Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
3. The Group of Homomorphisms
3.1. The Group HomR ( , )
Let R be a ring and let f : M ў® M be a homomorphism of R-modules.
 Consider the abelian groups HomR (X , M ў) and Hom R (X , M ) for some R-module X. Given
a
a О HomR (X , M ў) , then the sequence X ® M ў®f M  f o a О Hom R (X , M )  we have a
map HomR (X , M ў) ® HomR (X , M ) : a a f o a .
 Consider the abelian groups HomR (M ў,Y ) and Hom R (M ,Y ) , for some R-module Y. Given
f
b
b О Hom R (M ,Y ) , then the sequence M ў® M ® Y  b o f О HomR (M ў,Y )  we have a
map HomR (M ,Y ) ® HomR (M ў,Y ) : b a b o f
Both maps are group homomorphisms as seen by verifying the properties:
 f o (a 1 + a 2 ) = f o a 1 + f o a 2
 ( b1 + b 2 ) o f = ( b1 o f ) + ( b 2 o f )
g
f
It follows that a sequence M ў® M ® M ўў induces two sequences

Hom R (X , M ў) ® Hom R (X , M ) ® Hom R (X , M ўў)
aў

a
f o aў
Hom R (M ўў,Y ) ®
b ўў a
a
a
goa
Hom R (M ,Y ) ®
b ўўof
b
a
Hom R (M ў,Y )
b og
Proposition 1. Let M , M ў, and M ўў be R-module.
g
f
a) A sequence 0 ® M ў® M ® M ўў is exact iff the sequence
fo
go
0 ® Hom R (X , M ў) ® Hom R (X , M ) ® Hom R (X , M ўў)
is exact for any R-module X.
f
g
b) A sequence M ў® M ® M ўў® 0 is exact iff the sequence
of
0 ® Hom R (M ўў,Y ) ®
og
Hom R (M ,Y ) ®
14
Hom R (M ў,Y )
Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
is exact for any R-module Y.
Proof. a) Assuming the first sequence is exact. To show fo is 1-1, let a ўО ker fo  f o a ў= 0 on X
 for all x О X , 0 = ( f o a ў)x = f (a ў(x )) . Since f is 1-1, a ў(x ) = 0 for all x О X .
ker go = Im fo : Let a О Im fo , g o a = g o fo(a ў) = g o f o a ў= 0 since g is zero on Imf. If a О ker go 
g o a = 0 on X  Im a Н ker g = Im f . Since f is 1-1, f - 1 : Im f ® M ў an isomorphism defined on
Im a  a ў= f - 1 o a О HomR (X , M ў) so that fo(a ў) = f o a ў= = f o ( f - 1 o a ) = a .
X
f
0
f
g
M
M
M
1

The rest of the proof is left as an exercise.
f
g
NOTE In general, an exact sequence 0 ® M ў® M ® M ўў® 0 dose not induce an exact sequence of
the form:
0 ® HomR (X , M ў) ®
HomR (X , M ) ®
HomR (X , M ўў) ® 0
(1)
0 ® HomR (M ўў,Y ) ®
HomR (M ,Y ) ®
HomR (M ў,Y ) ® 0
(2)
The possible failure is with exactness on the right i.e. the last induced map may not be surjective, see
Exercise 2. This fact is usually stated as Hom R ( , ) is left exact.
ґ 2
Exercise 2. Use the sequence 0 ® ў ® ў ® ў 2 ® 0 of ў -modules with a suitable ў -module for X
or Y so that the induced sequence is not exact.
It turns out that right exactness in (1) can be guaranteed for cretin types of R-modules X.
f
g
Proposition 3. If F is a free R-module then any exact sequence 0 ® M ў® M ® M ўў® 0 , induces an
exact sequence 0 ® HomR (F , M ў) ®
HomR (F , M ) ®
Proof. As exactness of 0 ® HomR (F , M ў) ®
HomR (F , M ўў) ® 0 .
HomR (F , M ) ®
1, it remains to show the exactness of the sequence
go
Hom R (F , M ) ®
HomR (F , M ўў) is guaranteed by
Hom R (F , M ўў) ® 0
15
Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
Let a ўўО Hom R (F , M ўў) . Suppose a basis for F is given by the family {x i }i ОI , and write {a ўў(x i )}i ОI
for their images in M ўў under a ўў. Because g is onto, then there exists a family {m i }i ОI in M so that
g(m i ) = a ўў(x i ) . Now there exists a unique R-homomorphism a : F ® M so that a (x i ) = m i for all i.
The image of a is go (a ) = g o a which satisfies (g o a )(x i ) = g(a (x i )) = = g(m i ) = a ўў(x i ) for all
i О I . g o a and a ўў are equal on a basis, by uniqueness property g o a = a ўў.

Theorem 4. Let M 1, M 2, N be R-modules. Then there is an isomorphism of abelian groups
a. Hom R (M 1 Е M 2, N ) @ Hom R (M 1, N ) ґ Hom R (M 2, N )
b. Hom R (N , M 1 ґ M 2 ) @ Hom R (N , M 1 ) ґ Hom R (N , M 2 )
i1
f
i2
f
Proof. a. For f : M 1 Е M 2 ® N , we use M 1 ® M 1 Е M 2 ® N and M 2 ® M 1 Е M 2 ® N to define
y
: Hom R (M 1 Е M 2, N ) ® Hom R (M 1, N ) ґ Hom R (M 2, N )
y (f ) =
The map
y
( f o i1, f o i 2 )
is a group homomorphism: For if f , g О Hom R (M 1 Е M 2, N ) ,
y ( f + g) = ((f + g )o i1, (f + g )o i2 ) = ( f o i1 + g o i1, f o i2 + g o i2 )
= ( f o i1, f o i 2 ) + (g o i1, g o i 2 ) = y ( f ) + y (g)
To show
y
is onto: Let ( f1, f2 ) О Hom R (M 1, N ) ґ Hom R (M 2, N ) , if we define
f : M 1 Е M 2 ® N : f (m 1, m 2 ) = f1(m 1 ) + f2 (m 2 ) ,
then one easily checks that f an R-homomorphism with
y ( f ) = ( f o i1, f o i2 ) = ( f1, f2 ) .
Finally, y is 1-1: For if f : M 1 Е M 2 ® N is an R-homomorphism so that ( f oi1, f oi 2 ) = (0, 0) О
Hom R (M 1, N ) ґ Hom R (M 2, N ) , then for all (m 1, m 2 ) О M 1 Е M 2 ,
f ((m 1, m 2 ) ) = f ((m 1, 0) + (0, m 2 ) ) = f ((m 1, 0) ) + f ((0, m 2 ) )
= f o i1(m 1 ) + f o i 2 (m 2 ) = 0 + 0 = 0
which means f is the zero map.
p1
f
f
p2
b. For f : N ® M 1 ґ M 2 , we use N ® M 1 ґ M 2 ® M 1 and N ® M 1 ґ M 2 ® M 2 to define
j
: Hom R (N , M 1 ґ M 2 ) ® Hom R (N , M 1 ) ґ Hom R (N , M 2 )
j
The proof that
j
( f ) = ( p1 o f , p 2 o f )
is an isomorphism is similar to a and is left to the reader.
16

Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
Corollary 5. Let M 1, M 2,K , M n , and N be R-modules. Then there is an isomorphism of abelian
groups:
a. Hom R (Е in= 1 M i , N ) @P in= 1 Hom R (M i , N )
b. Hom R (N , P in= 1M i ) @P in= 1 Hom R (N , M i )
Proof. Follows directly from Theorem4 by induction.

NOTES.
(i)
The proofs that y and j are isomorphisms can be done using universal properties of
direct sums and products of. Indeed for y : Starting from the homomorphisms
Hom R (M 1, N )
Hom R (M 2, N )
1
2
Hom R (M 1
M 2, N )
From the property of the direct Hom R (M 1, N ) ґ Hom R (M 2, N ) , there exists a unique homomorphism
y such that the following diagram is commutative
Hom R (M 1, N )
Hom R (M 2, N )
1
2
Hom R (M 1, N )
Hom R (M 2, N )
2
1
Hom R (M 1
M 2, N )
Since such y is unique it must have the form given in a. The fact that it is an isomorphism follows
from the property of direct sum M 1 Е M 2 : For any f1 : M 1 ® N and f2 : M 2 ® N , there exists a
unique f : M 1 Е M 2 ® N so that f o i1 = f1 and f o i 2 = f2 which shows y is onto. Now, y is 1-1
since 0 : M 1 Е M 2 ® N is the unique map with 0 o i1 = 0 and 0 o i 2 = 0 .
(ii)
Theorem 4 is also true in the infinite case, i.e.
a. Hom R (Е i ОI M i , N ) @P i ОI Hom R (M i , N )
b. Hom R (N , P i ОI M i ) @P i ОI Hom R (N , M i )
When R is commutative, recall that the group of homomorphisms is an R-module.
17
Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
Theorem 6. Let M be an R-module, R commutative. Then there is an isomorphism of R-modules
Hom R (R , M ) @M .
Proof. Consider the map Q : Hom R (R , M ) ® M : Q( f ) = f (1) . The fact that  is an Rhomomorphism follows directly from the properties
( f + g)(1) = f (1) + g(1)
(rf )(1) = rf (1)
 is 1-1, since Q( f ) = 0 Ы f (1) = 0  for all r О R , f (r ) = f (r 1) = rf (1) = 0 ,  f = 0. Now, we
show  is onto. For any m О M , then with {1} as a basis for R, let f : R ® M be the unique
homomorphism so that f (1) = m . Hence  ( f ) = m.

Corollary 7. Let M be an R-module, R is commutative. Then there is an isomorphism of R-modules
Hom R (Е ni = 1 R , M ) @P ni = 1M .
Proof. Use Corollary5 with M i = R for all i = 1,2,K , n .

NOTE. It is also true that if I is any indexing set, then Hom(Е i ОI R , M ) @P i ОI M . (Compare with the
previous note).
Proposition 8. Let 0 ® M ў®f M ®g M ўў® 0 be an exact sequence of R-modules. The following
conditions are equivalent:
a) There exists a homomorphism j : M ўў® M such that g o j = idM ўў.
b) There exists a homomorphism y : M ® M ў such that y o f = idM ў.
If these conditions are satisfied, then we isomorphisms:
M @Im f Е ker y ,
M @ker g Е Im j ,
M @M ўЕ M ўў
Proof. Assume a) then we have
g
ѕ ѕѕѕ® M ўў® 0
M¬
j
Let x О M , then x - j (g (x ))О ker g  M = ker g + Im j . ker g + Im j @ker g Е Im j : If
x О ker g З Im j Ю x = j (y ) Ю 0 = g (x ) = g (j (y )) = y Ю x = 0 .
Ю M = ker g Е Im j . ker g = Im f @M ў and Im j @M ўў  M @M ўЕ M ўў.
Assume b)
f
ѕ® M
0 ® M ў¬ѕ ѕѕ
y
18
Math 540, Lecture Notes
3. The Group of Homomorphisms
Dr. Alshammari
Let x О M , then x - f (y (x )) О ker y  M = ker y + Im f . ker y + Im f @ker y Е Im f : If
x О ker y З Im f Ю x = f (y ) Ю 0 = y (x ) = y (f (y )) = y Ю x = 0 .
Ю M = ker y Е Im f . Im f @M ў and ker y @M ўў: The restriction of g to ker y Н M is a
homomorphism, we again call it g : ker y ® M ўў. Since M = ker y + Im f and g (M ) = M ўў 
M ўў= g (M ) = g (ker y + Im f ) = g (ker y ). If x О ker y and g (x ) = 0 , then x О ker g = Im f . But
Im f З ker y = 0 Ю x = 0 . We proved that the restriction is an isomorphism ker y @M ўў. Hence
M @M ўЕ M ўў.
It remains to the equivalence of a) and b). To do this it suffices to show if M @M ўЕ M ўў, then both a)
and b) are true. If this isomorphism is given by h : M ® M ўЕ M ўў. It follows from the universal
property of direct products and sums that the required maps are given by j := h - 1 o i2 and y := p1 o h
where p1, p 2 , i1, i 2 are the canonical maps.
Mў
i1
®
i2
¬
M ўЕ M ўў
M ўў
Ї h- 1 [
]
f
M
p1 oh
[
Mў ¬
p1
h - 1 oi 2
g
Їh
]
M ўґ M ўў
®p
M ўў
2
Definition 9. If the above conditions are satisfied, we say that the exact sequence splits.
Example 10. The exact sequence of ў -modules 0 ® ў 2 ® ў 4 ® ў 2 ® 0 dose not split. Since if it
did, then we would have ў 4 @ ў 2 Е ў 2 .
Exercise 11. Prove that Hom ў (ў m , ў n ) @ ў (m ,n ) , where
(m , n ) denotes the greatest common
m
m
divisor of m and n. (Hint: Use the exact sequence 0 ® ў ® ў ® ў m ® 0 and apply Hom ў (
induce the exact sequence
o mm
0 ® Hom ў (ў m , ў n ) ® Hom ў ( ў , ў n ) ® Hom ў ( ў , ў n )
Note that o mm is also multiplication by m on Hom ў ( ў , ў n ) @ ў n ).
19
, ў n ) to