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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.

Tensor product is right exact

Let $R$ be a ring .

Fix a left $R$ -module $N$ . Then the functor

(-)\otimes_R N : (\mathrm{Mod}\text{-}R) \to \mathrm{Ab}

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

(Equivalently, if $M$ is a fixed right $R$ -module, then $M\otimes_R(-): R\text{-}\mathrm{Mod}\to\mathrm{Ab}$ is also right exact.)

Statement (right exactness)

A' \xrightarrow{u} A \xrightarrow{v} A'' \to 0

is an exact sequence of right $R$ -modules, then

A'\otimes_R N \xrightarrow{u\otimes 1} A\otimes_R N \xrightarrow{v\otimes 1} A''\otimes_R N \to 0

is exact.

Equivalently: tensoring preserves cokernels and epimorphisms.

Failure of left exactness and Tor

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

0 \to A' \to A \to A'' \to 0,

there is a natural exact sequence

\mathrm{Tor}^R_1(A'',N)\to A'\otimes_R N \to A\otimes_R N \to A''\otimes_R N \to 0,

where $\mathrm{Tor}^R_1$ is defined in Tor and arises from the long exact sequence in Tor .

In particular, $N$ is flat iff $(-)\otimes_R N$ is exact (iff $\mathrm{Tor}^R_1(-,N)=0$ ).

Examples

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

Consider the injective map of $\mathbb Z$ -modules

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z

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

0 \to \mathbb Z\otimes \mathbb Z/n \xrightarrow{\times n} \mathbb Z\otimes \mathbb Z/n.

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

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

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

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

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0

with $\mathbb Q$ :

0 \to \mathbb Q \xrightarrow{\times n} \mathbb Q \to (\mathbb Z/n)\otimes \mathbb Q \to 0.

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

Example 3: Tensor with a free module is exact

If $N\cong R^{\oplus r}$ is free, then

A\otimes_R N \cong A^{\oplus r},

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