1.
a. 0x = (0 + 0)x = 0x + 0x, so 0x = 0.
c. Using (a), x + (− 1)x = (1 + (− 1))x = 0x = 0 = x + (− x) . Now “subtract” x from both sides by adding −x to both sides.
2.
a. If
α :
M →
N is onto and
R-linear, and if
M =
Rx1 +
+
Rxn, then
N =
Rα(
x1) +
+
Rα(
xn) . Since some of the
α(
xi) may be zero, the result follows.
c. Let
K =
Rx1 +
+
Rxm and let
M/
K =
R(
y1 +
K) +
+
R(
yn +
K) where the
xi and
yj are in
M. If
x M let
with
ri R for each
i. Then
x − (
r1y1 +
+
rnyn) is in
K, so
where
sj R for each
j. Hence {
x1, . . .,
xm,
y1, . . .,
yn} generates
M.
3. If
Re is an ideal then
er (
Re)
r ⊆
Re for any
r R, say
er =
se,
s R. Hence
ere =
se2 =
se =
er, that is
er(1 −
e) = 0 . As
r was arbitrary, this proves
eR(1 −
e) = 0 . Conversely, if
eR(1 −
e) = 0 then
er =
ere for each
r R. Hence
er Re for all
r, so
Rer ⊆
Re, proving that
Re is a right ideal. Hence it is an ideal.
5. Recall that
ki K,
n ≥ 1}.
a. Given
x = Σ
aiki in
AK and
r R, then
rx = Σ
r(
aiki) = Σ(
rai)
ki AK because
A is a left ideal.
c. If
x (
A +
B)
K then
This proves that (A + B)K ⊆ AK + BK. The reverse inclusion is similar.
7. Let
K and
N be submodules of
M. We use the module isomorphism theorem.
a. Define
α :
N → (
K +
N)/
K by
α(
n) =
n +
K for all
n N. Then
α is
R-linear and ker
α = {
n n +
K =
K} =
K ∩
N. So by Theorem 1 it suffices to show that
α is onto. But each element of (
K +
N)/
K has the form (
k +
n) +
K =
n +
K =
α(
n), as required.
8. Let
R be an integral domain. Given
RM let
T(
M) = {
x M x is torsion}.
a. We must show
T(
M) is a submodule of
M. Let
x T(
M), say
ax = 0 for 0 ≠
a R. If also
y T(
M), say
by = 0 where 0 ≠
b R, then (since
R is commutative)
ab(
x +
y) =
b(
ax) +
a(
by) = 0 + 0 = 0 . Since
ab ≠ 0 (
R is a domain), this shows that
s +
t T(
M) . Similarly, if
r R then
so
rx T(
M).
9. If
M =
P ⊕
Q are modules, we must show that
M/
P ≅
Q and
M/
Q ≅
P. Define
π :
M =
P ⊕
Q →
Q by
π(
p +
q) =
q for all
p P and
q Q. This is well defined because if
p +
q =
p1 +
q1 then
p −
p1 =
q −
q1 P ∩
Q = 0, whence
q −
q1. It is easy to see that
π is an onto
R-morphism, and ker
π =
P. Hence the isomorphism theorem gives
M/
P ≅
Q. A similar argument shows
M/
Q ≅
P.
11.
a. Yes. (m, n) = (n, n) + (m − n, 0) shows that M = K+ X ; clearly K ∩ X = 0.
c. Yes. (
m,
n) = (3
m − 2
n, 3
m − 2
n) + (2(
n −
m), 2(
n −
m)) shows that
M =
K +
X. If (
m,
n)
K ∩
X, then (
m,
n) = (
k,
k) and (
m,
n) = (2
l, 3
l) so
k = 2
l and
k = 3
l. Hence
k = 0, proving that
K ∩
X = 0.
13. Let (
k1 +
+
ks) +
m2 +
+
mr = 0 in
K1 ⊕
⊕
Ks ⊕
M2 ⊕
⊕
Mr. Then (
k1 +
+
ks) = 0 and
m2 =
=
mr = 0 because
is direct. Then
k1 =
=
ks = 0 because
K1 +
+
Ks is direct.
15. As in the Hint, let 1 =
e +
f where
e A and
f B. If
a A then
because
a −
ae A and
af B (using the fact that
A and
B are left ideals). Hence
a =
ae for all
a A so, since
e A, we obtain
e2 =
e and
A ⊆
Re. But
Re ⊆
A because
A is a left ideal and
e A, so
A =
Re. Now observe that
f = 1 −
e satisfies
f2 =
f, so
B =
Rf =
R(1 −
e) follows in the same way.
16.
a. We have
M =
π(
M) + ker
π because
m =
π(
m) + (
m −
π(
m)) for each
m M and
π[
m −
π(
m)] =
π(
m) −
π2(
m) = 0 . If
m π(
M) ∩ ker
π, let
m =
π(
m1) with
m1 M. Then 0 =
π(
m) =
π2(
m1) =
π(
m1) =
m, so
π(
M) ∩ ker
π = 0.
17. Suppose
with
αβ = 1
M. If
m M observe that
because
αβ = 1
M. Hence the fact that
m =
βα(
m) + [
m −
βα(
m)] shows that
M =
β(
M) + ker (
α) . But if
x β(
M) ∩ ker (
α) and we write
x =
β(
m), then 0 =
α(
x) =
α[
β(
m)] =
m, whence
x =
β(
m) =
β(0) = 0. Hence
and we have proved that N = β(M) ⊕ ker (α).
18. We use Corollary 3 of Theorem 3 several times.
a. We have
K ≅
X as both are isomorphic to
However
because
while
because
19. Since gcd (
m,
n) = 1 let 1 =
xn +
ym,
. Then if
Hence
G =
Gm +
Gn. If
g Gm ∩
Gn, then
mg = 0 =
ng so
Thus G = Gm ⊕ Gn.
21. Here
A is an ideal of a ring
R, and
RW is a module.
a. To see that
is well defined, let
we must show that
We have
say
where each
ai A. Because
α is
R-linear, we obtain
Hence
as required.
23. Let α : M → N where M and N are simple. If α ≠ 0, then α(M) is a nonzero submodule of N. By hypothesis, α(M) = N, that is α is onto. Again, α ≠ 0 implies that ker (α) ≠ M. But then ker (α) = 0, again because M is simple, and we have shown that α is one to one. Hence α is an isomorphism.
24. (2) ⇒ (3). By (2), we can identify
P with a summand of a finitely generated free module
F, say
F =
P ⊕
Q, an internal direct sum. Define
π :
F →
P by
π(
p +
q) =
p for all
p P and
q Q. Let {
x1,
xn} be a basis of
F. Given the diagram we have
βπ :
F →
N, so since
α is onto, choose
mi M such that
α(
mi) =
βπ(
xi) for each
i. By Theorem 6, there exists an
R-homomorphism
θ :
F →
M such that
θ(
xi) =
mi for each
i. Then
αθ(
xi) =
α(
mi) =
βπ(
xi) for each
i, so
αθ =
βπ because the
xi span
F. But then, if
p P, we obtain
αθ(
p) =
βπ(
p) =
β(
p) because
π(
p) =
p. Hence the restriction
γ :
P →
M given by
γ(
p) =
θ(
p) for
p P, satisfies all our requirements.
25. As in the Hint:
i. is divisible. (If
and
then
nx =
q where
n =
ii. If
Q =
K ⊕
N is divisible, then
K is divisible. (If
and
k K, let
nx =
k,
x G. If
x =
y +
z,
y K,
z N, then
nx =
ny +
nz so
nx =
ny because
K +
N is direct. Hence
ny =
k,
y K .)
iii. is not divisible. (
and
but 2
x = 3 has no solution in
Now assume that
is free and let {
bi i I} be a basis. Then
and
It follows that
is divisible, a contradiction.