Download 43 Projective modules

43
Projective modules
43.1 Note. If F is a free R-module and P ⊆ F is a submodule then P need
not be free even if P is a direct summand of F .
Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules and we
have an isomorphism of Z/6Z-modules:
Z/6Z ∼
= Z/2Z ⊕ Z/3Z
Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands of
the free module Z/6Z.
43.2 Definition. An R-module P is a projective module if there exists an Rmodule Q such that P ⊕ Q is a free R-module.
43.3 Examples.
1) If R is a ring with identity then every free R-module is projective.
2) Z/2Z and Z/3Z are non-free projective Z/6Z-modules.
43.4 Definition. Let
fi+1
fi
. . . −→ Mi −→ Mi+1 −→ Mi+2 −→ . . .
be a sequence of R-modules and R-module homomorphisms. This sequence is
exact if Im(fi ) = Ker(fi+1 ) for all i.
43.5 Definition. A short exact sequence is an exact sequence of R-modules the
form
f
g
0 −→ N −→ M −→ K −→ 0
(where 0 is the trivial R-module).
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43.6 Note.
f
g
1) A sequence 0 → N −→ M −→ K → 0 is short exact iff
• f is a monomorphism
• g is an epimorphism
• Im(f ) = Ker(g).
2) If M 0 is a submodule of M then we have a short exact sequence
0 −→ M 0 −→ M −→ M/M 0 −→ 1
Morever, up to an isomorphism, every short exact sequence is of this form:
/
0
∼
=
/
0
f
N
/
g
M
/
=
/
Ker(g)
/
M
/
K
0
∼
=
/
M/ Ker(g)
0
43.7 Definition. A short exact sequence
f
g
0 −→ N −→ M −→ K −→ 0
∼
=
is split exact if there is an isomorphism ϕ : M −→ N ⊕K such that the following
diagram commutes:
0
/
N
=
0
/
N
f
/M
g
/
/
ϕ ∼
=
N ⊕K
/
0
/
0
K
/
=
K
f
g
43.8 Proposition. Let R be a ring and let 0 → N −→ M −→ K → 0 be a
short exact sequence of R-modules. The following conditions are equivalent.
1) The sequence is split exact.
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2) There exists a homomorphism h : K → M such that gh = idK .
3) There exists a homomorphism k : M → N such that kf = idN
Proof. Exercise.
43.9 Theorem. Let R be a ring with identity and let P be an R-module. The
following conditions are equivalent.
1) P is a projective module.
2) For any homomorphism f : P → N and an epimorphism g : M → N there
is a homomorphism h : P → M such that the following diagram commutes:
}
h
M
~}
}
}
}
}
}
}
P
f
g
/
f
N
g
3) Every short exact sequence 0 → N −→ M −→ P → 0 splits.
Proof.
(1) ⇒ (2) Let Q be a module such that P ⊕ Q is a free module, and let
B = {bi }i∈I be a basis of P ⊕ Q. Since g is an epimorphism for every i ∈ I we
can find mi ∈ M such that g(mi ) = f (bi ). Define
h̃ : P ⊕ Q −→ M
by
!
h̃
X
r i bi
:=
i
X
i
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ri mi
Check: since B is a basis of P ⊕ Q the map h̃ is a well defined R-module
homomorphism and g h̃ = f . Then we can take h = h̃|P .
(2) ⇒ (3) We have a diagram
P
idP
g
/P
M
Since g is an epimorphism there is h : P → M such that gh = idP . Therefore
f
g
by (43.8) the sequence 0 → N −→ M −→ P → 0 splits.
(3) ⇒ (1) We have the canonical epimorphism of R-modules:
M
f:
R→P
p∈P
This gives a short exact sequence
0 −→ Ker(f )−→
M
f
R −→ P −→ 0
p∈P
By assumption on P this sequence splits. so we obtain
M
P ⊕ Ker(f ) ∼
R
=
p∈P
and thus P is a projective module.
43.10 Corollary. If R is a ring with identity, P is a projective R-module and
f : M → P is an epimorphism of R-modules then M ∼
= P ⊕ Ker(f ).
Proof. We have a short exact sequence
f
0 −→ Ker(f )−→M −→ P −→ 0
which splits by Theorem 43.9.
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44
Projective modules over PIDs
44.1 Theorem. If R is a PID, F is a free R-module of a finite rank, and M ⊆ F
is a submodule then M is a free module and rank M ≤ rank F .
44.2 Corollary. If R is a PID then every finitely generated projective R-module
is free.
Proof. If P is a finitely generated projective R-module then we have an epimorphism f : Rn → P for some n > 0. By Corollary 43.10 we have an isomorphism
P ⊕ Ker(f ) ∼
= Rn
Therefore we can identify P with a submodule of Rn , and thus by Theorem 44.1
P is a free module.
44.3 Note. Theorem 44.1 is true also for infinitely generated free modules over
PIDs. As a consequence Corollary 44.2 is true for all (non necessarily finitelly
generated) projective modules over PIDs.
Proof of Theorem 44.1 (compare with the proof of Theorem 13.6).
We can assume that F = Rn .
We want to show: if M ⊆ Rn then M is a free R-module and rank M ≤ n.
Induction with respect to n:
If n = 1 then M ⊆ R, so M is an ideal of R. Since R is a PID we have M = hai
for some a ∈ R. If a = 0 then M = {0} is a free module of rank 0. Otherwise
we have an isomorphism of R-modules
∼
=
f : R −→ M,
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f (r) = ra
and so M is a free module of rank 1.
Next, assume that for some n every submodule of Rn is a free R-module of rank
≤ n, and let M ⊆ Rn+1 . Take the homomorphism of R-modules
f : Rn+1 → R,
We have:
f (r1 , . . . , rn+1 ) = rn+1
Ker(f ) = {(r1 , . . . , rn , 0) | ri ∈ R} ∼
= Rn
We have an epimorphism:
f |M : M → Im(f |M )
Since Im(f |M ) ⊆ R, thus Im(f |M ) is a free R-module, and so by Corollary 43.10
we have
M∼
= Im(f |M ) ⊕ Ker(f |M )
We also have:
Ker(f |M ) = Ker(f ) ∩ M
It follows that that Ker(f |M ) is a submodule of Ker(f ), and since Ker(f ) is a
free R-module of rank n by the inductive assumption we get that Ker(f |M ) is a
free R-module of rank ≤ n. Therefore
M∼
= Im(f |M ) ⊕ Ker(f |M )
| {z }
| {z }
free
rank ≤ 1
free
rank ≤ n
and so M is a free R-module of rank ≤ n + 1.
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45
The Grothendieck group
Recall. If R is a ring with IBN and F is a free, finitely generated R-module then
rank F = number of elements of a basis of F
Goal. Extend the notion of rank to finitely generated projective modules.
Idea.
1) Rank should be additive: rank(P ⊕ Q) = rank P + rank Q.
2) Rank of a module need not be an integer. Each ring determines a group
K0 (R) such that for each finitely generated projective module rank of P is an
element [P ] ∈ K0 (R).
Recall. A commutative monoid is a set M together with addition
M × M → M,
(x, y) 7→ x + y
and with a trivial element 0 ∈ M such that the addition is associative, commutative and 0 + x = x for all x ∈ M .
45.1 Example. Let ProjfRg be the set of isomorphism classes of finitely generated
projective R-modules. For a projective finitely generated R-module P denote
[P ] = the isomorphism class of P
The set ProjfRg is a commutative monoid with addition given by
[P ] + [Q] := [P ⊕ Q]
The identity element in ProjfRg is [0], the isomorphism class of the zero module.
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45.2 Theorem. Let M be a commutative monoid. There exists an abelian
group Gr(M ) and a homomorphism of monoids
αM : M → Gr(M )
that satisfies the following universal property. If G is any abelian group and
f : M → G is a homomorphism of monoids then there exists a unique homomorphism of groups f¯: Gr(M ) → G such that the following diagram commutes:
f
M
z
αM
z
z
z
z
z f¯
z
z
/G
z=
Gr(M )
Moreover, such group Gr(M ) is unique up to isomorphism.
45.3 Note. Let CMono denote the category of commutative monoids. We have
the forgetful functor
U : Ab −→ CMono
Theorem 45.2 is equivalent to the statement that this functor has a left adjoint
Gr : CMono → Ab,
M 7→ Gr(M )
45.4 Definition. Let M be a commutative monoid. The group Gr(M ) is called
the group completion or the Grothendieck group of the monoid M .
Proof of Theorem 45.2.
Construction of Gr(M ). Let M be a commutative monoid. Define
Gr(M ) := M × M/ ∼
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where
(x, y) ∼ (x0 , y 0 ) iff x + y 0 + t = x0 + y + t for some t ∈ M
Check: ∼ is an equivalence relation on M × M . Notation:
[x, y] := the equivalence class of (x, y)
(Intuitively: [x, y] = x − y)
Note: for any x ∈ M we have [x, x] = [0, 0] since x + 0 = 0 + x.
Addition in Gr(M ):
[x, y] + [x0 , y 0 ] = [x + x0 , y + y 0 ]
Check: this operation is well defined, it is associative, and it has [0, 0] as the
identity element.
Additive inverses in Gr(M ):
−[x, y] = [y, x]
Indeed: [x, y] + [y, x] = [x + y, y + x] = [0, 0]
Construction of the homomorphism αM : M → Gr(M ).
Define
αM : M → Gr(M ),
x 7→ [x, 0]
The universal property of Gr(M ).
Let G be an abelian group and let f : M → G be a homomorphism of commutative monoids. Define
f¯: Gr(M ) → G,
f¯([x, y]) := f (x) − f (y)
Check:
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1) f¯ is a well defined group homomorphism.
2) f¯αM = f
3) f¯ is the only homomorphism Gr(M ) → G satisfying conditions 1)-2).
Uniqueness of Gr(M ) follows from the universal property.
45.5 Examples.
1) Gr(N) ∼
=Z
2) Let M = N ∪ {∞} with n + ∞ = ∞ for all n ∈ M . Then Gr(M ) is the
trivial group. Indeed, for any m, n ∈ M we have
[m, n] = [∞, ∞]
since m + ∞ = ∞ + n.
3) If G is an abelian group then Gr(G) ∼
= G.
45.6 Definition. If R is a ring then K0 (R) := Gr(ProjfRg )
45.7 Notation. For [P ], [Q] ∈ ProjfRg denote
[P ] − [Q] :=[P, Q] ∈ K0 (R)
[P ] :=[P, 0]
−[Q] :=[0, Q]
45.8 Proposition. Let R be a ring with identity. If P , Q are finitely generated
projective R-modules then [P ] = [Q] in K0 (R) iff there exists n ≥ 0 such that
P ⊕ Rn ∼
= Q ⊕ Rn .
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Proof.
(⇒) If P ⊕ Rn ∼
= Q ⊕ Rn then in K0 (R) we have
[P ] + [Rn ] = [P ⊕ Rn ] = [Q ⊕ Rn ] = [Q] + [Rn ]
and so [P ] = [Q]
(⇒) If [P ] = [Q] in K0 (R) then
P ⊕S ∼
=Q⊕S
for some finitely generated projective R-module S.
Exercise: If S is a finitely generated projective R-module then there is a finitely
generated projective R-module T such that S ⊕ T ∼
= Rn for some n ≥ 0.
We obtain
P ⊕ Rn ∼
=P ⊕S⊕T ∼
=Q⊕S⊕T ∼
= Q ⊕ Rn
45.9 Definition. Let R be a ring with identity. We say that R-modules M , N
are stably isomorphic if M ⊕ Rn ∼
= N ⊕ Rn for some n ≥ 0.
45.10 Definition. Let R be a ring with identity. We say an R-module M is
stably free if M ⊕ Rn ∼
= Rm for some m, n ≥ 0.
45.11 Note. Let R be a ring with identity. We have a homomorphism ϕ : Z →
K0 (R) given by
(
[Rn ]
for n ≥ 0
ϕ(n) :=
−n
−[R ] for n < 0
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45.12 Proposition. Let R be a ring with identity, and let ϕ : Z → K0 (R) be
the homomorphism as in (45.11).
1) ϕ is 1-1 iff R has IBN.
2) ϕ is an epimorphism iff every finitely generated projective R-module is
stably free.
Proof.
1) By Proposition 45.8 for n ≥ 0 we have n ∈ Ker(ϕ) iff Rn ⊕ Rm ∼
= 0 ⊕ Rm for
some m ≥ 0. If R has IBN this is possible only if n = 0, and so Ker(ϕ) = {0}.
Conversely, assume that R does not have IBN. Then Rn ∼
= Rm for some n > m.
This gives Rn−m ⊕ Rm ∼
= 0 ⊕ Rm , and so ϕ(n − m) ∈ Ker(ϕ).
2) (⇒) Assume that ϕ is an epimorphism. Then for every finitely generated
projective R-module P we have [P ] = [Rn ] for some n ≥ 0 or [P ] = −[Rn ] for
some n ≥ 0.
If [P ] = [Rn ] then by Proposition 45.8 we have P ⊕ Rm ∼
= Rn ⊕ Rm and so P
is a stably free module.
If [P ] = −[Rn ] then
[0] = [P ] + [Rn ] = [P ⊕ Rn ]
Again by Proposition 45.8 this gives 0 ⊕ Rm ∼
= P ⊕ Rn ⊕ Rm , and again we
obtain that P is stably free.
(⇐) The group K0 (R) is generated by elements [P ] where P is a finitely generated projective R-module, so it is enough to show that for any such P we have
[P ] = ϕ(k) for some k ∈ Z.
Since P is stably free we have P ⊕ Rn ∼
= Rm for some n, m ≥ 0. This gives
[P ] + [Rn ] = [P ⊕ Rn ] = [Rm ]
Therefore [P ] = [Rm ] − [Rn ] = ϕ(m) − ϕ(n) = ϕ(m − n).
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45.13 Example.
Here is an example of a stable free module that is not free. For details see:
R. G. Swan, Vector bundles and projective modules, Transactions AMS 105 (2)
(1962), 264-277.
Let B be a compact, normal, topological space and let p : E → B be a real
vector bundle over B. Define:
C(B) = {f : B → R | f - continuous }
C(B) is a ring (with pointwise addition and multiplication). Let Γ(p) be the set
of all continuous sections of p:
Γ(p) = {s : B → E | ps = idB }
Note: Γ(p) is an C(B)-module with poinwise addition and pointwise multiplication by elements of C(B).
Fact 1. The module Γ(p) is free iff p is a trivial vector bundle.
Fact 2. If p : E → B and q : E 0 → B are real vector bundles over B then we
have an isomorphism of C(B)-modules:
Γ(p ⊕ q) ∼
= Γ(p) ⊕ Γ(q)
Upshot. If p : E → B, q : E 0 → B are bundles such that p is non-trivial, but
both q and p ⊕ q are trivial bundles then Γ(p) is stably free C(B)-module that
is not free.
Indeed, in such case we have:
Γ(p) ⊕ Γ(q) ∼
= Γ(p ⊕ q)
|{z} | {z }
free
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free
Fact 3. It is possible to find vector bundles as above. Take e.g. p : T S 2 → S 2
to be the tangent bundle of the 2-dimensional sphere, and q : S 2 × R1 → S 2 to
be the 1-dimensional trivial bundle over S 2 .
Note: one can also show that a C(B)-module M is finitely generated projective
module iff M ∼
= Γ(p) for some vector bundle p : E → B.
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