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**Example text**

Iv) The ideals (d1 ), . . , (dr ) depend only on the pair Y ⊂ X. Proof. 2. 4(i) implies that there is another basis e1 , . . , en of X and non-zero elements d1 , . . , dr of R such that d1 | · · · | dr (and r ≤ n) for which Y is generated by d1 e1 , . . , dr er . There is no R-linear relation between these r elements of X, hence they form a basis of Y (which is then free of rank r). This proves (i) and (ii). The statement (iii) is an immediate consequence of (ii). 7 below (the reader can check that our reasoning is not circular).

15 (for R = Z). (3) There are canonical isomorphisms of R-modules D(M/aM ) D(M )[a] and D(M [a]) D(M )/aD(M ). 14) Exercise. Let R be a ring, let f : Rn −→ Rn be a homomorphism of R-modules given by a matrix A ∈ Mn (R) (n ≥ 1). (1) f is surjective ⇐⇒ f is bijective ⇐⇒ det(A) ∈ R∗ . (2) f is injective ⇐⇒ det(A) does not divide 0 in R. 15) Exercise. Let R be a ring, M a finitely generated R-module and f ∈ EndR (M ) = HomR (M, M ) an endomorphism of M . If m1 , . . , mn ∈ M is a set of generators of M , then there exists a matrix n A = (aij ) ∈ Mn (R) (not necessarily unique) such that f (mi ) = j=1 aij mj (1 ≤ i ≤ n).

Proof. Denote by pr : M −→ M/N the canonical projection. 1(i)) and M/N (indeed, for any submodule X ⊂ M/N the submodule pr−1 (X) ⊂ M is finitely generated, hence so is X = pr(pr−1 (X))). Conversely, if N and M/N are noetherian and M1 ⊂ M2 ⊂ · · · ⊂ M is a chain of submodules, then the chains of submodules M1 ∩ N ⊂ M2 ∩ N ⊂ · · · ⊂ N, pr(M1 + N ) ⊂ pr(M2 + N ) ⊂ · · · ⊂ M/N must stabilise: there exists j such that Mk ∩ N = Mj ∩ N and pr(Mk + N ) = pr(Mj + N ) for all k ≥ j. 3 below then implies that Mk = Mj for k ≥ j.