Suppose and are rings with unity (but not necessarily commutative), that we have ring homomorphism and that is an module. Then is an module by “restricting scalars” so that .
Any -vector space is a vector space.
Any vector space over is a module over .
There is a ring map sending
and so any left module over can be viewed as a complex vector space. There are other elements in satisfying , and so there are lots of ways to view a left module over as a -vector space.
Extension of scalars
Suppose that is a map of rings with unity, and is an -module. Is there a way to make into an -module? Maybe a better way to say it is: can we find a “smallest” -module together with a map ?
Example: Suppose that is a finite dimensional real vector space. Choose a for . So is isomorphic to using this basis. So we can think of as inside . If we chose a different basis, we’d get a different map , but the two maps would be related by a change of basis transformation so in some sense these are “isomomorphic”.
Extension of scalars (More examples)
Example: Suppose that is an abelian group (hence a -module). Can we embed in a -module? Sometimes, yes: if is for some , then embeds in . On the other hand if is finite, there are no maps from for any . If is a mixture of free and torsion parts, we can embed the free part of in but not the torsion part.
Example: If is an abelian group, can we embed into a vector space over – here the map is the map ? Sometimes yes – if is -torsion, we can do it, but for general no.
Universal approach
To study this in general we take a (left) -module and a ring map and ask: how can we make an -module out of and the map ?
An -module structure on means we need a map
satisfying the axioms
for .
Extension of scalars via tensor product
Our strategy is to make an abelian group whose elements are pairs modulo the relations above. The equivalence classes of this abelian group are written (or, in cases where we need more context, or even ). THe group itself is called . So the following rules hold:
.
.
A typical element of is a sum of the form . It is an -module via multiplication by on the first factor.
We have a map given by .
Important: The elements belong to a quotient group and so the representation of an element as a sum of “simple tensors” need not be unique! In fact it’s quite possible for to be zero.
Some examples
Suppose that and is . Then consists of sums of elements . But where , so we can write . is isomorphic to .
Suppose that is a finite group of order . Then any element of can be written where . But . Therefore
so is the zero module.
The universal property
The idea is that is the smallest module containing , where the action of comes via the map . In other words, if is any other module and there is an -module map , then there is a unique -module map so that this triangle commutes:
More on the universal property
If is the map , then maps injectively in . This is the “largest” quotient of which embeds into an -module.
Example: Let be a finitely generated abelian group. Then is isomoprhic to where is a finite group. If we want to map to a -vector space, the kernel has to include . And in fact the kernel of is . Further, the -dimension of the vector space is the rank of the free part of .
More examples
Example: Let be an module and let be the quotient map. Then is isomorphic to . First notice that if , then , so is in the kernel of . Therefore we have a map . We have a map in the opposite direction given by . So if is a finite abelian group, then is which is zero if has no -torsion.
Example: If is a vector space over of dimension , and is a field extension, then is an -dimensional vector space over .
Tensor products of modules
The commutative case
Assume for the moment that is a commutative ring with unity. Suppose that and are -modules. If is yet another -module, a bilinear map
is a map that is linear in each variable separately and also satisfies for . The tensor product of and is the free abelian group on pairs modulo the relations:
The equivalence class of the pair is written . is an module via the action and
Universal property
If is bilinear, then we can defined by . This is well defined and it converts a bilinear map into a module homomorphism. To go in the other direction, there is a map which is bilinear, sending . This is the ``universal” bilinear map. The universal property says that, if is any bilinear map, there is a unique module homomorphism such that
Tensor product of vector spaces
Suppose that is a field and , , are vector spaces over of dimensions and respectively. Let be a basis for and a basis for .
If is another -vector space, then a bilinear map is determined by its values on all pairs .
The tensor product is an -vector space and is spanned by the tensors .
Now construct a bilinear map by setting
By the universal property we have unless and in which case it is one.
Tensor product of vector spaces continued
Suppose that
Then for all pairs and therefore all ; in other words, the are linearly independent. Thus is an dimensional -vector space.
Endomorphisms
Let be a vector space and let be its dual. Then there is an isomorphism
where .
The noncommutative case
Now suppose that is a noncommutative ring. If and are left modules, then we have a problem defining a bilinear map where is also a left module. Namely, on the one hand, we would need:
but on the other hand
and since and are different we can’t define this consistently.
The noncommutative case continued
In the non-commutative case (with unity) we have to make some compromises:
First, we assume is a right module and is a left module.
Next, we are only going to construct an abelian group, not a module, from and .
Finally, we are going to consider maps , where is an abelian group, that are balanced, meaning that for .
We create an abelian group spanned by subject to the relations together with the additivity and similarly for .
This is the tensor product of the modules and – remember it is an abelian group, NOT an -module in general.
Universal property
still satisfies a universal property. Call a map , where is a right -module, is a left -module, and is an abelian group, balanced if is additive in and separately and satisfies for all .
Then given such a balanced map from to , there is a unique map of abelian groups such that the following diagram commutes.
Here the map is the expected one: .
As is always the case, the universal property characterizes the tensor product up to isomorphism.
Bimodules
Now suppose that and are rings with unity and that is simultaneously a left -module and a right module, so that . Such an object is called an -bimodule.
For example, suppose , , and is the space viewed as row vectors with acting on the right as matrix multiplication and on the left as scalar multiplication.
Tensor product of bimodules
If is a left module, we can form the tensor product which is an abelian group; but we can furthermore let act by .
This makes into a left -module. (If is an -bimodule so that acts on the left and on the right, then is a right -module.)
If is commutative, and is a left module, it is also a right -module via . So it is automatically an -bimodule. This is how is automatically an -module if is commutative.
General Properties
Tensor product of maps
If and are maps of right/left -modules, then defined by is a well defined group homomorphism.
If and are bimodules and and are -module homomorphismsm then is an -module homomorphism. (If is commutative all this is automatic).
Further, provided everything makes sense,
The Kronecker Product
If and are matrices giving linear maps from to and to .
Then the tensor product of these maps is a linear map from to .
The standard bases of give us bases of and . Thus the tensor product is represented as an matrix. This is called the Kronecker Product of the matrices and .
Associativity
The tensor product is associative in the sense that is isomorphic to provided that is a right module, is an bimodule, and is a left -module. If and are commutative this is automatic.
First check that both versions of the tensor product make sense.
Notice that is a left -module and is a right -module.
For fixed , the map is -balanced, so there is a well-defined map
This gives a well-defined map
and by the universal property this translates to a map
You can also reverse this construction to create the inverse map.
If is an bimodule then both constructions yield left -modules. Then is an bimodule and is a left -module.
Commutativity
If is commutative, the tensor product is commutative in the sense that is isomorphic to .
Distributive law
is isomorphic to and similarly if is a direct sum. By induction this extends to finite direct sums. With care it holds for infinite direct sums.
The proof of this uses the fact that there is a well-defined balanced map
defined by
and so we have a map
On the other hand we have balanced maps sending to and similarly for . These give a map from which is inverse to the map above.
Tensor product of algebras
If and are algebras where is commutative, then is an algebra with multiplication . (remember that an algebra is a ring in which is mapped into the center of the ring).