Following the earlier article on tensor products of vector spaces, we will now look at tensor products of modules over a ring *R*, not necessarily commutative. It turns out we have to distinguish between left and right modules now. Indeed recall the isomorphism from earlier: for vector spaces. If we replace *V* and *W* by left *R*-modules, then *V** becomes a right module, so this hints that we should consider a tensor product between a right module and a left one.

Definition. Let M be a right R-module, N a left R-module and X an abelian group. Abilinearmap of modules is a map such that

- when we fix n∈N, B(-, N): M→X is additive;
- when we fix m∈M, B(m, -): N→X is additive;
- B(mr, n) = B(m, rn) for any r∈R, m∈M, n∈N.
The

tensor productof M and N, denoted is an abelian group together with a bilinear map such that the following universal property holds:

- for any bilinear map there is a unique additive map such that
As before, the element for any is called a

pure tensor.

The universal property again guarantees that the tensor product is unique if it exists.

**Proof of Existence**

The proof is identical to earlier. Let *T* be the free abelian group with basis Take the subgroup *U* generated by elements of the form:

for . Now our desired group is *T*/*U*, and is given by so *m*⊗*n* is the image of in *T*/*U*. ♦

**Note**

A bilinear map *M* × *N* → *X* corresponds to a homomorphism of right modules [ Recall that since *N* is a left *R*-module, Hom(*N*, *X*) becomes a right *R*-module. ] Hence, the above universal property can also be written as:

Often, we would like the tensor product to be a module instead of merely an abelian group:

Proposition. Let R, S be rings. If M is an (R, S)-bimodule and N is a left S-module, then is a left R-module satisfyingThe bilinear map is also R-linear in M, i.e.

**Proof**

By definition is an abelian group. For each *r*∈*R*, consider the map

The map is *R*-bilinear (e.g. *f _{r}*(

*ms*,

*n*) = (

*r*(

*ms*))⊗

*n*= (

*rm*)

*s*⊗

*n*=

*rm*⊗s

*n*=

*f*(

_{r}*m*,

*sn*)). So it induces taking One now checks that this turns into a left

*R*-module such that is

*R*-linear in

*M*. ♦

Properties.

- For any left R-module M, as left R-modules.
- For any right R-module M, (R, S)-bimodule N, and left S-module P, we have
- For any right R-module M, and left R-modules , we have

**Sketch of Proof**.

We’ll briefly sketch the proof of the second claim. Fixing *m*∈*M*, the map taking is bilinear over *S*, so it induces a group homomorphism which is easily checked to be left *R*-linear. Check that this is right *R*-linear in *r*, so we obtain a group homomorphism taking (*m*⊗*n*)⊗*p* to *m*⊗(*n*⊗*p*). ♦

## Base Extension

Suppose *S* is an *R*-algebra and *M* a left *R*-module. Since *S* is an (*S*, *R*)-bimodule, the tensor product is now a left *S*-module. This is called the **base extension** of *M*. Since we have:

Proposition. Base extension commutes, i.e. ifTis anS-algebra, then asT-modules.

Base extension satisfies the following universal property.

Universal Property of Base Extension. The canonical map taking is R-linear.For any left S-module N and R-module homomorphism there is a unique S-module homomorphism such that

Thus for any S-module N:

**Proof**

The first statement is obvious. For the second, let us prove existence of *f*. The map mapping is *R*-bilinear so it induces satisfying This is *S*-linear since

and it satisfies Uniqueness follows from the fact that *M* generates *M _{S}* as an

*S*-module.

**Example**

Suppose *R* = **R**, the real field and *V* is the space of polynomials with real coefficients and degree ≤ 2. Then is naturally identified with the space of polynomials with complex coefficients and degree ≤ 2.

## For Group Representations

Tensor products are useful in group representations in two different ways. First, suppose *V* and *W* are *K*[*G*]-modules for a field *K* and finite group *G*. Then (note: the tensor product over *K*, not the group ring!) becomes a *K*[*G*]-module as well. Indeed, each *g* induces isomorphisms *V*→*V* and *W*→*W* and thus a bilinear map *V*×*W*→*V*⊗*W* taking (*v*, *w*) to *gv*⊗*gw*. This in turn gives us a linear map *V*⊗*W* → *V*⊗*W* taking *v*⊗*w* to *gv*⊗*gw*. [ It is an isomorphism since *g*^{-1} induces the inverse. ]

Another use of tensor product is via the **induced representation**. If *H* ⊆ *G* is a subgroup and *V* is a *K*[*H*]-module, then one can define: to obtain a representation of *G*. The universal property of base extension gives us: for any *K*[*G*]-module *W*,

This is the **Frobenius reciprocity theorem** in representation theory and is often written as:

## Tensor Product is Right Exact

We have the following

Theorem. If is an exact sequence of left R-modules, then for any right R-module M, we get an exact sequence of abelian groups:

**Proof**

We use the property from above:

Now, for any abelian group *X*, since is exact by condition, the following is exact since Hom is left-exact:

Again since Hom is left-exact, we have:

which is precisely the same as saying:

is exact for all *X*. Thus, is exact. ♦

**Example**

Let *I* be a two-sided ideal of *R*, so *R*/*I* becomes an *R*-algebra via the canonical map *R*→*R*/*I*. We claim that for any left *R*-module *M*, base extension gives

Indeed, consider the map taking (*r*+*I*, *m*) to *rm*+*IM*. This map is well-defined on *R*/*I* and *R*-bilinear so it induces a map Conversely, take the map where Everything in *IM* maps to 0, so this factors through It’s easy to check that both maps are mutually inverse.

**Note**

From our example above, it is easy to find examples where the tensor product is not left-exact. For example, consider 0 → 2**Z** → **Z**. Tensoring with **Z**/2 is the same as taking *M* to *M*/2*M*; so we obtain 0 → 2**Z**/4**Z** → **Z**/2**Z** which is not exact since the second map takes everything to 0.