Exactness properties of Hom and tensor - Imported Knowlpedia Content

Let $R$ be a ring and consider the functors on $R$ -modules:

$\operatorname{Hom}_R(M,-)$ (covariant in the second variable),
$-\otimes_R N$ ,
$\operatorname{Hom}_R(-,N)$ (contravariant in the first variable).

Basic exactness statements

Hom is left exact

For any fixed $M$ , the functor $\operatorname{Hom}_R(M,-)$ is left exact: from a short exact sequence $0\to A\to B\to C\to 0$ one gets an exact sequence

0 \to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C),

but surjectivity of $\operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C)$ can fail in general.

Projectivity criterion. $\operatorname{Hom}_R(M,-)$ is exact (i.e. also right exact) iff $M$ is projective.

Tensor is right exact

For any fixed $N$ , the functor $-\otimes_R N$ is right exact: from $A\to B\to C\to 0$ exact one gets

A\otimes_R N \to B\otimes_R N \to C\otimes_R N \to 0

exact, but injectivity of $A\otimes_R N\to B\otimes_R N$ can fail.

Flatness criterion. $-\otimes_R N$ is exact (i.e. also left exact) iff $N$ is flat.

Contravariant Hom is left exact (in the contravariant sense)

For fixed $N$ , the functor $\operatorname{Hom}_R(-,N)$ sends short exact sequences $0\to A\to B\to C\to 0$ to exact sequences

0 \to \operatorname{Hom}_R(C,N)\to \operatorname{Hom}_R(B,N)\to \operatorname{Hom}_R(A,N),

again with possible failure of surjectivity on the right.

Injectivity criterion. $\operatorname{Hom}_R(-,N)$ is exact (i.e. also right exact in this contravariant direction) iff $N$ is injective.

Link to Ext and Tor

Because Hom and tensor are only one-sided exact in general, their derived functors measure the obstruction:

$\operatorname{Ext}$ are derived from Hom,
$\operatorname{Tor}$ are derived from tensor.

See Ext and Tor as derived functors and derived-functor definitions.

Examples

Example 1 (tensor is not left exact over $\mathbb Z$ )

In $\mathbf{Ab}$ , take the injection $0\to \mathbb Z \xrightarrow{\cdot n} \mathbb Z$ . Tensor with $\mathbb Z/n$ :

\mathbb Z\otimes \mathbb Z/n \xrightarrow{\cdot n} \mathbb Z\otimes \mathbb Z/n

becomes

\mathbb Z/n \xrightarrow{0} \mathbb Z/n,

which is not injective. The “missing” left exactness is measured by

\operatorname{Tor}_1^\mathbb Z(\mathbb Z/n,\mathbb Z/n)\cong \mathbb Z/n

(see Tor).

Example 2 (flat module: $\mathbb Q$ over $\mathbb Z$ )

The $\mathbb Z$ -module $\mathbb Q$ is flat (localization of $\mathbb Z$ ). Hence tensoring any short exact sequence of abelian groups with $\mathbb Q$ preserves exactness. In particular,

\operatorname{Tor}_1^\mathbb Z(\mathbb Q, A)=0

for every abelian group $A$ .

Example 3 (Hom is not right exact unless the source is projective)

Consider the short exact sequence

0\to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \to \mathbb Z/n\to 0.

Apply $\operatorname{Hom}_\mathbb Z(\mathbb Z/n,-)$ . The resulting map

\operatorname{Hom}(\mathbb Z/n,\mathbb Z)\to \operatorname{Hom}(\mathbb Z/n,\mathbb Z/n)

fails to be surjective (indeed $\operatorname{Hom}(\mathbb Z/n,\mathbb Z)=0$ , while $\operatorname{Hom}(\mathbb Z/n,\mathbb Z/n)\cong \mathbb Z/n$ ). The obstruction is exactly

\operatorname{Ext}^1_\mathbb Z(\mathbb Z/n,\mathbb Z)\cong \mathbb Z/n,

illustrating that $\mathbb Z/n$ is not projective over $\mathbb Z$ (see Ext).