Math 811, Ping Li

Exercise 3.2.5: Suppose that ϕ : F → M is any surjection, where F is finitely generated and M is finitely presented. Use the Snake Lemma to show that ker(ϕ) is finitely generated.

Proof: Let M is finitely presented, ϕ : F → M is surjection, then

R^{m} →^{α} R^{n} → M → 0

and

0 → ker(ϕ) → F →^{ϕ} M → 0

are exact.

Let h = id_{M}. Since R^{n} is projective, there is g : R^{n} → F

\xymatrix{R^{n} \ar[r] \ar@{.>}[d]_{g} & M \ar[d]^{h} \\ F \ar[r]^{ϕ} & M \ar[r] & 0}

and gα maps R^{m} into ker(ϕ). So there is a map f : R^{m} → ker(ϕ), and the diagram

\xymatrix{ & R^{m} \ar[r]^{α} \ar[d]_{f} & R^{n} \ar[r] \ar[d]_{g} & M \ar[r] \ar[d]_{h}^{≅} & 0 \\ 0 \ar[r] & ker(ϕ) \ar[r] & F \ar[r]^{ϕ} & M}

is commutative. Applying the Snake Lemma, we have

ker(h) → coker(f) → coker(g) → coker(h)

is exact. Since ker(h) = 0 and coker(h) = 0, then coker(f) ≅ coker(g), i.e.,

\frac{ker(ϕ)}{im(f)} ≅ \frac{F}{im(g)}

Note that F, im(f) and im(g) are finitely generated, so ker(ϕ)/im(f) is finitely generated, and then ker(ϕ) is finitely generated.