FUNCTORS
RALF FRÖBERG
1. Module homomorphisms
A map f : M → N between R-modules is called an R-module homomorphism if
f (m1 + m2 ) = f (m1 ) + f (m2 ) and
f (rm) = rf (m),
where m1 , m2 , m ∈ M and r ∈ R.
Examples If I is an ideal in R, then the natural map R → R/I is an R-module homomorhism. If N is an R-submodule of M , then the inclusion N → M is an R-module
homomorphism.
A homomorphism which is bijective is called an isomorphism.
r t
, r, t ∈ R}. Show that R is a (commutative) ring and that it is
Exercise Let R = {
t r
isomorphic to C as R-module.
1.1. Exact sequences. A sequence of module homomorphisms
fn−1
fn
· · · −→ Mn−1 −→ Mn −→ Mn+1 −→ · · ·
is said to be exact at Mn if Im(fn−1 ) = Ker(fn ). A short exact sequence, SES is a sequence
f
g
0 −→ N −→ M −→ K −→ 0
which is exact at N, M and K. This means that f is injective, Im(f ) = Ker(g), and g is
surjective.
Example If I is an ideal in R, then 0 −→ I −→ R −→ R/I −→ with natural maps is
exact. More generally, if N is a submodule of M , then 0 −→ N −→ M −→ M/N −→ is
exact.
A long exact sequence can be divided into several SES
0 −→ ker(fn ) −→ Mn −→ Im(fn ) −→ 0 a.s.o.
f
g
Theorem 1.1. Let 0 −→ N −→ M −→ K −→ 0 be a SES of R-modules.. Then M is
Noetherian if and only if both N and K are Noetherian.
1
2
RALF FRÖBERG
Proof Suppose M is Noetherian. A submodule of N is a submodule of M , and thus
finitely generated, so N is Noetherian. Let H be a submodule of K. The inverse image
of H in M is a finitely generated module, generated by {m1 , . . . , mr } say. The images of
{m1 , . . . , mr } generate H, so H is finitely generated and K Noetherian. Now suppose M 0
be a submodule of M . The image of M 0 in K is finitely generated. The inverse image in
M of these generators together with the generators of N generates M 0 . But N is finitely
generated, so M 0 is finitely generated, and M Noetherian.
Corollary 1.1. If R is Noetherian, then Rn is Noetherian.
Proof Use induction and the SES 0 −→ Rn−1 −→ Rn −→ R −→ 0.
Corollary 1.2. A finitely generated module over a Noetherian ring is Noetherian.
f
Proof Suppose M has n generators. 0 −→ Ker(F ) −→ Rn −→ M −→ 0 is exact. Ker(f )
is a submodule of Rn , which is Noetherian
2. Functors
Definition A covariant functor on modules is a map F such that it associates to each
R-module M a module F (M ), and associates to each R-homomorphism f : M → N a
homomorphism F (f ) : F (M ) → F (N ) such that the following two conditions hold:
F (idM ) = idF (N ) for every M
F (g ◦ f ) = F (g) ◦ F (f ) for all homomorphisms f : M → N and g : N → P.
A contravariant functor on modules is a map F such that it associates to each R-module
M a module F (M ) , and associates to each R-homomorphism f : M → N a homomorphism
F (f ) : F (N ) → F (M ) such that the following two conditions hold:
F (idM ) = idF (N ) for every M
F (g ◦ f ) = F (f ) ◦ F (g) for all homomorphisms f : M → N and g : N → P.
Let 0 → N → M → P → 0 be a SES.
Then a covariant functor F is said to be
left-exact if 0 → F (N ) → F (M ) → F (P ) is exact,
right-exact if F (N ) → F (M ) → F (P ) → 0 is exact,
exact if 0 → F (N ) → F (M ) → F (P ) → 0 is exact.
If G is a contravariant functor, G is said to be
left-exact if 0 → G(P ) → G(M ) → G(N ) is exact,
right-exact if G(P ) → G(M ) → G(N ) → 0 is exact,
exact if 0 → G(P ) → G(M ) → G(N ) → 0 is exact.
Exercise Show that a covariant functor F is left-exact if 0 → N → M → P is exact
implies that 0 → F (N ) → F (M ) → F (P ) is exact. Formulate the corresponding results
for right-exact covariant and right- or left-exact contravariant functors.
Exercise For an R-module M , let Mtor = {m ∈ M ; rm = 0 for some r 6= 0}. Let F (M ) =
Mtor and if f : M → N let F (f ) be the induced map. Show that F is a left exact covariant
functor.
FUNCTORS
3
3. The Hom-functors
If M and N are R-modules, the set of R-module homomorphisms M → N is denoted HomR (M, N ). HomR (M, N ) is an R-module where (f + g)(m) = f (m) + g(m)
and (rf )(m) = r(f (m)).
Theorem 3.1. If f : N → N 0 is an R-module homomorphism, then for any M there is an
induced map g : HomR (M, N ) → HomR (M, N 0 ) defined by φ → f ◦φ. This gives a covariant
functor Hom(M, −). There is also an induced map h : HomR (N 0 , M ) → HomR (N, M )
defined by ψ → ψ ◦ f . This gives a contravariant functor Hom(−, M ).
Exercise Prove this.
Theorem 3.2. Hom(M, −) and Hom(−, M ) are left exact. These functors are exact in
the case of vector spaces.
Proof We will prove this only for Hom(−, M ). So, what we will prove is that if 0 −→
f
g
A−→B −→ C −→ 0 is exact, then
F (g)
F (f )
0 −→ HomR (C, M ) −→ HomR (B, M ) −→ HomR (A, M )
is exact. We have to prove that F (g) is injective, that Im(F (g)) ⊆ Ker(F (f )), and that
Ker(F (f )) ⊆ Im(F (g)). Let φ ∈ Ker(F (g)). Then F (g)(φ) = φ ◦ g = 0 so φ(g(m)) = 0
for all m ∈ M . But g is surjective, so every n ∈ N is of the form g(m), so φ(n) = 0 for
all n ∈ N , so φ = 0. Thus Ker(F (g)) = 0. Im(f ) =Ker(g), so g ◦ f = 0, so F (gf ) =
F (f )F (g) = 0, so Im(F (g)) ⊆Ker(F (f )). Take φ ∈Ker(F (f )), so F (φ) = φ ◦ f = 0,
so Ker(g) =Im(f ) ⊆Ker(φ). Then there is an induced map φ̄ : B/Ker(g) → M . (Let
φ̄(b + Ker(g)) = φ(b), this is well defined and a homomorphism.) Since g is surjective, we
have that M/Ker(g) is isomorphic to C. Call the isimorphism ψ. Let λ = φ̄◦ψ −1 : C → M .
Then F (g)(λ) = λ ◦ g = φ̄ ◦ ψ −1 ◦ g = φ̄ ◦ p = φ, where p : M → M/Ker(g) is the projection.
Thus φ ∈ Im(F (g)). Now suppose we have vector spaces. We will show that F (f ) is
surjective. We have that f is injective, so A is isomorphic to the submodule Im(F ) of B.
That F (f ) is surjective means that every φ ∈ HomR (A, M ) can be written as φ = ψ ◦ f
for some ψ ∈ HomR (B, M ), which means that φ can be extended to B But this is true for
vector spaces.
Exercise Show that F (f ) is not always surjective for modules. Hint: Consider
2·
0 → Z → Z → Z/2 → 0.