Definition 8.1:
Let
be modules. The direct product of
is the infinite cartesian product
![{\displaystyle \prod _{\alpha \in A}M_{\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31f78b98a60f9b5b630b83d836c5790e05711c3e)
together with component-wise addition, module operation and thus zero and additive inverses.
Theorem 8.2:
In the category of modules, the direct product constitutes a product.
Proof:
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
Lemma 8.4:
Let
be modules. Their direct sum is a submodule of the direct product.
Proof:
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.
Lemma 8.5:
For each
, there is a canonical morphism
.
Proof:
.![{\displaystyle \Box }](https://wikimedia.org/api/rest_v1/media/math/render/svg/029b77f09ebeaf7528fc831fe57848be51f2240b)
Lemma 8.6:
.
Proof:
Consider the morphism
.
We claim that this is an isomorphism, so we check all points.
1. Well-defined:
Both
and
are morphisms (with suitable domains and images), so
is as well.
2. Injective:
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
.
3. Surjective:
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
.![{\displaystyle \Box }](https://wikimedia.org/api/rest_v1/media/math/render/svg/029b77f09ebeaf7528fc831fe57848be51f2240b)
Theorem 8.7:
direct sum is coproduct in category of modules
To be then used to construct the tensor product.
Definition 8.8:
Let
be a ring and
modules over that ring. Consider the set of all pairs
![{\displaystyle (m,n),m\in M,n\in N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd19ce7162a737e9b4db3025735e1719d59453bb)
and endow this with multiplication and addition by formal linear combinations, producing elements such as
![{\displaystyle \sum _{k=1}^{l}r_{k}(m_{k},n_{k})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31cb31dbad129143e706b242d07235249f52c5e7)
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:
Proof:
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.
Indeed, let
,
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
![{\displaystyle M\times N{\overset {\otimes }{\rightarrow }}M\otimes N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa08341fbdd14329bd6e57c995eea2697595bda)
is a universal arrow. Hence, we call the result of theorem 8.9 the universal property of the tensor product.
Lemma 8.10:
Let
be a ring and
be an
-module. Recall that using canonical operations,
is an
-module over itself. We have
.
Proof:
Define the morphism
,
extend it to all formal linear combinations via summation
![{\displaystyle \sum _{k=1}^{l}s_{k}(r_{k},m_{k})\mapsto \sum _{k=1}^{l}s_{k}r_{k}m_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a1724b798cb05659b8a5496123aa479a447e720)
and then observe that
![{\displaystyle \varphi :R\otimes _{R}M\to M,\varphi \left(\left[\sum _{k=1}^{l}s_{k}(r_{k},m_{k})\right]\right):=\sum _{k=1}^{l}s_{k}r_{k}m_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7413ab86a1d67e551cd6e86543b70a160c82c71a)
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
![{\displaystyle \left[\sum _{k=1}^{l}s_{k}r_{k}m_{k}-\sum _{j=1}^{r}s_{j}'r_{j}'m_{j}'\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7966a1bd3aab917eba482dc5cd9e484b303850a5)
since
.
Lemma 8.11:
Let
be
-modules. Then
.
Proof:
For
fixed, define the bilinear function
.
Applying theorem 8.9 yields
![{\displaystyle g_{m}:N\otimes K\to (M\otimes N)\otimes K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4cae2a0ab46fb2526628060c319c4c476831272)
such that
. Then define
.
This function is bilinear (linearity in
from
)
and thus theorem 8.9 yields a morphism
![{\displaystyle G:M\otimes (N\otimes K)\to (M\otimes N)\otimes K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca6844b1fdea13ad18f7a3b49c4c9184fd890822)
such that
.
An analogous process yields a morphism
![{\displaystyle H:(M\otimes N)\otimes K\to M\otimes (N\otimes K)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4e50625fcd323fdeeb513deb4aeebc554377518)
such that
.
Since addition within tensor products commutes with equivalence class formation,
and
are inverses.
Lemma 8.12:
Let
be
-modules, let
be an
-module. Then
.
Proof:
We define
.
This is bilinear (since formation of equivalence classes commutes with summation and module operation), and hence theorem 8.9 yields a morphism
![{\displaystyle g:M\otimes \left(\bigoplus _{\alpha \in A}N_{\alpha }\right)\to \bigoplus _{\alpha \in A}(M\otimes N_{\alpha })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/977a51ea05a9c7467b54471cb31dc666f45689cf)
such that
.
This is obviously surjective. It is injective because
![{\displaystyle {\begin{aligned}&g\left(\left[\sum _{j=1}^{l}r_{j}(m_{j},(n_{\alpha }^{j})_{\alpha \in A})\right]\right)=g\left(\left[\sum _{q=1}^{p}s_{q}(x_{q},(y_{\alpha }^{q})_{\alpha \in A})\right]\right)\\\Leftrightarrow &\left(\left[\sum _{j=1}^{l}r_{j}(m_{j},n_{\alpha }^{j})\right]\right)_{\alpha \in A}=\left(\left[\sum _{q=1}^{p}s_{q}(x_{q},y_{\alpha }^{q})\right]\right)_{\alpha \in A}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8dd05542e971fe0cc5834011ff5742e3d0d58e9)
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.
Lemma 8.13:
.
Proof:
Linear extension of
![{\displaystyle [(m,n)]\mapsto [(n,m)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17c6fca249c48d96f3aa99c9f0d8cb56a8f35dc6)
defines a morphism which is well-defined due to symmetry, linear by definition and bijective because of the obvious inverse.
We have proven:
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
.
Proof:
Set
.
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.
Further set
.
By theorem 8.9 and thinking outside the box, we get a map
![{\displaystyle \theta :\operatorname {Hom} (M,\operatorname {Hom} (N,K))\to \operatorname {Hom} (M\otimes N,K)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6691f6093d33ec50d695d854a6b975f14897634f)
such that
.
Then
and
are inverse morphisms, since
is determined by what it does on elements of the form
.
Proof:
The map
![{\displaystyle \phi :M\times K\to N\otimes K,\phi (m,k)=\theta (m)\otimes k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aca506f1bcbb3913221b2f65b49cbc5077f5006d)
is bilinear, and hence induces a map
![{\displaystyle \varphi :M\otimes K\to N\otimes K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77003c9e977b4bff54941f637c2ab828ae7acafe)
such that
.
Similarly, the map
![{\displaystyle \phi _{-1}N\times K\to M\otimes K,\phi (n,k)=\theta ^{-1}(n)\otimes k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9367c46682cd77a524b27209b7bc1c36421cb70a)
induces a map
![{\displaystyle \varphi ^{-1}:N\otimes K\to M\otimes K}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ed31c9b04d779a3d9b7f02a6b2a36b03908b8b7)
such that
.
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.