Download HW 3

Math 210B: Algebra, Homework 3
Ian Coley
January 29, 2014
Problem 1.
Let R = Z[X]. Show that the sequence of R-module homomorphisms
f
h
0→R→R→Z→0
where f (g) = X · g and h(g) = g(0) is exact. Does it split? Does it split as a sequence of
abelian groups? (We view Z as an R-module via X · 1 = 0.)
Solution.
P
P
Let g ∈ R[X], and write g = ni=0 ai X i . Then f (g) = ni=0 ai X i+1 , so it has no constant
term. Hence h(f (g)) = 0, so im f ⊂Pker h. Now, it is also evident that every polynomial
which satisfies g(0) = 0 is precisely Pni=0 ai 0i = a0 = 0 (where we use 00 = 1 to simplify
notation). Therefore if g ∈ ker h, g = ni=1 ai X i . Therefore we have
!
n−1
n−1
n
X
X
X
f
ai+1 X i = X ·
ai+1 X i =
ai X i = g,
i=0
i=0
i=1
so ker f ⊂ im h. Hence ker f = im h so the sequence is exact at the middle term. Further, f
is injective since Z[X] is a domain, so left multiplication by any element is injective. Finally,
h is surjective since the constant polynomials g = a have g(0) = a for every a ∈ Z. Therefore
the sequence is exact.
The sequence is split over R. The requisite map q : Z → R so that h ◦ q = 1Z . It is clear
that q(a) = f (ga ) + a for some ga ∈ R, since h(q(a)) = a implies that q(a) has constant term
a. Since we require
q(ab) = q(a)q(b) = f (ga )f (gb ) + b · f (ga ) + a · f (gb ) + ab
= b · q(a) = b · f (ga ) + ab,
we see that deg ga = 0 for every a ∈ Z so that deg q(a)q(b) = deg q(a). Therefore we have
q(a) = a, the constant polynomial. However X · q(a) = aX 6= 0 = q(X · a), so this sequence
cannot be split.
Additionally, the sequence is split as abelian groups, i.e. as Z-modules. Since Z is a free
Z-module, it is projective, so there exists a splitting q : Z → R. Therefore the sequence is
split.
1
Problem 2.
(a) Let I be an ideal of a ring R. Show that for every left R-module M , the factor group
M/IM has a natural structure of a left R/I-module (here IM is the subgroup in M
generated by xm for all x ∈ I and m ∈ M ). Show that if M is a free R-module, then
M/IM is a free R/I-module.
(b) Show that if the free modules Rn and Rm over a nonzero commutative ring R are
isomorphic, then n = m.
Solution.
(a) We see the action of R/I on M/IM by
(r + I) · (m + IM ) = rm + IM.
The distributive laws hold since they do for M as an R-module. To see this is well
defined, suppose r + I = r0 + I, and r = r0 + i. Then
(r+I)·(m+IM ) = (r0 +i+I)·(m+IM ) = r0 m+im+IM = r0 m+IM = (r0 +I)·(m+IM ).
Suppose that m + IM = m0 + IM and m = m + in. Then
(r + I) · (m + IM ) = (r + I) · (m0 + in + IM )
= rm0 + rin + IM = rm0 + IM = (r + I) · (m0 + IM )
since ri ∈ I. Therefore the action of R/I on M/IM is well defined.
Now suppose M is a free R-module, thus M ∼
= Rn . Then by the reasoning in part (b),
M/IM ∼
= (R/I)n . Therefore M/IM is a free R/I module.
(b) Since R is commutative, there exists a maximal ideal m ⊂ R. We have
mRn = {m(r1 , . . . , rn ) : m ∈ m, ri ∈ R} = mn
simply by taking the generators (0, . . . , 1, . . . , 0). We similarly have mRm = mm . Then
we claim that
∼
ϕ : Rn /mn → (R/m)n via (r1 , . . . , rn ) + mn 7→ (r1 + m, . . . , rn + m).
This map is injective since ϕ((r1 , . . . , rn ) + mn ) = (m, . . . , m) implies ri ∈ m for each i,
so that (r1 , . . . , rn ) + mn = mn . This map is surjective since
(r1 , . . . , rn ) + mn ∈ ϕ−1 (r1 + m, . . . , rn + m)
for every element in (R/m)n . We have a similar result for Rm /m ∼
= (R/m)m . Let
F = R/m, which is a field. Then
Rn ∼
= Rm =⇒ Rn /mn ∼
= Rm ∼
= mm ⇐⇒ F n = (R/m)n ∼
= (R/m)m = F m .
Since F n , F m are vector spaces, in particular they are free R modules, and are isomorphic if and only if their bases have the same cardinality, i.e. n = m. This completes
the proof.
2
Problem 3.
Prove that if every module over a domain R is free, then R is a field.
Solution.
Suppose that I ⊂ R is a nonzero, proper ideal. Then R/I is a nontrivial R-module, which
by assumption is free. Then I annihilates R/I, which would make R/I a torsion module,
which is a contradiction since R is a domain. Therefore R has no nonzero, proper ideals, so
R is a field.
Problem 4.
(a) Show that a direct sum of projective modules is projective.
(b) Show that a direct product of injective modules is injective.
Solution.
(a) Recall that one characterisation of projective modules is the diagram
P
h
M
g
0
N
f
`
Then let P = Pα be a direct sum of projective modules. Then for each gα : Pα → N ,
there exists hα making the
` above diagram
Q commute. Then supposeQwe have a map
∼
g : P → N . Since Hom(
Pα , N ) =
Hom(Pα , N ), we have g =
gα . Therefore
Q
there is a unique h = hα making the diagram commute. Therefore P is a projective
module.
(b) The proof is analogous to the above. We need the diagram
0
X
f
Y
g
h
Q
Q
Q
Q
to commute. Suppose that Q =
Qβ . We use Hom(X, Qβ ) ∼
Hom(X, Qβ ).
=
Given injective modules Qβ and maps gβ : X → Q
Qβ , there are unique lefts hβ : Y →
QQβ .
Therefore since a map g : X → Q, we have g = gβ , so there is a unique lift h = hβ ,
so Q is an injective module.
3
Problem 5.
Show that Q is not a projective Z-module.
Solution.
If Q were a projective Z-module, it would be a direct summand of a free Z-module F . Then
we have the projection map p : F → Q and an inclusion map i : Q → F such that p ◦ i = 1Q .
Then
n
X
i(1) =
aj ej
j=1
where ej are basis elements of F and aj ∈ Z. Let N = 1 + maxj=1,...,n |aj |. Then
i(1) = N · i(1/N ) =
n
X
aj ej ,
j=1
which implies that N | aj for all j. But since N > |aj | for all j, this implies aj = 0 for all j.
But this implies that
p(i(1)) = p(0) = 0 6= 1
which is a contradiction. Hence Q is not a projective Z-module.
Problem 6.
Find an example of a non-free module N ⊂ M of a free module M over some ring R.
Solution.
Let R = Z[X] and let M = R. Then submodules of M are precisely ideals of R. Suppose
that a free submodule I ⊂ R had a basis of more than two elements, and let a, b be those
elements. Then since −b · a + a · b = 0, we have a nontrivial relation between basis elements,
which is a contradiction. Hence any free submodule of R must be cyclic, i.e. it must be a
principal ideal.
Now let N = (2, X). We claim that N is not principal, so that N cannot have a basis
and hence is not a free module. Suppose that N = (p) for p ∈ Z[X]. Then we would need
a · p = 2 for some a ∈ Z[X]. Therefore since Z is a domain, we must have deg p = 0, i.e.
p ∈ Z. Since 2 is prime, we have p = ±1 or ±2. Clearly p = ±1 is inappropriate since
(p) = M 6= N . Additionally, if p = ±2, then we cannot obtain b · 2 = X for any choice of
b ∈ Z[X]. Therefore N is not principal, and we are done.
Problem 7.
Prove that M ⊕ N is a projective (resp. injective) R-module, then M and N are also
projective (resp. injective).
Solution.
Since M ⊕ N is projective, there exists P 0 such that M ⊕ N ⊕ P 0 ∼
= Rn is a free module.
But then M ⊕ (N ⊕ P 0 ) ∼
= Rn , so M and N are projective too.
= Rn and N ⊕ (M ⊕ P 0 ) ∼
If M ⊕ N is injective, then for every injective homomorphism f : X → Y and map
g : X → M ⊕ N , there exists a unique lift h : Y → M ⊕ N . Then since M ⊕ N has
projection maps pM and pN onto M and N , we have:
4
0
X
f
Y
g
h
M ⊕N
pM , pN
M, N
which shows that M and N are injective too.
Problem 8.
Let a1 , . . . , an be elements of a commutative ring R generating the idealPR. Show that
the submodule M in Rn consisting of all n-tuples (x1 , . . . , xn ) such that ni=1 ai xi = 0 is
projective.
Solution.
P
Consider the map ϕ : Rn → R given by (x1 , . . . , xn ) 7→ ni=1 ai xi . Then M = ker ϕ, and ϕ
is surjective since the ai generate R. Hence by the first isomorphism theorem, R ∼
= Rn /M ,
i.e. Rn ∼
= M ⊕ R. Therefore M is a direct summand in a free module, so it is projective.
Problem 9.
Prove that every Z/6Z-module is projective and injective. Find a Z/4Z-module that is
neither projective nor injective.
Solution.
First, note that if every Z/6Z-module is projective, then every Z/6Z-module is injective. To
see this, we would like to show that for a module A and exact sequence
0 → A → B → C → 0,
it splits for any valid choices B and C. But since C is projective, it does split, so A is
injective. Therefore we would like to show that every Z/6Z-module is projective. From
the previous homework, we know that since Z/6Z ∼
= Z/2Z ⊕ Z/3Z, every Z/6Z module is
expressible as a direct sum of Z/2Z and Z/3Z modules. Since Z/2Z and Z/3Z are fields,
every module over them is free, and a fortiori projective. By Problem 4(a), the direct sum
of projective modules if projective, so every module over Z/6Z is projective. Hence every
Z/6Z-module is injective and projective.
Now consider Z/2Z as a Z/4Z-module, where the action is given by left multiplication
modulo 2. Then we have an exact sequence
f
g
0 → Z/2Z → Z/4Z → Z/2Z → 0
where f (0) = 0 and f (1) = 2 and g(0) = g(2) = 0 and g(1) = g(3) = 1. Then if Z/2Z were
either projective or injective, this sequence would be split. However, Z/4Z ∼
6= Z/2Z ⊕ Z/2Z.
Hence Z/2Z is neither projective nor injective as a Z/4Z-module.
5
Problem 10.
√
√
Prove that the ideal I ⊂ Z[ −5] generated by 2 and 1 + −5 is a projective R-module. Is
I a free module?
Solution. √
√
Let R = Z[ −5] and let f : R2 → I be given by (r, s) 7→ 2r + (1 + −5)s. Then we have a
short exact sequence
0 → ker f → R2 → I → 0.
We claim this sequence is split. Let j : I → R2 be given by
√
1 − −5
α .
j(α) = 2α, −
2
This is only valid if
we have
√
1− −5
α
2
√
∈ R for any α ∈ I. Since we may represent α = 2x+(1+ −5)y,
√
√
√
−5
(2x + (1 + −5)y) = −(1 − −5)x − 3y ∈ R.
−
2
So this is well defined. Then we see
√
√
√
1 − −5
1 − −5
α 7→ 2α, −
α 7→ 4α − (1 + −5)
α = α.
2
2
1−
Thus the sequence splits, so I ⊕ ker f ∼
= R2 , hence I is projective. But I is not principal,
so it cannot
be a free module. To see it is not principal, suppose that I = (α). Then since
√
2, 1 + −5 ∈ I, we would need
√
N (α) | N (2) = 4, N (α) | N (1 + −5) = 6,
where N is the usual complex
implies that N (α) = 2 (since N (α) = 1 is
√ norm. 2But this
2
nonsense). However N (a + b −5) = a + 5b , so N (α) = 2 is impossible. Therefore we are
done.
6