Z /2$ . The sequence does not split: a section $μ : Z /2 \to Z /4$ needs $μ (1)$ of additive order dividing $2$ , hence $μ (1) \in {0, 2}$ , and then $φ (μ (1)) = 0 \neq = 1$ . Consistently, $Z /4 ≆ Z /2 \oplus Z /2$ — the left side has an element of order $4$ . The direct sum $Z /2 \oplus Z /2$ is a different extension of the same two modules, so the ends of a short exact sequence do not determine the middle.

Splitting

The simplest extension is the direct sum $B = A \oplus C$ , in which $C$ sits inside $B$ as a complement to $A$ . An extension of this shape is split.

For modules the two one-sided conditions are equivalent, unlike for groups, because the underlying groups are abelian: where a split extension of groups is only a semidirect product, a split extension of modules is a genuine direct sum. Not every sequence splits. The sequence $0 \to Z n Z \to Z / n Z \to 0$ , with the first map multiplication by $n$ , has no section, since there is no nonzero homomorphism $Z / n Z \to Z$ . It is a nontrivial extension of $Z / n Z$ by $Z$ .

A split sequence admits a section

μ

picking out a complement, so

B ≅ A \oplus C

; the non-split sequence at right (multiplication by

n

) has no section back from the quotient.

Projective, injective, and flat modules

Applying $Hom_{R} (D, -)$ or $D \otimes_{R} -$ to a short exact sequence usually produces a sequence that is exact only at one end. The modules $D$ for which exactness is fully preserved are the projective, injective, and flat ones.

Free modules are projective (they are their own instance of the last condition), and the lifting property is the working characterization: because $P$ has enough freeness to have no relations obstructing a lift, a map defined into a quotient can be raised to a map into the module above.⁵

The lifting property defining a projective module: given a surjection onto

N

and any map from

P

into

N

, the map lifts to

P \to M

making the triangle commute.

Injective modules are the dual: $Q$ is injective if every map into $Q$ from a submodule extends to the whole module, equivalently if $Hom_{R} (-, Q)$ preserves exactness. Baer's criterion tests this on ideals alone, and over a principal ideal domain injective coincides with divisible ( $r Q = Q$ for all $r \neq = 0$ ): $Q$ and $Q / Z$ are injective $Z$ -modules, while $Z$ is not.

Flat modules measure the tensor product instead. The functor $A \otimes_{R} -$ is always right exact; $A$ is flat when it is also left exact, so that tensoring with $A$ carries injections to injections. Projective modules are flat; $Q / Z$ is injective but not flat, and $Z \oplus Q$ is flat but neither projective nor injective, so the three classes are genuinely different.

Module class	Preserves exactness of	Fails at	PID / field example
Projective	$Hom_{R} (P, -)$	— (right end)	free modules; over a field, all
Injective	$Hom_{R} (-, Q)$	— (right end)	$Q$ , $Q / Z$ divisible
Flat	$A \otimes_{R} -$	— (left end)	$Q$ ; over a field, all

Over a field all three collapse to everything, since every vector space is free and every short exact sequence of vector spaces splits. The distinctions are visible only over rings with genuine ideal structure, and measuring the failure of exactness in general is the entry point to homological algebra.

Dummit & Foote, Abstract Algebra, 3rd ed., §10.4 — the tensor product of modules, constructed as a quotient of the free abelian group on $M \times N$ by the balancing relations. ↩
Dummit & Foote, §10.4 — extension and restriction of scalars, and Corollary 9: $N / ker t$ is the largest quotient of $N$ embeddable in an $S$ -module, with the $Z \to Q$ examples. ↩
Dummit & Foote, §10.4, Theorem 10 and Corollary 12 — the universal property of the tensor product with respect to balanced maps, and the bilinear form of it over a commutative ring. ↩
Dummit & Foote, §10.5, Definition and Corollary 23 — exact and short exact sequences, and Proposition 25/26 on splitting via sections and retractions. ↩
Dummit & Foote, §10.5, Proposition 30 and Corollary 31 — the equivalent characterizations of a projective module, including the lifting property and being a direct summand of a free module; injective (Proposition 34, Baer's criterion) and flat modules are treated later in the same section. ↩