Algebra, X, p. 187
Let C be an A-complex, M an A-module. The extension module of M by C (resp. of C by M) is called the graded k-module
ExtA(C,M)=H(HomgrA(C,I(M)))(resp. ExtA(M,C)=H(HomgrA(L(M),C))). If
f:C→C′ is a morphism of complexes and
g:M→M′ a homomorphism of
A-modules, one sets
ExtA(f,g)=H(Homgr(f,I(g))),
ExtA(g,f)=H(Homgr(L(g),f)). a) If
f:C→C′ is an homologism, the
k-homomorphisms
ExtA(f,1) and
ExtA(1f) are bijective. If
C is projective and bounded on the right (resp. injective and bounded on the left), the graded
k-module
ExtA(C,M) (resp.
ExtA(M,C)) is isomorphic to
H(HomgrA(C,M)) (resp.
H(HomgrA(M,C))). b) Let
0→C′→C→C′′→0 be an exact sequence of
A-complexes and
0→M′→M→M′′→0 an exact sequence of
A-modules. Prove that there exist exact sequences
⋯→ExtAn(C′′,M)→ExtAn(C,M)→ExtAn(C′,M)→ExtAn+1(C′′,M)→⋯⋯→ExtAn(C,M′)→ExtAn(C,M)→ExtAn(C,M′′)→ExtAn+1(C,M′)→⋯⋯→ExtAn(M,C′)→ExtAn(M,C)→ExtAn(M,C′′)→ExtAn+1(M,C′)→⋯⋯→ExtAn(M′′,C)→ExtAn(M,C)→ExtAn(M′,C)→ExtAn+1(M′′,C)→⋯ c) Prove that there exists a spectral sequence
"E (X, pp. 175-177, exercises 14 to 17) converging to
ExtA(C,M) such that
"E2pq=Hq(ExtAp(C,M)). If
C is bounded on the right, prove that there exists a spectral sequence
′E, converging to
ExtA(C,M), such that
′E2pq=ExtAp(Hq(C),M) (consider the bicomplexe
HomgrA(C,I(M))).