Bourbaki

Exercise 24, § 2

Algebra, II, p. 188

Let M be an A-module, I its injective envelope (II, p. 186, exerc. 18), a a two-sided ideal of A, Q the submodule of I formed by the z∈Iz \in I such that az={0}az = \{0\}; Q is naturally endowed with a (A/a)(A/a)-module structure. Let N be the submodule of M formed by the x∈Mx \in M such that ax={0}ax = \{0\}; if N is considered as a (A/a)(A/a)-module, now prove that Q is isomorphic to an injective envelope of N.

25 a) For an A-module Q to be injective, it is enough that for every projective A-module P and every submodule P’ of P, every linear mapping from P’ into Q extends to a linear mapping from P into Q (use the fact that every A-module is a quotient of a projective A-module). b) For an A-module P to be projective, it is enough that for every injective A-module Q and every quotient module Q” of Q, every homomorphism from P into Q” is of the form φ ∘ u, where φ : Q → Q” is the canonical homomorphism and u is a homomorphism from P into Q (use the fact that every A-module is a submodule of an injective A-module).

26) For every Z\mathbf{Z}-module G and every integer n>0n > 0, denote by nG_n G the kernel of the endomorphism x↦nxx \mapsto nx of G. If 0→G′→G→G′′→00 \to G' \to G \to G'' \to 0 is an exact sequence, define a canonical homomorphism d:nG′′→G′/nG′d : _n G'' \to G'/nG' such that the sequence

0⟶nG′⟶nG⟶nG′′→dG′/nG′⟶G/nG⟶G′′/nG′′⟶0 0 \longrightarrow _n G' \longrightarrow _n G \longrightarrow _n G'' \xrightarrow{d} G'/nG' \longrightarrow G/nG \longrightarrow G''/nG'' \longrightarrow 0

is exact. If nG_n G and G/nGG/nG are finite, put

hn(G)=Card(G/nG)−Card(nG); h_n(G) = \mathrm{Card}(G/nG) - \mathrm{Card}(_n G);

now prove that nG′,nG′′,G′/nG′,G′′/nG′′_n G', _n G'', G'/nG', G''/nG'' are finite and that one has hn(G)=hn(G′)+hn(G′′)h_n(G) = h_n(G') + h_n(G'').

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.