Knowlpedia

Tensor product is right exact

For fixed N, the functor -⊗_R N preserves cokernels (exactness on the right); its failure to be left exact is measured by Tor.

Let RR be a .

Fix a left RR-module NN. Then the functor

(−)⊗RN:(Mod-R)→Ab(-)\otimes_R N : (\mathrm{Mod}\text{-}R) \to \mathrm{Ab}

from right RR-modules to abelian groups is the tensor product functor (see ).

Equivalent characterizations

(Equivalently, if MM is a fixed right RR-module, then M⊗R(−):R-Mod→AbM\otimes_R(-): R\text{-}\mathrm{Mod}\to\mathrm{Ab} is also right exact.)

Equivalently: tensoring preserves cokernels and epimorphisms.

Statement (right exactness)

If

A′→uA→vA′′→0A' \xrightarrow{u} A \xrightarrow{v} A'' \to 0

is an exact sequence of right RR-modules, then

A′⊗RN→u⊗1A⊗RN→v⊗1A′′⊗RN→0A'\otimes_R N \xrightarrow{u\otimes 1} A\otimes_R N \xrightarrow{v\otimes 1} A''\otimes_R N \to 0

is exact.

Failure of left exactness and Tor

Tensor need not preserve kernels (i.e. it need not preserve injections). For a

0→A′→A→A′′→0,0 \to A' \to A \to A'' \to 0,

there is a natural exact sequence

Tor1R(A′′,N)→A′⊗RN→A⊗RN→A′′⊗RN→0,\mathrm{Tor}^R_1(A'',N)\to A'\otimes_R N \to A\otimes_R N \to A''\otimes_R N \to 0,

where Tor1R\mathrm{Tor}^R_1 is defined in and arises from .

In particular, NN is iff (−)⊗RN(-)\otimes_R N is exact (iff Tor1R(−,N)=0\mathrm{Tor}^R_1(-,N)=0).

Examples

Example 1: Tensor is not left exact (over Z\mathbb Z)

Consider the injective map of Z\mathbb Z-modules

0→Z→×nZ0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z

with cokernel Z/n\mathbb Z/n. Tensor with Z/n\mathbb Z/n:

0→Z⊗Z/n→×nZ⊗Z/n.0 \to \mathbb Z\otimes \mathbb Z/n \xrightarrow{\times n} \mathbb Z\otimes \mathbb Z/n.

Since Z⊗Z/n≅Z/n\mathbb Z\otimes \mathbb Z/n \cong \mathbb Z/n and multiplication by nn on Z/n\mathbb Z/n is the zero map, the induced map is not injective. Concretely,

0→Z/n→0Z/n0 \to \mathbb Z/n \xrightarrow{0} \mathbb Z/n

fails exactness on the left, and the defect is detected by

Tor1Z(Z/n,Z/n)≅Z/n≠0.\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,\mathbb Z/n)\cong \mathbb Z/n \neq 0.

Example 2: Tensor with a flat module is exact (localization)

Over Z\mathbb Z, the module Q\mathbb Q is a localization and hence flat. Tensor the short exact sequence

0→Z→×nZ→Z/n→00 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0

with Q\mathbb Q:

0→Q→×nQ→(Z/n)⊗Q→0.0 \to \mathbb Q \xrightarrow{\times n} \mathbb Q \to (\mathbb Z/n)\otimes \mathbb Q \to 0.

Here ×n:Q→Q\times n:\mathbb Q\to\mathbb Q is an isomorphism, so (Z/n)⊗Q=0(\mathbb Z/n)\otimes \mathbb Q=0 and the tensored sequence remains exact.

Example 3: Tensor with a free module is exact

If N≅R⊕rN\cong R^{\oplus r} is free, then

A⊗RN≅A⊕r,A\otimes_R N \cong A^{\oplus r},

so (−)⊗RN(-)\otimes_R N is a finite direct sum of copies of the identity functor and is therefore exact.