Download HOMEWORK 1 SOLUTIONS Solution.

HOMEWORK 1 SOLUTIONS
MATH 121
Problem (10.1.2). Prove that
R×
and
M
group action of the multiplicative group
satisfy the two axioms in Section 1.7 for a
R×
on the set
M.
Solution. For the rst axiom, we have to check that for
r1 , r2 ∈ R
and
m ∈ M,
we
have
For the second axiom, we have
r1 (r2 m) = (r1 r2 )m.
to check that for m ∈ M , 1m = m.
Both of these are
included as part of the denition of a module.
Problem (10.1.4). Let
M
Rn described in Example 3 and let I1 , I2 , . . . , In
following are submodules of M :
be the module
R. Prove that the
{(x1 , x2 , . . . , xn ) | xi ∈ Ii }.
{(x1 , x2 , . . . , xn ) | xi ∈ R and x1 + x2 + · · · + xn = 0}.
be left ideals of
(a)
(b)
for
N . We have to check that N is a subgroup, and that
x ∈ N , rx ∈ N . Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ N .
(a) Call the set
Solution.
r ∈R
and
Then
x + y = (x1 + y1 , . . . , xn + yn ) ∈ N
since the Ii 's are ideals and hence closed under sums. Now, let
N
and
r ∈ R.
x = (x1 , . . . , xn ) ∈
Then
rx = (rx1 , . . . , rxn ) ∈ N,
again because the
(b) Call this set
P.
Ii 's
are ideals. Hence
under scalar multiplication.
r ∈ R.
N
is a submodule.
P is a subgroup and is closed
x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ P and
As in part (a), we check that
Let
Then
x + y = (x1 + y1 , . . . , xn + yn ).
From the above, we know that this is in N , so we need to check that (x1 +
y1 ) + · · · + (xn + yn ) = 0. This is true because x1 + · · · + xn = y1 + · · · + yn = 0.
Similarly,
rx = (rx1 , . . . , rxn ) ∈ N,
and
so
Date
(rx1 ) + · · · + (rxn ) = r(x1 + · · · + xn ) = 0,
rx ∈ P . Hence P is a submodule.
: 10 January, 2011.
1
2
MATH 121
Problem (10.1.9). If
{r ∈ R | rn = 0
ideal of R.
be
Solution. Let
and
n ∈ N.
N
is a submodule of
for all
ann(N )
n ∈ N }.
M,
the annihilator of
N in R is dened to
N in R is a 2-sided
Prove that the annihilator of
N in R. Suppose r ∈ ann(N ), a ∈ R,
ar, ra ∈ ann(N ). We have
be the annihilator of
We have to show that
(ar)(n) = a(rn) = 0
because
for some
rn = 0,
and
n0 ∈ N .
(ra)(n) = r(an) = rn0 = 0
ann(N ) is a 2-sided ideal of R.
Hence
I in M is dened to be
{m ∈ M | am = 0 for all a ∈ I}. Prove that the annihilator of I in M is a submodule
of M .
Problem (10.1.10). If
I
is a right ideal of
ann(I) be the annihilator
a ∈ I . Then we have
Solution. Let
r ∈ R,
and
R, the annihilator
of
I
in
M.
of
Suppose that
m, m0 ∈ ann(I),
a(m + m0 ) = am + am0 = 0,
so
m + m0 ∈ ann(I).
Furthermore,
a(rm) = (ar)m = a0 m = 0
for some
a0 ∈ I .
Hence
Problem (10.1.11). Let
rm ∈ ann(I).
M
Thus
ann(I)
is a submodule of
be the abelian group (i.e.,
M.
Z-module) Z/24Z × Z/15Z ×
Z/50Z.
(a) Find the annihilator of
(b) Let
I = 2Z.
M
in
Z
(i.e., a generator for this principal ideal).
Describe the annihilator of
I
in
M
as a direct product of cyclic
groups.
a ∈ Z annihilates M if and only if am = 0 for all
m in a set of generators of M . One choice of generators of M is (1, 0, 0),
(0, 1, 0), and (0, 0, 1). Hence, ann(M ) is the intersection of the annihilators
Solution.
(a) An element
of the submodules generated by these elements. Clearly, the annihilators of
(24), (15), and (50), respectively, so their intersection is
(lcm(24, 15, 50)) = (600).
An element m = (m1 , m2 , m3 ) of M is annihilated by I if and only if
these submodules are
(b)
(2m1 , 2m2 , 2m3 ) = 0.
This happens if and only if
50Z/50Z,
25Z/50Z.
2m1 ∈ 24Z/24Z, 2m2 ∈ 15Z/15Z, and 2m3 ∈
m1 ∈ 12Z/24Z, m2 ∈ 15Z/15Z, and m3 ∈
or, equivalently, if
Hence
ann(I) = 12Z/24Z × 15Z/15Z × 25Z/25Z.
HOMEWORK 1 SOLUTIONS
Problem (10.2.5). Exhibit all
Z-module
3
homomorphisms from
Z/30Z
to
Z/21Z.
Solution. A
Z-module homomorphism from a cyclic module to any module is determined by where a generator is sent. (See Problem 10.2.9 below.) Let φ : Z/30Z →
Z/21Z be a Z-module homomorphism. Then we must have 30φ(1) = 0. The elements
y ∈ Z/21Z so that 30y = 0 are y = 7k (mod 21) for k = 0, 1, 2, so there are three
such homomorphisms, given by 1 7→ 7, 1 7→ 14, and 1 7→ 0.
HomR (R, M ) and M are
isomorphic as left R-modules. [Show that each element of HomR (R, M ) is determined
by its value on the identity of R.]
Problem (10.2.9). Let
R
be a commutative ring. Prove that
Solution. We dene a map
ψ : HomR (R, M ) → M given by φ 7→ φ(1). Let's show
R-modules. For φ1 , φ2 ∈ HomR (R, M ), we have
that this is a homomorphism of left
ψ(φ1 + φ2 ) = (φ1 + φ2 )(1) = φ1 (1) + φ2 (1) = ψ(φ1 ) + ψ(φ2 ),
and for
φ ∈ HomR (R, M )
and
r ∈ R,
we have
ψ(rφ) = (rφ)(1) = rφ(1) = rψ(φ),
as desired. Now let's show that
that
φ(1) = 0.
Now, let
r ∈ R.
ψ
is injective. Suppose that
ψ(φ) = 0.
This means
We have
φ(r) = rφ(1) = r0 = 0
φ is an R-module homomorphism. Hence φ = 0, as desired. Finally, we show
ψ is surjective. Pick m ∈ M . We need to nd φ ∈ HomR (R, M ) so that
φ(1) = m. We dene φ(r) = rm, but we need to check that this is actually an
R-module homomorphism. Let r1 , r2 ∈ R. We have
since
that
φ(r1 + r2 ) = (r1 + r2 )m = r1 m + r2 m = φ(r1 ) + φ(r2 )
and
Hence
φ
is indeed an
φ(r1 r2 ) = (r1 r2 )m = r1 (r2 m) = r1 φ(r2 ).
R-module homomorphism, and so ψ is an
Problem (10.2.10). Let
R
be a commutative ring. Prove that
isomorphism.
HomR (R, R)
and
R
are
isomorphic as rings.
HomR (R, R) and R are isomorphic as
R-modules, so it suces to check that our map ψ from that problem is in fact a ring
map. Hence, we have to check that ψ(φ1 ◦ φ2 ) = ψ(φ1 )ψ(φ2 ) and ψ(1) = 1. For the
Solution. We showed in Problem 10.2.9 that
rst one, we have
ψ(φ1 ◦ φ2 ) = (φ1 ◦ φ2 )(1) = φ1 (φ2 (1)) = φ1 (1)φ2 (1) = ψ(φ1 ◦ φ2 ).
For the second, we have
ψ(id) = id(1) = 1,
as desired. Thus
ψ
is a ring isomorphism.
4
MATH 121
Problem (10.2.11). Let
for each
i = 1, 2, . . . , n.
A1 , A2 , . . . , An
be
R-modules and let Bi
be a submodule of
Ai
Prove that
(A1 × · · · × An )/(B1 × · · · × Bn ) ∼
= (A1 /B1 ) × · · · × (An /Bn ).
[Recall Exercise 14 in Section 5.1.]
Solution. In Problem 5.1.4, we showed that there is an isomorphism
(A1 × · · · × An )/(B1 × · · · × Bn ) ∼
= (A1 /B1 ) × · · · × (An /Bn )
of groups.
We must now show that it is compatible with the
R-module
structure.
Recall that the map is given by
mod B1 × · · · × Bn ) = (a1
φ((a1 , . . . , an )
Now, let
r∈R
and
(a1 , . . . , an ) ∈ A1 × · · · × An .
mod B1 , . . . , an
We have
mod B1 × · · · × Bn ) = φ((ra1 , . . . , ran ) mod B1 × · · · × Bn )
φ(r(a1 , . . . , an )
= (ra1
mod B1 , . . . , ran
= r(a1
mod B1 , . . . , an
= rφ((a1 , . . . , an )
Problem (10.2.13). Let
37, Section 7.3), let
homomorphism.
then
ϕ
M
I
mod Bn )
mod B1 × · · · × Bn )
R (cf. Exercise
ϕ : M → N be an R-module
φ : M/IM → N/IN is surjective,
be a nilpotent ideal in a commutative ring
and
N
be
R-modules
and let
Show that if the induced map
n ∈ N.
By hypothesis, we can nd some
ϕ(m) = n +
ai ∈ I
and
n0i ∈ N .
For each
i,
X
X
m∈M
so that
ai n0i ,
we can nd some
ϕ(m0i ) = n0i +
Hence
mod Bn )
is surjective.
Solution. Let
where
mod Bn ).
m0i
so that
a0ij n00ij .
X
X
ϕ m−
ai m0i = n −
ai a0ij n00ij ∈ N + I 2 N.
i,j
M → N/I 2 N
is surjective. Continuing in
the same manner, we can see that all the induced maps
r
Since some I = 0, this shows that the original map ϕ :
M → N/I k N are surjective.
M → N is surjective.
Hence we've shown that the induced map
Note that we didn't use the fact that
I
is nilpotent until the last step. Consider the
R = Z, M = N = Z,
I = (3), and suppose that ϕ : M → N is multiplication by 5. Now, the induced map
M/IM → N/IN is surjective, which means that every integer is a multiple of 5 up
to a multiple of 3. For example, if n = 4, then n is a multiple of 5 up to a multiple
of 3 because 4 + 2 × 3 = 10 is a multiple of 5. The argument above shows that we
more down-to-earth example that sets aside that hypothesis. Let
HOMEWORK 1 SOLUTIONS
5
can improve the situation to saying that 4 is a multiple of 5 up to a multiple of 9,
4 + 4 × 9 = 40 is a multiple of
27, because 4 + 3 × 27 = 85 is
which is true because
5. Similarly, 4 is a multiple of 5
up to a multiple of
a multiple of 5, and so on. The
only dierence here is that we cannot conclude that
no power of
I
ϕ
is actually surjective because
is 0.
R-module M is called a torsion module if for each m ∈ M
element r ∈ R such that rm = 0, where r may depend on m (i.e.,
Problem (10.3.4). An
there is a nonzero
M = Tor(M )
in the notation of Exercise 8 of Section 1).
abelian group is a torsion
that is a torsion
Z-module.
Z-module.
Solution. Suppose that
A
Give an example of an innite abelian group
is a nite abelian group (or
by Lagrange's Theorem, for each
Prove that every nite
a ∈ A, na = 0.
Hence
example of an innite abelian group that is a torsion
Z-module) of order n. Then
A is a torsion Z-module. An
Z-module
is
∞
Y
(Z/2Z).
i=1
R-module M is called irreducible if M 6= 0 and if 0 and M are
the only submodules of M . Show that M is irreducible if and only if M 6= 0 and M is
Problem (10.3.9). An
a cyclic module with any nonzero element as its generator. Determine all irreducible
Z-modules.
Solution. Suppose that
of
M.
Then
Hence,
M
Rm
M
is an irreducible
is a submodule of
M,
R-module.
Let
m
be a nonzero element
so by hypothesis, we must have
is cyclic, and any nonzero element of
M
si a generator.
M = Rm.
For the other
M is a cyclic module, and that any nonzero element is a generator.
N is a nonzero submodule of M . Then for any nonzero n ∈ N , Rn
of M , because n is also a nonzero element of M . As any submodule of
some Rn, M must be irreducible. The irreducible Z-modules, then, are
direction, suppose
Now suppose that
must be all
M
contains
exactly the cyclic groups of prime order.
Problem (10.3.11). Show that if
nonzero
M
M1
and
M2
are irreducible
R-modules,
then any
R-module homomorphism from M1 to M2 is an isomorphism. Deduce that if
EndR (M ) is a division ring (this result is called Schur's Lemma.)
is irreducible then
[Consider the kernel and image.]
Solution. Let
φ : M1 → M2
R-module
homomorphism between two
Then ker φ and im φ are submodules of M1 and M2 , respecker φ = 0 and im φ = M2 . Hence φ is an isomorphism. Now,
let φ ∈ EndR (M ) be nonzero. By the previous argument, φ is an automorphism,
so it has an inverse. Since φ was arbitrary, every nonzero element of EndR (M ) is
invertible, so EndR (M ) is a division ring.
irreducible
R-modules.
be a nonzero
tively, so we must have
6
MATH 121
Problem (10.3.20). Let
R-module.
I
be a nonempty index set and for each
Mi
The direct product of the modules
i∈I
let
Mi
be an
is dened to be their direct product
R componentwise
Mi is dened to be the restricted direct
as abelian groups (cf. Exercise 15 in Section 5.1) with the action of
multiplication. The direct sum of the modules
product of the abelian groups
R
Mi
(cf. Exercise 17 in Section 5.1) with the action of
QMi 's is the
subset of the direct product, prodi∈I Mi , which consists of the elements
i∈I mi such
that only nitely many of the components mi are nonzero; the action of R on the
Q
Q
direct product or direct sum is given by r
i∈I mi =
i∈i rmi (cf. Appendix I for
componentwise multiplication. In other words, the direct sum of the
the denition of Cartesian products of innitely many sets). The direct sum will be
denoted by
L
i∈i
Mi .
(a) Prove that the direct product of the
of the
Mi 's
Mi 's
is an
R-module
is a submodule of their direct product.
R = Z, I = Z+ and Mi is the cyclic group of order
(b) Show that if
i,
and the direct sum
then the direct sum of the
Mi 's
i
for each
is not isomorphic to their direct product.
[Look at torsion.]
Solution.
(a) We have
(mi )i∈I + (m0i )i∈I = (mi + m0i )i∈I ∈
Y
Mi
and
Y
r(mi )i∈I = (rmi )i∈I ∈
Mi ,
Q
L
Q
so
Mi is an R-module. To show that
Mi is a submodule of
Mi , we
0
must show that if all but nitely many of the mi and mi are zero, then this
0
is also true of the mi + m + i and rmi . For the sum, there are only nitely
0
many indices i so that at least one of mi and mi is nonzero, so there can only
be nitely many indices in which the sum is nonzero. Similarly, for the scalar
product, there are only nitely many
the only positions in which
rmi
i
in which
mi
can be nonzero.
is nonzero, and these are
Hence the direct sum is
indeed a submodule of the direct product.
Q
M
is a torsion group, whereas
Mi is
i
L
not. To see that
Mi is torsion, let (mi ) be any element of Mi , and suppose
that n is large enough so that mi = 0 for i > n. Then n!(mi ) = 0, so (mi ) is
L
a torsion element. Since (mi ) was arbitrary,
Mi is a torsion group. Q
Now,
Q
let's illustrate a non-torsion element of
Mi . Take x = (1, 1, 1, . . .) ∈= Mi .
th place, n
Suppose that nx = 0. Then in order to annihilate the 1 in the i
+
must be a multiple of i. Hence, n must be a multiple of i for each i ∈ Z .
The only integer with this property is 0, so x is torsion-free.
L
Mi is countable (as a set), whereas
Q Another way to do this is to note that
Mi is uncountable.
(b) One way to do this is to observe that
L
HOMEWORK 1 SOLUTIONS
7
R be a Principal Ideal Domain, let M be a torsion R-module
p be a prime in R (do not assume M is nitely generated, hence
it need not have a nonzero annihilator cf. Exercise 5). The p-primary component
of M is the set of all elements of M that are annihilated by some positive power of p.
(a) Prove that the p-primary component is a submodule. [See Exercise 13 in
Problem (10.3.22). Let
(cf. Exercise 4) and let
Section 1.]
(b) Prove that this denition of
in Exercise 18 when
(c) Prove that
as
p
M
M
p-primary
component agrees with the one given
has a nonzero annihilator.
is the (possibly innite) direct sum of its
runs over all primes of
p-primary components,
R.
M [p] denote the p-primary component of M . Suppose that
m, n ∈ M [p] and r ∈ R. Suppose that m and n are annihilated by pa and pb ,
Solution.
(a) Let
respectively. Then
pa+b (m + n) = pa+b m + pa+b n = 0,
so
m + n ∈ M [p].
Also,
pa (rm) = r(pa m) = r0 = 0,
so
rm ∈ M [p].
Hence
M [p]
is a submodule of
(b) In this case, any element of
M [p]
M.
is annihilated by
p αp ,
so the denitions
coincide.
isomorphism.
φ : M →
L
M [p], as follows. Let m ∈ M . Then m is
contained in a nitely generated R-submodule of M , say N = Rm. Then
N has an annihilator ann(N ), so by Exercise 10.3.18, N is the direct sum
∼ L
of its p-primary components, θ : N →
N [p]. Now, we embed each N [p]
inside the corresponding M [p] in the natural way: ιp : N [p] ,→ M [p]. Now,
L
we dene φ(m) = (
ιp ◦ θ)(m). One can check that φ is in fact an R-module
(c) We dene a map