Tensor product of modules

In mathematics, the tensor product construction may be carried out, not only for vector spaces (see tensor product), but for any pair of modules over a commutative ring, with result s third module, and for a pair of a left-module and a right-module over any ring, with result an abelian group. (These constructions may not be accessible unless you understand the vector space case). Such tensor products are important in areas of abstract algebra, homological algebra and algebraic geometry.

The universal property of the tensor product of vector spaces extends to more general situations in abstract algebra. It allows the study of bilinear or multilinear operations via linear operations. In this general setting it is the only approach to 'tensors' that can act as a guide.

In general, we start with a ring R, not necessarily commutative. Let M_R be a right R-module and _RN be a left R-module. We will denote the Cartesian product of M and N, that is, the product as sets not modules, by M×N. For any abelian group Z, we define:

Bilin_R(M,N;Z) = the set of φ : M × N → Z such that

1. for all m, m′ in M, for all n in N, φ(m + m′,n) = φ(m,n) + φ(m′,n),
2. for all m in M, for all n, n′ in N, φ(m,n + n′) = φ(m,n) + φ(m,n′),
3. for all m in M, for all n in N, for all r in R, φ(m r,n) = φ(m,r n)}

Observe that

• The set Bilin_R(M,N;Z) is an abelian group under addition; i.e., if φ, ψ are in Bilin_R(M,N;Z), then φ + ψ is in Bilin_R(M,N;Z).
• The map Z → Bilin_R(M,N;Z) is a functor from the category of abelian groups to the category of sets.

The tensor product M⊗_RN is defined to be a representing object for the functor Z → Bilin_R(M,N;Z). This is equivalent to the universal mapping property given above.

To construct the tensor product, we proceed in exactly the same way as for vector spaces; the same construction carries over without any changes.

This construction is unrelated to the direct product of M and N. The direct product need not even be defined when R is not commutative. Even if it is, the only map M×N→Z which is both linear and bilinear is the zero map!

It is possible to generalize the definition to a tensor product of any number of spaces. For example, the universal property of M₁⊗M₂⊗M₃ is that every trilinear map on M₁×M₂×M₃→Z corresponds to a unique linear map M₁⊗M₂⊗M₃→Z. The binary tensor product is associative: (M₁⊗M₂)⊗M₃ is naturally isomorphic to M₁⊗(M₂⊗M₃). The tensor product of all three may therefore be identified with either of those.

The tensor product, as defined, is an abelian group, not an R-module. In general, it is impossible to put an R-module structure on the tensor product. However, if M is an (S,R)-bimodule, then M⊗N is a left S-module, and similarly, if N is an (R,T)-bimodule, then M⊗N is a right T-module. If M and N each have bimodule structures as above, then M⊗N is an (S,T)-bimodule. In particular, if R is a commutative ring, then M⊗N will always be an R-module.

If {m_i}_i∈I and {n_j}_j∈J are generating sets for M and N, respectively, then {m_i⊗n_j}_i∈I,j∈J will be a generating set for M⊗N. Because the tensor product is left exact, not exact, this may not be a minimal generating set, even if the original generating sets are minimal. However, if the tensor products are taken over a field, then we are in the case of vector spaces as above, and if the two given generating sets are bases, we will get a basis for M⊗N.

If S and T are commutative R-algebras, then S⊗T will be a commutative R-algebra as well, with the multiplication map defined by (m₁⊗m₂)(n₁⊗n₂)=(m₁n₂⊗m₂n₂) and extended by linearity. In this setting, the tensor product become a fibered coproduct in the category of R-algebras. Note that any commutative ring is a Z-algebra, so we may always take M⊗_ZN.

It is also possible to generalize the definition to tensor products of modules over the same ring. If the ring is non-commutative, we'll need to be careful about distinguishing right modules and left modules. We will write _RM for a left module, and M_R for a right module. If a module M has both a left module structure over a ring R and a right module structure over a ring S, and in addition for every m in M, r in R and s in S we have r(ms) = (rm)s, then we will say M is a bimodule, and will denote it by _RM_S. Note that every left-module is a bimodule with Z acting by mn = m + m + ... + m as the right ring, and vice versa.

When defining the tensor product, we need to be careful about the ring: most modules can be considered as modules over several different rings or over the same ring with a different actions of the ring on the module elements.

The most general form of the tensor product definition is as follows: let M_R and _RN be a right and a left module, respectively. Their tensor product over R is an abelian group P together with an R-bilinear operator T: M × N → P such that for every R-bilinear operator B: M × N → O there is a unique group homomorphism L: P → O such that L o T = B. P need not be a module over R. However, if _S1M_R is an S₁-R-bimodule, then there is a unique left S₁-module structure on P which is compatible with T. Similarly, if _RM_S2 is an R-S₂-bimodule, then there is a unique right S₂-module structure on P which is compatible with T. If M and N are both bimodules, then P is also a bimodule, again in a unique way. (P, T) are unique up to a unique isomorphism, and are called the "tensor product" of M and N.

If R is a ring, _RM is a left R-module, and the commutator rs-sr of any two elements r and s of R is in the annihilator of M, then we can make M into a right R module by setting mr = rm. Note that in this situation the action of R on M factors through an action of the commutative ring R/Z(R), R modulo its center. In this case the tensor product of M with itself over R is again an R-module. If M and N are both R-modules satisfying this condition, then their tensor product is again an R-module. This is a very common technique in commutative algebra.

Example

Consider the rational numbers Q and the integers modulo n Z_n. Both can be considered as modules over the integers, Z. Let B: Q × Z_n → M be a Z-bilinear operator. Then B(q, i) = B(q/n, ni) = B(q/n, 0) = 0, so every bilinear operator is identically zero. Therefore, if we define P to be the trivial module, and T to be the zero bilinear function, then we see that the properties for the tensor product are satisfied. Therefore, the tensor product of Q and Z_n is {0}.