# Algebra 2 [Lecture notes] by Jan Nekovar By Jan Nekovar

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Example text

3. 1) Definition. An R-module M is noetherian if it satisfies the following equivalent conditions. (i) Every submodule N of M is finitely generated. (ii) Ascending chain condition. Every ascending chain of submodules M1 ⊂ M2 ⊂ · · · ⊂ M of M stabilises: there is an index j such that Mk = Mj for all k ≥ j. , there is no P ∈ S such that P P ). ∞ Proof (of the fact that the three conditions are equivalent). (i) =⇒ (ii) The union N = i=1 Mi is a submodule of M , hence generated by a finite set of elements n1 , .

The set of all k × k minors of A (up to elements of R∗ ) does not change if we replace A by A = P AQ. 1) with d1 | · · · | dr the gcd of the k × k minors is equal to d1 · · · dk , for all k ≤ r. 5) Theorem on elementary divisors. Let R be a PID, let X be a free module of rank n over R and Y ⊂ X a submodule. (i) Y ⊂ X is free of rank r ≤ n. (ii) There exist non-zero elements d1 , . . , dr of R such that d1 | · · · | dr and a basis e1 , . . , en of X such that d1 e1 , . . , dr er is a basis of Y .

The statement (1) (resp. (2)) follows from the fact that K[X] is a PID and (1) = Ker(evα ) = (0) n−1 (resp. is automatic). The statement (3) is a consequence of (2) and the fact that the elements 1, X, . . , X (where X = X (mod f ) ∈ K[X]/(f )) form a basis of K[X]/(f ) as a K-vector space, since evα (X) = α and K[X]/(f ) = K[X]. 11 below). 9(ii). The second part of (4) follows from the fact that f | g, hence g is a constant multiple of f (since both are irreducible). 11 below, of degree deg(f ) over K (as in (3)).