MATH 601: Abstract Algebra II 2nd Homework Solutions Additional exercises (The functor Ext)

Jonathan Rosenberg

assignment due Monday, February 12, 2001

If R is a ring, M is an R-module and Hom denotes Hom in the category of R-modules, then as pointed out in Hungerford and in class, the functor Hom(M, ) is only left exact. In other words, if

0 → A ⟶^{α} B ⟶^{β} C → 0

is an exact sequence of R-modules, then

0 → Hom(M, A) ⟶^{α_{∗}} Hom(M, B) ⟶^{β_{∗}} Hom(M, C)

is exact but the map Hom(M, B) → Hom(M, C) is not necessarily onto, unless M is projective. But one can extend the sequence to the right as follows. Choose a free (or projective) R-module P_{1} mapping onto M and another free (or projective) R-module P_{2} mapping onto the kernel of the map P_{1} → M, etc. That gives an exact sequence

P_{3} ⟶^{ε} P_{2} ⟶^{δ} P_{1} ⟶^{γ} M → 0.

Define Ext_{R}^{1}(M, A) to be the kernel of

Hom(P_{2}, A) ⟶^{ε^{∗}} Hom(P_{3}, A)

divided by the image of

Hom(P_{1}, A) ⟶^{δ^{∗}} Hom(P_{2}, A).

1. Show that Ext_{R}^{1}(M, A) = 0 if M is R-projective. Solution. If M is R-projective, we can simply take P_{1} = M, γ = id_{M}, and P_{2} = P_{3} = 0. Then clearly the above definition (assuming it’s well defined) gives Ext_{R}^{1}(M, A) = 0. □

2. Show that one gets an exact sequence

0 → Hom(M, A) ⟶^{α_{∗}} Hom(M, B) ⟶^{β_{∗}} Hom(M, C) → Ext_{R}^{1}(M, A).

The proof is a long diagram chase.

Solution. We already know that α^{∗} is injective and that ker β_{∗} = im α^{∗}. So the problem is to define the map Hom(M, C) → Ext_{R}^{1}(M, A) and to prove exactness at Hom(M, C). Given f: M → C, compose with γ to get f ∘ γ: P_{1} → C. By projectivity of P_{1}, this lifts to a map g: P_{1} → B, with β ∘ g = f ∘ γ. In other words, we have a commuting diagram with exact rows:

[figure: P_{3}; ε; P_{2}; δ; h; P_{1}; γ; g; M; f; 0; 0; A; α; B; β; C; 0]

and try to fill in with a map h: P_{2} → A as shown. Indeed, since γ ∘ δ = 0, 0 = f ∘ γ ∘ δ = β ∘ g ∘ δ, so g ∘ δ: P_{2} → B lands in ker β = im α, and since α is injective, we can fill in with h. Furthermore, h ∘ ε = 0, since α ∘h∘ε = g ∘δ ∘ε = 0, so h ∈ ker(Hom(P_{2}, A) ⟶^{ε^{∗}} Hom(P_{3}, A)) and defines a class in Ext_{R}^{1}(M, A). We can also see that this class is well defined, because any two choices for g differ by a map P_{1} → ker β = im α, and result in h changing by something in the image of Hom(P_{1}, A) ⟶^{δ^{∗}} Hom(P_{2}, A).

Now we only need to check exactness. Suppose h is trivial in Ext_{R}^{1}(M, A), i.e., lies in the image of δ^{∗}. That means we have k: P_{1} → A with h = k ∘ δ. Then since α ∘ h = g ∘ δ, (g − α ∘ k) ∘ δ = 0. In other words, g − α ∘ k is zero on im δ = ker γ, and thus we have a map ℓ: M → B with g − α ∘ k = ℓ ∘ γ. Then

β ∘ ℓ ∘ γ = β ∘ (g − α ∘ k) = β ∘ g − (β ∘ α) ∘ k = f ∘ γ − 0 = f ∘ γ,

so this shows f = β ∘ ℓ, i.e., the kernel of the map Hom(M, C) → Ext_{R}^{1}(M, A) is the image of Hom(M, B) → Hom(M, C). □

3. Show that Ext_{R}^{1}(M, A) is independent of the choice of P_{1}, P_{2}, and P_{3}, so that the notation Ext_{R}^{1}(M, A) makes sense. Ext_{R}^{1} is the simplest example of what is called a derived functor; there are many other examples in algebra. Solution. Suppose given two choices

P_{3} ⟶^{ε} P_{2} ⟶^{δ} P_{1} ⟶^{γ} M → 0.

and

P′_{3} ⟶^{ε′} P′_{2} ⟶^{δ′} P′_{1} ⟶^{γ′} M → 0.

for a “projective resolution” of M. We will first show that we can construct maps f_{1}, f_{2}, f_{3}, g_{1}, g_{2}, g_{3} making the diagram

[figure: P_{3}; ε; f_{3}; P_{2}; δ; f_{2}; P_{1}; γ; f_{1}; M; 0; P′_{3}; ε′; g_{3}; P′_{2}; δ′; g_{2}; P′_{1}; γ′; g_{1}; M; 0; P_{3}; ε; P_{2}; δ; P_{1}; γ; M; 0]

commute. We construct f_{1} first, using the fact that P_{1} is projective, which means that we can fill in the diagram

\xymatrix{ & P_{1} \ar[d]^{γ} \ar@{.>}[dl]_{f_{1}} \\ P′_{1} \ar[r]^{γ′} & M \ar[r] & 0.}

Then we construct f_{2} and f_{3} the same way, using projectivity of P_{2} and P_{3}, respectively, and also construct the g’s by the same method, with the P ’s and P′’s interchanged. Then f_{1}, f_{2}, f_{3}, g_{1}, g_{2}, and g_{3} give rise to a commuting diagram

\xymatrix{Hom(P_{1}, A) \ar[r]^{δ^{∗}} & Hom(P_{2}, A) \ar[r]^{ε^{∗}} & Hom(P_{3}, A) \\ Hom(P′_{1}, A) \ar[r]^{_{δ′_{∗}}} \ar[u]^{_{f_{1}_{∗}}} & Hom(P′_{2}, A) \ar[r]^{_{ε′_{∗}}} \ar[u]^{_{f_{2}_{∗}}} & Hom(P′_{3}, A) \ar[u]^{_{f_{3}_{∗}}} \\ Hom(P_{1}, A) \ar[r]^{δ^{∗}} \ar[u]^{_{g_{1}_{∗}}} & Hom(P_{2}, A) \ar[r]^{ε^{∗}} \ar[u]^{_{g_{2}_{∗}}} & Hom(P_{3}, A), \ar[u]^{_{g_{3}_{∗}}}}

which induces maps

\xymatrix{ker ε′^{∗}/ im δ′^{∗} \ar[r]^{f^{∗}} & ker ε^{∗}/ im δ^{∗}. \ar[l]^{g^{∗}}}

We want to show these give isomorphisms. Since the P ’s and P′’s play symmetrical roles, we only need to show that f^{∗}∘g^{∗} = id on ker ε^{∗}/ im δ^{∗}. To see this, observe that γ∘(g_{1}∘f_{1}) = γ, so that γ∘(g_{1}∘f_{1}−id_{P_{1}}) = 0, i.e., im(g_{1} ∘ f_{1} − id_{P_{1}}) ⊆ ker γ = im δ. Since P_{1} is projective, that implies there is a map k_{1} : P_{1} → P_{2} with δ ∘k_{1} = g_{1} ∘f_{1} −id_{P_{1}}. Then (g_{1} ∘f_{1} −id_{P_{1}})∘δ = δ ∘(g_{2} ∘f_{2} −id_{P_{2}}) = δ ∘k_{1} ∘δ, so δ ∘(g_{2} ∘f_{2} −id_{P_{2}} −k_{1} ∘δ) = 0 and im(g_{2} ∘f_{2} −id_{P_{2}} −k_{1} ∘δ) ⊆ ker δ = im ε. Since P_{2} is projective, that implies there is a map k_{2} : P_{2} → P_{3} with ε ∘ k_{2} = g_{2} ∘ f_{2} − id_{P_{2}} − k_{1} ∘ δ.

Now suppose ϕ: P_{2} → A lies in ker ε^{∗}. Then

f^{∗} ∘ g^{∗}(ϕ) = ϕ ∘ g_{2} ∘ f_{2} = ϕ ∘ (ε ∘ k_{2} + id_{P_{2}} + k_{1} ∘ δ) = ϕ + ϕ ∘ (ε ∘ k_{2} + k_{1} ∘ δ).

This represents the same class as ϕ, since ϕ ∘ ε = 0 and ϕ ∘ k_{1} ∘ δ lies in im δ^{∗}. So f^{∗} ∘ g^{∗} induces the identity on ker ε^{∗}/ im δ^{∗}, and by symmetry, similarly with g^{∗} ∘ f^{∗}. □

4. Let R = ℤ, M = ℤ/k, and choose P_{1} = P_{2} = ℤ. Show that Ext_{ℤ}^{1}(ℤ/k, A) is just A/kA, and check the exact sequence

0 → Hom(ℤ/k, A) ⟶^{α_{∗}} Hom(ℤ/k, B) ⟶^{β_{∗}} Hom(ℤ/k, C) → Ext_{R}^{1}(ℤ/k, A)

directly. Solution. Take P_{3} = 0 and P_{2} ⟶^{δ} P_{1} to be ℤ ⟶^{k} ℤ. Since Hom(ℤ, A) is naturally isomorphic to A, Hom(P_{1}, A) ⟶^{δ^{∗}} Hom(P_{2}, A) is just A ⟶^{k} A, whose kernel is the k-torsion in A (usually denoted _{k}A), and whose cokernel is A/kA. So Ext_{ℤ}^{1}(ℤ/k, A) = A/kA. Also note that Hom(ℤ/k, A) = _{k}A, since a homomorphism ℤ/k → A is uniquely determined by the image of the coset of 1, which can be any k-torsion class in A. So the exact sequence simplifies in this case to

0 → _{k}A ⟶^{α_{∗}} _{k}B ⟶^{β_{∗}} _{k}C → A/kA.

To check the exact sequence, we only need to see how to define a map _{k}C → A/kA and check exactness at _{k}C. Given a k-torsion element x in C, lift it to an element y ∈ B. Then ky is not necessarily 0, but β(ky) = kβ(y) = kx = 0. So ky ∈ ker β = im α and ky = α(z) for a unique z ∈ A. Note that z depended on a choice, as we had to choose y mapping to x, and we are free to modify y by any element of ker β = im α. In other words, y is only well determined up to an element of A, so ky is only well determined up to an element of kA,and z is well determined in A/kA. So we get a map _{k}C → A/kA. The kernel of this map consists of elements of C that lift to k-torsion elements of B, or in other words, precisely the image of _{k}B ⟶^{β_{∗}} _{k}C. □