Math 593 Assignment # 6 Solutions

Bryden R. Cais November 8, 2002

1. Let k be a field and A_{i} a finite dimensional vector space over k. Consider the exact sequence:

0 ⟶^{f_{−1}} A_{0} ⟶^{f_{0}} A_{1} ⟶^{f_{1}} A_{2} ⟶^{f_{2}} … ⟶^{f_{n−2}} A_{n−1} ⟶^{f_{n−1}} A_{n} ⟶^{f_{n}} 0

The familiar rank nullity theorem tells us that

dim im f_{i} + dim ker f_{i} = dim A_{i}.

However, since the sequence is exact we have im f_{i} = ker f_{i+1} so that

dim A_{i} = dim im f_{i} + dim im f_{i−1}

for 0 ≤ i ≤ n. Hence

∑_{i=0}^{n}(−1)^{i} dim A_{i} = ∑_{i=0}^{n}(−1)^{i} (dim im f_{i} + dim im f_{i−1})

= ∑_{i=0}^{n}(−1)^{i} dim im f_{i} + ∑_{i=0}^{n}(−1)^{i} dim im f_{i−1}

= ∑_{i=0}^{n}(−1)^{i} dim im f_{i} − ∑_{i=0}^{n}(−1)^{i} dim im f_{i} + (dim im f_{−1} − dim im f_{n})

= (dim im f_{−1} − dim im f_{n})

= 0,

since the image of 0 under f_{−1} is 0 and the image of A_{n} under f_{n} is 0.

2. (a) Suppose that the sequence

0 ⟶ N ⟶^{u} M ⟶^{v} L ⟶ 0  (1)

splits, that is, we have the following commutative diagram

\begin{aligned}0 & ⟶ & N & ⟶^{u} & M & ⟶^{v} & L & ⟶ & 0 \\ & ^{ϕ_{N}} ↓ & & ^{φ}↓ & & ↓^{ϕ_{L}} \\ 0 & ⟶ & N & ⟶_{ι_{N}} & N ⊕ L & ⟶_{π_{L}} & L & ⟶ & 0 \end{aligned}  (2)

where ι_{N}(n) = (n, 0) is the natural inclusion map, π_{L}(n, l) = l the natural projection map, and ϕ_{N}, ϕ_{L}, φ are isomorphisms. Define

j : M ⟶ N

by

j := ϕ_{N}^{−1} ∘ π_{N} ∘ φ,

where π_{N}(n, l) = n is the natural projection map (and ϕ_{N}^{−1} is well defined since ϕ_{N} is an isomorphism). It is obvious that j is a homomorphism since it is the composition of homomorphisms. Then we have

j ∘ u = ϕ_{N}^{−1} ∘ π_{N} ∘ φ ∘ u = id_{N}

since the diagram above is commutative. Observe that we need not assume that ϕ_{N} = id_{N}. Similarly, define

s : L ⟶ M

by

s := φ^{−1} ∘ ι_{L} ∘ ϕ_{L},

where ι_{L}(l) = (0, l) is the natural inclusion map. Again, it is obvious that s is a homomorphism and we have

v ∘ s = v ∘ φ^{−1} ∘ ι_{L} ∘ ϕ_{L} = id_{L}

since the diagram commutes. This shows that (a) implies (b) and (c).

(b) Suppose that there is a homomorphism j : M ⟶ N such that j ∘ u = id_{N}. Define

φ : M ⟶ N ⊕ L

by

φ(m) = (j(m), v(m)).

Since j, v are homomorphisms, it is clear that φ is a homomorphism. Now suppose that φ(m) = 0. Then j(m) = (0, 0) so that m ∈ ker j = im u by exactness. It follows that m = u(n) for some n ∈ N, and hence that j(m) = j ∘ u(n) = n = 0. Therefore, m = u(0) = 0 so that φ is injective. Let (n, l) ∈ N ⊕ L be arbitrary. Since v is surjective, there exists m′ ∈ M such that v(m′) = l. Set

m = u(n) + m′ − u ∘ j(m′).

Then

v(m) = v(u(n) + m′ − u ∘ j(m′)) = v(m′) = l

since im u = ker j, and

j(m) = j(u(n) + m′ − u ∘ j(m′)) = j ∘ u(n) + j(m′) − j ∘ u ∘ j(m′) = n

since j ∘ u = id_{N}. Surjectivity of φ follows. Thus, φ is an isomorphism. Taking the obvious maps ϕ_{N} = id_{N} and ϕ_{L} = id_{L}, it is easy to see that the diagram (2) is commutative and hence that the sequence (1) splits. Similarly, suppose that we have a homomorphism s : L ⟶ M such that v ∘ s = id_{L}. Define

ψ : N ⊕ L ⟶ M by

ψ(n, l) = u(n) + s(l).

Since u, s are homomorphisms, it is clear that ψ is a homomorphism. Now suppose that ψ(n, l) = 0. Then certainly v(u(n) + s(l)) = l = 0 since ker v = im u and v ∘ s = id_{L} so that u(n) = 0. Since u is injective, it follows that (n, l) = (0, 0). Now let m ∈ M be arbitrary. Observe that

m = (m − s ∘ v(m)) + s ∘ v(m)

and that

v(m − s ∘ v(m)) = v(m) − v ∘ s ∘ v(m) = v(m) − v(m) = 0

since v ∘ s = id_{L}. It follows that (m − s ∘ v(m)) ∈ ker v = im u so there exists n ∈ N such that u(n) = (m − s ∘ v(m)). Then set l = v(m) whence m = ψ(n, l) and ψ is surjective. Thus, ψ is an isomorphism and we let φ = ψ^{−1}, ϕ_{N} = id_{N} and ϕ_{L} = id_{L}. As before, it is easy to see that the diagram (2) is commutative and hence that the sequence (1) splits. We have thus shown that (b) and (c) imply (a). This completes the proof.

3. (a) Let F be a free R module with basis {x_{i}}_{i∈I}, and N an arbitrary R module. Let n_{i} ∈ N for i ∈ I. Because F is free, every element f ∈ F has a unique representation

f = ∑_{i∈I} a_{i}x_{i}  (3)

for a_{i} ∈ R. Define

ϕ : F ⟶ N

by ϕ(x_{i}) = n_{i} and extend R-linearity to all of F. By definition, ϕ is a R module homomorphism. Moreover, the uniqueness of (3) guarantees that ϕ is well defined. Since {x_{i}} is a basis for F, any homomorphism F → N agreeing with ϕ on {x_{i}} agrees with ϕ on all of F. Hence ϕ is unique.

(b) By part (a), there is a unique homomorphism

ϕ : F → ⨁_{i∈I} R

given by ϕ(x_{i}) = (0, 0, …, 0, 1, 0, …) where the 1 is i^{th} position. (OK, we are supposing that I is countable.) If

ϕ (∑_{i∈I} a_{i}x_{i}) = (0, 0 …),

then we clearly have a_{i} = 0 for all i. This shows that ϕ is injective. Surjectivity is also clear: if a = (a_{i})_{i∈I} ∈ ⊕_{i∈I}R then ϕ(∑_{i∈I} a_{i}x_{i}) = a. Thus ϕ is an isomorphism.

(c) Suppose that M ⊃ N are R modules and M/N ≃ F is free. Then we have an exact sequence

0 ⟶ N ⟶^{u} M ⟶^{v} F ⟶ 0.  (4)

Let {x_{i}} be a basis for F. Since v is surjective, there exist {m_{i}}_{i∈I} such that v(m_{i}) = x_{i}. Since F is free, by part (a) we have a unique homomorphism s : F → M given by s(x_{i}) = m_{i}. It is clear that v ∘ s = id_{L}, so that by problem 2, the sequence (4) splits and M ≃ F ⊕ N, as desired.

4. (a) Let P be a free module, and suppose we have an exact sequence M → N → 0 together with a homomorphism s : P → N. Let {x_{i}} be a generating set for P. Since u is surjective, for each i there exists m_{i} ∈ M such that u(m_{i}) = x_{i}. By part (a) of problem 3, there exists a unique homomorphism

s̄: P ⟶ M

given by s̄(x_{i}) = m_{i}. It follows that the diagram

P

[figure: s̄]

[figure: M; u; N; s; 0]

is commutative, and hence that P is projective.

(b) Suppose that P is projective and we have an exact sequence

0  \overline{}M  \overline{}N  \overline{}P \overline{0.}

Then since P is projective, there exists a homomorphism s̄ such that the following diagram is commutative:

P

[figure: s̄]

[figure: 0; M; N; u; P; id_{P}; 0]

We now apply problem 2(c), from which it follows that any such exact sequence splits. Thus (a) implies (b). Next, assume (b) and recall that every module is the quotient of some free module. We therefore have an exact sequence

0  \overline{}Q  \overline{}F  \overline{}P \overline{0,}

where F is free. Since we are assuming that any such sequence splits, we have an isomorphism

F ⟶ P ⊕ Q,

that is, P is a direct summand of a free module. Thus (b) implies (c). Finally, suppose (c) and let M ⟶^{u} N ⟶ 0 be exact. Suppose that we have a homomorphism s : P → N. Let π_{P} : P ⊕ Q → P be the natural projection map. Since there exists a module Q such that P ⊕ Q is free, by part (a) of this problem, we have a homomorphism s̄′ such that the diagram

P ⊕ Q
[figure: s̄′]
P\begin{matrix}π_{P} \\  \end{matrix}

[figure: M; u; N; s; 0]

commutes. We therefore have a map s̄: P ⟶ M given by

s̄ = s̄′ ∘ ι_{P}

where ι_{P} : P ⟶ P ⊕ Q is the natural inclusion map, such that

s = u ∘ s̄.

It follows that P is projective and hence (c) implies (a).

5. Until next time …