The following definitions are straightforward generalisations from linear algebra. We begin by repeating a definition we already saw in chapter 6.
Definition 6.1 (generators of modules):
Let
be a module over the ring
. A generating set of
is a subset
such that
.
We also have:
Definition 11.1:
Let
be an
-module. A subset
of
is called linearly independent if and only if, whenever
, we have
.
Theorem 11.3:
Let
be free modules. Then the direct sum
![{\displaystyle \bigoplus _{\alpha \in A}M_{\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d9414f0504c4a0ff138a1b4d03f4b8c583582f0)
is free.
Proof:
Let bases
of the
be given. We claim that
![{\displaystyle \left\{\left(0,\ldots ,0,\overbrace {e_{\beta _{\alpha }}} ^{\alpha {\text{-th place}}},0,\ldots ,0\right){\big |}\alpha \in A,\beta _{\alpha }\in B_{\alpha }\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9fc5c9e6a8c83b9c23e321858572ea5c2e52bab)
is a basis of
.
Indeed, let an arbitrary element
be given. Then by assumption, each of the
has a decomposition
![{\displaystyle m_{\alpha }=\sum _{j=1}^{n_{\alpha }}r_{j,\alpha }e_{\beta _{j,\alpha }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51b3aad77596d2aeff11fc138429889a8be80651)
for suitable
. By summing this, we get a decomposition of
in the aforementioned basis. Furthermore, this decomposition must be unique, for otherwise projecting gives a new composition of one of the particular
.
The converse is not true in general!
Theorem 11.4:
Let
be free
-modules, with bases
and
respectively. Then
![{\displaystyle M\otimes _{R}N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1946b1034181e2db2825dca869b8ef0753e4831f)
is a free module, with basis
,
where we wrote for short
![{\displaystyle e_{\alpha }\otimes f_{\beta }:=[(e_{\alpha },f_{\beta })]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2813954a588b7a76e283d3215743041935bd4fef)
(note that it is quite customary to use this notation).
Proof:
We first prove that our supposed basis forms a generating system. Clearly, by summation it suffices to show that elements of the form
, ![{\displaystyle m\in M,n\in N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f2a5720916496a4758d4f4c788e8bd4db83260a)
can be written in terms of the
. Thus, write
and
,
and obtain by the rules of computing within the tensor product, that
.
On the other hand, if
![{\displaystyle 0=\sum _{\alpha \in A,\beta \in B}t_{\alpha ,\beta }e_{\alpha }\otimes f_{\beta }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bedbca8be6ed6e152fbd436f8f6bfab09933a300)
is a linear combination (i.e. all but finitely many summands are zero), then all the
must be zero. The argument is this: Fix
and define a bilinear function
,
where
,
are the coefficients of
,
in the decomposition of
and
respectively. According to the universal property of the tensor product, we obtain a linear map
with
,
where
is the canonical projection on the quotient space. We have the equations
,
and inserting the given linear combination into this map therefore yields the desired result.
The following is a generalisation of free modules:
Theorem 11.6:
Every free module is projective.
Proof:
Pick a basis
of
, let
be surjective and let
be some morphism. For each
pick
with
. Define
where
.
This is well-defined since the linear combination describing
is unique. Furthermore, it is linear, since we have
,
where the right hand side is the sum of the linear combinations coinciding with
and
respectively, which is why
. By linearity of
and definition of the
, it has the desired property.
There are a couple equivalent definitions of projective modules.
Theorem 11.7:
A module
is projective if and only if there exists a module
such that
is free.
Proof:
: Define the module
![{\displaystyle L:=\bigoplus _{m\in M}R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3efd4dbb0d45863d37a5c6595316cb8459594687)
(this obviously is a free module) and the function
.
is a surjective morphism, whence we obtain a commutative diagram
;
that is,
.
We claim that the map
![{\displaystyle \varphi :M\oplus \ker f\to L,\varphi (m,k):=h(m)+k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fec28ed45197cfb55c54b32fe3806b706507b0f)
is an isomorphism. Indeed, if
, then
and thus also
(injectivity) and further
, where
, which is why
![{\displaystyle \varphi ((rm,-k))=(0,\ldots ,0,\overbrace {r} ^{m{\text{-th place}}},0,\ldots ,0)=(0,\ldots ,0,\overbrace {r} ^{m{\text{-th place}}},0,\ldots ,0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81943403d414846eccbaf9476089e0db39a8bf2a)
(surjectivity).
: Assume
is a free module. Assume
is a surjective morphism, and let
be any morphism. We extend
to
via
.
This is still linear as the composition of the linear map
and the linear inclusion
. Now
is projective since it's free. Hence, we get a commutative diagram
![](//upload.wikimedia.org/wikipedia/commons/thumb/2/20/Projective-proof2.svg/150px-Projective-proof2.svg.png)
where
satisfies
. Projecting
to
gives the desired diagram for
.
Definition 11.8:
An exact sequence of modules
![{\displaystyle 0\rightarrow K\rightarrow N\rightarrow M\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06ff516fd348c9c222afb54e813f2cfc613cba5b)
is called split exact iff we can augment it by three isomorphisms such that
![](//upload.wikimedia.org/wikipedia/commons/thumb/3/3c/Split_exact_sequence.svg/400px-Split_exact_sequence.svg.png)
commutes.
Theorem 11.9:
A module
is projective iff every exact sequence
![{\displaystyle 0\rightarrow K\rightarrow N\rightarrow M\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06ff516fd348c9c222afb54e813f2cfc613cba5b)
is split exact.
Proof:
: The morphism
is surjective, and thus every other morphism with codomain
lifts to
. In particular, so does the projection
. Thus, we obtain a commutative diagram
![](//upload.wikimedia.org/wikipedia/commons/thumb/c/c0/Projective-proof3.svg/400px-Projective-proof3.svg.png)
where we don't know yet whether
is an isomorphism, but we can use
to define the function
,
which is an isomorphism due to injectivity:
Let
, that is
. Then first
![{\displaystyle m=f(h(0,m))=f(h(0,m)+g(k))=f(0)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65112aed95b1e56db180e024e017dc7b06cb1673)
and therefore second
.
And surjectivity:
Let
. Set
. Then
![{\displaystyle h(0,m)-n\in \ker f=\operatorname {im} h}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f4fbf77265ee9627b81d1f05e78f9797c5951c1)
and hence
for a suitable
, thus
.
We thus obtain the commutative diagram
![](//upload.wikimedia.org/wikipedia/commons/thumb/4/40/Projective-proof4.svg/400px-Projective-proof4.svg.png)
and have proven what we wanted.
: We prove that
is free for a suitable
.
We set
, ![{\displaystyle f:K\to M}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f57c297634611f53536ff34a4e3f4e0bfed86777)
where
is defined as in the proof of theorem 11.7
. We obtain an exact sequence
![{\displaystyle 0\rightarrow \ker f{\overset {\iota }{\hookrightarrow }}K{\overset {f}{\rightarrow }}M\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d62770dc299750c61b7d53259fa985f29cc18f4)
which by assumption splits as
![](//upload.wikimedia.org/wikipedia/commons/thumb/a/a3/Projective-proof5.svg/400px-Projective-proof5.svg.png)
which is why
is isomorphic to the free module
and hence itself free.
Theorem 11.10:
Let
and
be projective
-modules. Then
is projective.
Proof:
We choose
-modules such that
and
are free. Since the tensor product of free modules is free,
is free. But
,
and thus
occurs as the summand of a free module and is thus projective.
Theorem 11.11:
Let
be
-modules. Then
is projective if and only if each
is projective.
Proof:
Let first each of the
be projective. Then each of the
occurs as the direct summand of a free module, and summing all these free modules proves that
is the direct summand of free modules.
On the other hand, if
is the summand of a free module, then so are all the
s.
The following is a generalisation of projective modules:
The morphisms in the right sequence induced by any morphism
are given by the bilinear map
.
Theorem 11.13:
The module
is a flat
-module.
Proof: This follows from theorems 9.10 and 10.?.
Theorem 11.14:
Flatness is a local property.
Proof: Exactness is a local property. Furthermore, for any multiplicatively closed
![{\displaystyle S^{-1}(M\otimes _{R}N)\cong S^{-1}M\otimes _{S^{-1}R}S^{-1}N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8945c8d189472c5da8007785d541d7b6874371ee)
by theorem 9.11. Since every
-module is the localisation of an
-module (for instance itself as an
-module via
), the theorem follows.
Theorem 11.15:
A projective module is flat.
Proof:
We first prove that every free module is flat. This will enable us to prove that every projective module is flat.
Indeed, if
is a free module and
a basis of
, we have
![{\displaystyle M\cong \bigoplus _{\alpha \in A}R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc93645fc6c8501226264b0fd83a08fd1a5732a)
via
,
where all but finitely many of the summands on the left are nonzero. Hence, by distributivity of direct sum over tensor product, if we are given any exact sequence
,
to show that the sequence
![{\displaystyle 0\rightarrow A\otimes M\rightarrow B\otimes M\rightarrow C\otimes M\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2286c7226433f9ab47e2b99a032fdeaaf7ff0bf)
is exact, all we have to do is to prove that
![{\displaystyle 0\rightarrow A\otimes M\rightarrow B\otimes M\rightarrow C\otimes M\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2286c7226433f9ab47e2b99a032fdeaaf7ff0bf)
is exact, since we may then augment the latter sequence by suitable isomorphisms
Theorem 11.16:
direct sum flat iff all summands are
Theorem 11.17:
If
are flat
-modules, then
is as well.
Proof:
Let
![{\displaystyle 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14cece4b06c269d1d7f461dc26ba155a5e5a1146)
be an exact sequence of modules.
The following is a generalisation of flat modules:
Lemma 11.19:
The torsion of a module is a submodule of that module.
Proof:
Let
,
. Obviously
(just multiply the two annihilating elements together), and further
if
(we used commutativity here).
We may now define torsion-free modules. They are exactly what you think they are.
Definition 11.20:
Let
be a module.
is called torsion-free if and only if
.
Theorem 11.21:
A flat module is torsion-free.
To get a feeling for the theory, we define
-torsion for a multiplicatively closed subset
.
Definition 11.22:
Let
be a multiplicatively closed subset of a ring
, and let
be an
-module. Then the
-torsion of
is defined to be
.
Theorem 11.23:
Let
be a multiplicatively closed subset of a ring
, and let
be an
-module. Then the
-torsion of
is precisely the kernel of the canonical map
.