Math 540
Practice Exercises
Solution
The Hom( , ) & Splitting Exact Sequences
1. Compute Hom ў ( ў , ў 7 ) , Hom ў ( ў 7 , ў ) , Hom ў ( ў 7 , ў 14 ) , and Hom ў ( ў 14 , ў 7 ) .
Sol. Hom ў ( ў , ў 7 ) @ ў 7 , by a theorem. Hom ў ( ў 7 , ў ) = 0 , since ў 7 is torsion and  is free.
Hom ў ( ў 7 , ў 14 ) @ Hom ў ( ў 7 , ў 7 Е ў 2 ) @ Hom ў ( ў 7 , ў 7 ) Е Hom ў ( ў 7 , ў 2 ) @ ў 7 Е 0 @ ў 7 .
Similarly, Hom ў ( ў 14 , ў 7 ) @ Hom ў ( ў 7 Е ў 2, ў 7 ) @ ў 7 .
2. Let R be commutative. Show that an R-module M is free of finite rank iff Hom R (R ,M) is free of
n
finite rank for all n і 1 .
Sol. Ю If M @ R m for some m і 1 , then for all n і 1 , Hom R (R n , R m ) @ Е ni = 1 Hom R (R , R m ) @
n
n
Е ni = 1 R m @ Е mn
i = 1 R . Ь Since Hom R (R ,M) @ M , then the result follows from n =1.
3. Let R be commutative and M an R.-module. Prove that Hom R (R , Hom R (R , M )) @ M .
Sol. Since Hom R (R , M ) @ M Ю Hom R (R , Hom R (R , M )) @ Hom R (R , M ) @ M .
4. Let R = F a field. Show that every exact sequence of R-modules splits.
Sol. Follows from the fact that all F-modules are vector spaces. Since any vector space has a basis, then
all F-modules are free, hence are projective.
5. Give an example of an exact sequence of -modules that dose not split.
ґ 2
Sol. The exact sequence abelian groups 0 ® ў ® ў ® ў 2 ® 0 does not split, since if it did, then
ў @ ў Е ў 2 which is impossible(why?).
Projective and Injective Modules
1. Show that ў n , n і 1 is both projective and injective ў n -module.
Sol. ў n is a projective ў n -module because it is free. We show ў n is an injective ў n -module, by
showing that any homomorphism f : J ® ў n , where J  ў n an ideal, extends to a map j : ў n ®
ў n . Note that ideal J of ў n has the form kў n , where k |n. Let f ([k ]) = [m ] . Since ( nk ) [k ] = [0 ],
then ( nk ) f ([k ]) = ( nk )[m ] = [0 ]  n |
j ([k ]) = [dk ] =
[mk
nm
k
 k | m . Let
m
k
= d and define j ([x ]) = [dx ] . Note that
k ] = [m ] = f ([k ]) , as desired.
2. Let M be an R-module and let N be a submodule. Show that if N and M N are projective then M is
projective.
Sol. Since M N is projective, then the exact sequence 0 ® N ® M ® M N ® 0 splits 
M @ N Е M N a direct sum of projectives, hence M is projective.
3. Prove that every R-module is projective iff every R-module is injective.
Sol. Ю Let Q be an R-module. Since all R-modules are projective, then any exact sequence of the form
0 ® Q ® M ® M ўў ® 0 splits as M ўў is projective. Therefore, Q is injective. Ь Let P be an Rmodule. Since all R-modules are injective, then any exact sequence of the form
0 ® M ў ® M ® P ® 0 splits as M ў is injective. Therefore, P is projective.
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Math 540
Practice Exercises
Solution
4. Let R be a ring. a О R is called idempotent if a 2 = a .
(a) Show that the ideal R a is a projective module over R.
(b) Show that if R J is projective, then J = R a for some idempotent a О R .
Sol. (a) Let T a ,T 1- a be multiplication by a and multiplication by 1 - a maps on R. Define
T : R ® R ґ R : T (r ) = (T a (r ),T 1- a (r )) = (r a , r (1 - a )) . T is clearly an R-homomorphism with
image R a ґ R (1 - a ). T is also 1-1 since T (r ) = 0 Ы r a = 0, r (1 - a ) = 0 
r a + r (1 - a ) = 0 Ю r (a + 1 - a ) = 0  r = 0. Hence, we have an isomorphism R 
R a Е R (1 - a ) and R a is a direct summand of free.
f
g
(b) If R J is projective, then the seq. 0 ® J ® R ® R J ® 0 splits. Then using the splitting maps
j : R J ® R , g o j = id
y : R ® J , y o f = id
Let e = y (1) , then f (e ) = e Ю e = ( y o f )(e ) = y ( f (e )) = y (e ) = y (e1) = e y (1) = ee . Now let
a О J , then a = y ( f (a )) = y (a ) = y (1a ) = ae  a О Re .
5. If R is ID, then its field of fractions F is injective.
Sol. It suffices to show that a map g : J ® F extends to a map j : R ® F , for any ideal J of R.
Given r О R , there exists a nonzero s О R such that rs = x О J , define j (r ) =
well-defined, since if also we have rs1 = r1 О J , then r =
g(s1x ) = g(sx1) Ю s1g(x ) = sg(x1) Ю
g(x )
s
=
g(x1 )
s1 .
x
s
=
x1
s1
g( x )
s
. Note that j is
 s1x = sx 1 О J and
Hence, j extends g.
6. Show that a quotient, direct summand, direct product, direct sum of divisible abelian groups are
divisible.
Sol. Straightforward.
Modules Over PID's
1. For each n і 1 , find a free -module M of rank n, with a proper submodule N of rank n.
Sol. M = Е ni= 1 ў and N = Е ni= 1(i + 1)ў .
2. Let R be a commutative ring so that every submodule of a free module is free. Show that R is a PID.
Sol. Since submodules of R are the ideals, then R must be an integral domain, as the principal ideals Ra
are free for a О R , a № 0 . Since we know any two elements of R are dependent, then a basis for every
ideal consists of a single element i.e. every ideal is principal.
3. Prove that over a PID, all projective modules are free.
Sol. A projective module is a direct summand, hence a submodule, of a free module. Over a PID, a
submodule of a free module is free.
4. Let R be PID and let F be its field of fractions, prove that if F is a projective R-module, then R is a
field. Use this to give another proof that  is not projective over .
Sol. If F is projective over R, a PID, then it is free. Since any two elements of F are linearly
independent over R, a basis for F consists of a single element. Hence, we have an isomorphism F  R
over R. Then R must be a field.
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Math 540
Practice Exercises
Solution
5. Prove or give a counter example.
a. Every submodule of a projective ў -module is projective.[Hint: Every submodule of
a free ў -module is free]
b. Every submodule of an injective ў -module is injective.
c. Every quotient of a projective ў -module is projective.
d. Every quotient of an injective ў -module is injective.
Sol. a. True. Since  is a PID, a projective module is free, its submodules are then free. Therefore they
are necessarily projective.
b. False.  is an injective -module and  is a submodule which is not injective.
c. False.  over itself is free, hence projective. The quotient ў 2 @ ў 2ў is not projective.
d. True. Since injective ў -modules are divisible abelian groups and their quotients are divisible.
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