Direct products and direct sums
Let be modules. The direct product of is the infinite cartesian product
together with component-wise addition, module operation and thus zero and additive inverses.
In the category of modules, the direct product constitutes a product.
Let be any index category, that contains one element for each , no other elements, and only the identity morphisms. Let be any other object such that
Let be a commutative ring, and let be modules over . The direct sum
is defined to be the module consisting of tuples where only finitely many of the s are nonzero, together with component-wise addition and component-wise module operation.
Let be modules. Their direct sum is a submodule of the direct product.
Both have the same elements and the same operations, and the direct product is a subset that is a module with those operations. Therefore we have a submodule.
For each , there is a canonical morphism
Consider the morphism
We claim that this is an isomorphism, so we check all points.
Both and are morphisms (with suitable domains and images), so is as well.
Assume . Then for any contained in we have
note that the sum is finite, since we are in the direct sum; this is necessary since infinite sums are not defined. Hence .
Let . Define
The latter sum is finite because and all but finitely many are nonzero. Thus this is well-defined as a function, and direct computation proves easily that it is -linear. Hence we have a morphism, and further
direct sum is coproduct in category of modules
To be then used to construct the tensor product.
The tensor product
Let be a ring and modules over that ring. Consider the set of all pairs
and endow this with multiplication and addition by formal linear combinations, producing elements such as
where the are in . We have obtained the vector space of formal linear combinations (call ). Set the subspace
the generated subspace. We form the quotient
This is called the tensor product. To indicate that are -modules, one often writes
The following theorem shows that the tensor product has something to do with bilinear maps:
Let be -modules and let be -bilinear. Then there exists a unique morphism such that the following diagram commutes:
Let be any -bilinear map. Define
where the square brackets indicate the equivalence class.
Once we proved that this is well-defined, the linearity of easily follows. We thus have to show that maps equivalent vectors to the same element, which after subtracting the right hand side follows from mapping to zero.
where all are one of the four types of generators of . By distinguishing cases, one obtains that each type of generator of is mapped to zero by because of bilinearity. Well-definedness follows, and linearity is clear from the definition and since addition and module operation interchange with equivalence class formation.
Note that from a category theory perspective, this theorem 8.9 states that for any two modules over the same ring, the arrow
is a universal arrow. Hence, we call the result of theorem 8.9 the universal property of the tensor product.
Let be a ring and be an -module. Recall that using canonical operations, is an -module over itself. We have
Define the morphism
extend it to all formal linear combinations via summation
and then observe that
is well-defined; again, by subtracting the right hand side, it's enough to show that is mapped to zero, and this is again done by consideration of each of the four generating types.
This is a morphism as shown by direct computation (using the rules for the module operation), it is clearly surjective (map ) and it is injective because if
- , then
Let be -modules. Then
For fixed, define the bilinear function
Applying theorem 8.9 yields
such that . Then define
This function is bilinear (linearity in from
and thus theorem 8.9 yields a morphism
An analogous process yields a morphism
Since addition within tensor products commutes with equivalence class formation, and are inverses.
Let be -modules, let be an -module. Then
This is bilinear (since formation of equivalence classes commutes with summation and module operation), and hence theorem 8.9 yields a morphism
This is obviously surjective. It is injective because
by the linearity of and component-wise addition in the direct sum, and equality for the direct sum is component-wise. We split the argument up into sums where only one component of the right direct sum matters, and observe equality since we divide out isomorphic spaces.
Linear extension of
defines a morphism which is well-defined due to symmetry, linear by definition and bijective because of the obvious inverse.
We have proven:
Let be a fixed ring. The set of all -modules forms a commutative semiring, where the addition is given by (direct sum), the multiplication by (tensor product), the zero by the trivial module and the unit by .
Note that we have more: From lemma 8.12 even infinite direct sums (uncountably many, as many as you like, ...) distribute over the tensor product. Incidentally, only finite direct sums are identical to the direct product. This may give hints for an infinite distributive law for infinitesimals.
Theorem 8.15 ("tensor-hom adjunction"):
Let be -modules. Then
Due to the equalities holding for elements of the tensor product and the linearity of , this is well-defined. Further, we obviously have linearity in since function addition and module operation are defined point-wise.
By theorem 8.9 and thinking outside the box, we get a map
Then and are inverse morphisms, since is determined by what it does on elements of the form .
Let be -modules isomorphic to each other (via ), and let be any other -module. Then
via an isomorphism
for all , .
is bilinear, and hence induces a map
Similarly, the map
induces a map
These maps are obviously inverse on elements of the type , , and by their linearity and since addition and equivalence classes commute, they are inverse to each other.