0.1
Modules
Exercise 1 Let R be a ring with 1 and M be a left R-module. Prove that 0R m = 0M ; r0M = 0M and
( 1) m = m for all m 2 M; r 2 R.
Solution 1 0R m = (0R + 0R ) m = 0R m + 0R m which implies 0R m = 0M . Similarly, r0M = r (0M + 0M ) =
r0M + r0M which implies r0M = 0M . 0M = 0R m by the previou result. So, 0M = 0R m = (1 1) m =
1:m + ( 1) m = m + ( 1) m which implies ( 1) m is the additive inverse of m in M . Hence, ( 1) m = m.
Exercise 2 Assume that rm = 0 for some r 2 R and some m 2 M with m 6= 0. Prove that r does not have
a left inverse (i.e. there is no s 2 R such that sr = 1).
Solution 2 Assume by contradiction, rm = 0 for some r 2 R, m 2 M with m 6= 0 and r has a left inverse
s then m = 1R m = (sr) m = s (rm) = s0M = 0M contradicts with the assumption m is non-zero.
Exercise 3 Give an example of an R-module M with non-zero elements r 2 R; x 2 M such that rx = 0.
Solution 3
1. Any abelian group is a Z module. So we can give as an example R = Z and M = Zm let
m 2 R and 1 2 M then m:1 = m:1 = m = 0.
2. Any ring R is an R-module. R = Zm and M = Zm where m is a composite number can be given as
an example. There exists integers 1 < a; b < m with m = ab. In other words, a 2 R and b 2 M are
non-zero elements but ab = ab = m = 0 2 M .
0 a
: a 2 R is a subring of M at2
0 0
two elements is zero. Just consider R as an R-module.
3. It is easy to see that R =
0
0
4. Similarly in the ring R = M at2
an R-module.
2
1
0
0
0
5
0
=
0
0
0
0
2
(R) moreover the product of any
:
(2Z4 ) the product of any two elements is zero. Just consider R as
2 0
0 2
2 0
0 2
=
0 0
0 0
:
Exercise 4 Let R and S be commutative rings and let ' : R ! S be a ring homomorphism. If M is an
S-module prove that M is also an R-module if we de…ne rm = ' (r) m for all r 2 R and m 2 M .
Solution 4 M is an abelian group and 8r 2 R; m 2 M; rm = ' (r) m 2 M since M is an S-module and
' (r) 2 S. Next we are to show that (r + p) m = rm + pm; ; 8r; p 2 R; m 2 M is satis…ed;
(r + p) m =
=
=
=
' (r + p) m; by de…nition of product
[' (r) + ' (p)] m since ' is a ring homomorphism,
' (r) m + ' (p) m since M is an S-module
rm + pm by de…nition of product.
Next we need to show that (rp) m = r (pm) ; 8r; p 2 R; m 2 M ;
(rp) m =
=
=
=
=
' (rp) m by de…nition of product
[' (r) ' (p)] m since ' is a ring homomorphism,
' (r) (' (p) m) since M is an S-module
' (r) (pm) by de…nition of product
r (pm) by de…nition of product.
As last step we need to show that r (m + n) = rm + rn; 8r 2 R; m; n 2 M is satis…ed;
r (m + n)
= ' (r) (m + n) by de…nition of product
= ' (r) m + ' (r) n since M is an S-module
= rm + rn by de…nition of product.
Hence, M is an R-module.
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Exercise 5 Find all Z-module homomorphisms from Z=30Z to Z=21Z. (In other words, …nd all group
homomorphisms from Z30 to Z21 )
Solution 5 The groups are additive cyclic groups generated by 1 so the homomorphism is determined by the
image of this generator. We claim that
'a
k
:
Z30 ! Z21
f
7! ka
is a well-de…ned Z-module homomorphism i¤ 30a 0 (mod 21). If 'a is well de…ned then 'a 0 = 'a 30
f = 30a
g which implies 30a 0 (mod 21). Conversely, if 30a 0 (mod 21) then 'a is well de…ned and
i.e 0a
f la
e = ' k +' l so it preserves addition. Similarly,
8k; l 2 Z30 ; 'a k + l = 'a k + l = (k^
+ l) a = ka+
a
a
g = z ka
f = z' k . Hence ' is a Z-module homomorphism
8k 2 Z30 and z 2 Z; 'a zk = 'a zk = zka
a
a
for a = 0; 7; 14.
Exercise 6 Give an example to a map from one R-module to another which is a group homomorphism but
not an R-module homomorhism.
Solution 6 Let R = Z [x] and M = Z [x] , N = 2Z [x] in other words consider R as an R-module and its
ideal as an R-module ' : N ! M be de…ned as ' (f (x)) = f x2 . ' is a group homomorphism from (N; +)
to (M; +) since
' (f (x) + g (x))
= ' ((f + g) (x)) = (f + g) x2
= f x2 + g x2 = ' (f (x)) + ' (g (x))
But it is not an R-module homomorphism. Assume by contradiction it is an R-module homomorphism so
' (rm) = r' (m) for all r 2 R; m 2 M . As a counter example; ' (x) = x2 on the other hand, ' (1) = 1 and
' (x) = ' (x:1) = x' (1) = x which implies x2 = x and that is a contradiction.
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