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By Piergiorgio Odifreddi

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Let R be a ring and M an R-module. We say that M is a projective module if it is a direct summand of a free R-module. We say that M has finite projective resolution of the length K if there exists a family {Mk }K k=1 of R-modules such that the sequence 0 −→ MK −→ MK−1 −→ · · · −→ M1 −→ M −→ 0 is exact. Note that every free module is projective and every projective module has a finite projective resolution – consisting of itself. 27) Exercise. e. generated by one element. Show that every module M over a ring R which is a principal ideal domain is projective.

Thus the homotopy Wt : C → C is d-compact hence deg W = deg W1 = deg W0 , where W0 (z) = a0 z n . Let at (0 ≤ t ≤ 1) be a path joining in C \ 0 the points a0 and 1. Then Ht (z) = at z n is a d-homotopy from W0 to the map z n hence deg W0 = n. 28) Exercise. Prove that the degree of a complex polynomial in each root is positive and equals the multiplicity of this root. Hint. Let z0 be a root of a polynomial W (z) of multiplicity r. Then W (z) = V (z)(z −z0 )r where V (z) is a polynomial satisfying V (z0 ) = v0 = 0.

1) deg (id Rm ) = 1. In fact id∗ (z0 ) = 1 · z0 . 2) Let f: R → R be given by f(t) = at. If a = 0, then f −1 (0) = {0} is compact. The homomorphism f∗ : H1 (R, R \ 0) → H1 (R, R \ 0) is given by f∗ (z) = sgn (a) · z which implies deg (f) = sgn (a). For a = 0, the map f is not d-compact, hence the degree is not defined. 3) Let f: C → C be given by f(z) = z k for k ∈ Z. We will show that deg (f) = k. Since f −1 (0) = 0, we have to show that f∗ : H2(C, C \ 0) → H2 (C, C\0) = Z is the multiplication by k.

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