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. A bilinear map 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 product of 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. ♦
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 satisfying
The bilinear map is also R-linear in M, i.e.
By definition is an abelian group. For each r∈R, consider the map
The map is R-bilinear (e.g. fr(ms, n) = (r(ms))⊗n = (rm)s⊗n = rm⊗sn = fr(m, sn)). So it induces taking One now checks that this turns into a left R-module such that is R-linear in M. ♦
- 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). ♦
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. if T is an S-algebra, then as T-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:
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 MS as an S-module.
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:
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. ♦
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.
From our example above, it is easy to find examples where the tensor product is not left-exact. For example, consider 0 → 2Z → Z. Tensoring with Z/2 is the same as taking M to M/2M; so we obtain 0 → 2Z/4Z → Z/2Z which is not exact since the second map takes everything to 0.