Tensor product of modules

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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 a 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, algebraic topology 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.

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[edit] Multilinear mappings

In general, we start with a ring R, not necessarily commutative. Let MR 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:

BilinR(M,N;Z) = the set of φ : M × NZ 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

[edit] Definition

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

[edit] Construction

To construct the tensor product, we can proceed as for vector spaces; the 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.

[edit] Several modules

It is possible to generalize the definition to a tensor product of any number of spaces. For example, the universal property of

M1M2M3

is that every trilinear map on

M1×M2×M3Z

corresponds to a unique linear map

M1M2M3Z.

The binary tensor product is associative: (M1M2)⊗M3 is naturally isomorphic to M1⊗(M2M3). The tensor product of all three may therefore be identified with either of those.

[edit] Structure on the tensor product

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 MN is a left S-module, and similarly, if N is an (R,T)-bimodule, then MN is a right T-module. If M and N each have bimodule structures as above, then MN is an (S,T)-bimodule. In particular, if R is a commutative ring, then MN will always be an R-module.

If {mi}iI and {nj}jJ are generating sets for M and N, respectively, then {minj}iI,jJ will be a generating set for MN. Because the tensor product is right exact, not in general 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 MN.

If S and T are commutative R-algebras, then ST will be a commutative R-algebra as well, with the multiplication map defined by (m1m2)(n1n2)=(m1n1m2n2) 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 MZN.

It is also possible to generalize the definition to tensor products of modules over the same ring. If the ring is non-commutative, distinguish right modules and left modules. We will write RM for a left module, and MR 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 RMS. 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, the ring in question should be specified: 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 MR 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 × NP such that for every R-bilinear operator B: M × NO there is a unique group homomorphism L: PO such that L o T = B. P need not be a module over R. However, if S1MR is an S1-R-bimodule, then there is a unique left S1-module structure on P which is compatible with T. Similarly, if RMS2 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 M and N are both R-modules over a commutative ring, then their tensor product is again an R-module. If R is a ring, RM is a left R-module, and the commutator

rssr

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.

The action of R on M factors through an action of a quotient commutative ring. In this case the tensor product of M with itself over R is again an R-module. This is a very common technique in commutative algebra.

[edit] Example

Consider the rational numbers Q and the integers modulo n Zn. Both can be considered as modules over the integers, Z. Let B: Q × ZnM 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 Zn is {0}.

[edit] See also

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