Bourbaki

Exercise 7, § 1

Commutative Algebra, I, 40

Let U, V, WU,\ V,\ W be three commutative groups and f:U→V, g:V→Wf : U \to V,\ g : V \to W homomorphisms. (a) Consider the diagram

0⟶U→αU×V→βV⟶0f↓h↓−g↓0⟶V→γW×V→δW⟶0 \begin{array}{ccccccccc} 0 & \longrightarrow & U & \xrightarrow{\alpha} & U \times V & \xrightarrow{\beta} & V & \longrightarrow & 0 \\ & & f \downarrow & & h \downarrow & & -g \downarrow & \\ 0 & \longrightarrow & V & \xrightarrow{\gamma} & W \times V & \xrightarrow{\delta} & W & \longrightarrow & 0 \end{array}

where α(u)=(u,f(u)), β(u,v)=v−f(u), γ(v)=(g(v),v), δ(w,v)=w−g(v), h(u,v)=(g(f(u)),v)\alpha(u) = (u, f(u)),\ \beta(u, v) = v - f(u),\ \gamma(v) = (g(v), v),\ \delta(w, v) = w - g(v),\ h(u, v) = (g(f(u)), v). Show that this diagram is commutative and that its rows are exact. (b) Deduce from (a) and Proposition 2 of no. 4 an exact sequence

0→Ker⁡(f)→Ker⁡(gof)→Ker⁡(g)→Coker⁡(f)→Coker⁡(gof)→Coker⁡(g)→0.0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(g_{of}) \to \operatorname{Ker}(g) \to \operatorname{Coker}(f) \to \operatorname{Coker}(g_{of}) \to \operatorname{Coker}(g) \to 0.

Give a direct definition of this exact sequence.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.