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4) are symmetric functions of the roots x1 , . . , xn in the sense that they do not change if we permute the roots. 4) by breaking this symmetry. In this section we recall the classical approach to solving equations of degree n ≤ 4. After that we reformulate it using Lagrange’s idea of resolvents and show that this method does not permit to solve equations of degree n ≥ 5. The history of the subject can be found in [Ti 2]. 2) Quadratic equations. 1) is solved by completing the square: p 2 Equivalently, one can write x = u + v, which yields 0= x+ 2 +q− p 2 2 .

Xn )). 4) Example. The polynomials x1 x2 x3 and x71 + x72 + x73 ∈ R[x1 , x2 , x3 ] are symmetric, but x21 x2 + x22 x3 + x23 x1 is not. 5) Symmetrised monomials. For any set I = (i1 , . . , in ) of integers i1 ≥ i2 ≥ · · · ≥ in ≥ 0 we define f ∈ R[x1 , . . ,in = where AI = {σ · (xi11 · · · xinn ) | σ ∈ Sn }. f ∈AI We often omit the values ik = 0. For example, s1 = σ1 , s1,1 = σ2 , s1,1,1 = σ3 , ... 4) and sk = xk1 + · · · + xkn , s2,1 = x21 x2 + x1 x22 + x21 x3 + x1 x23 + · · · + x2n−1 xn + xn−1 x2n .

1(ii). 3) Lemma. If X ⊂ Y ⊂ M are submodules such that X ∩ N = Y ∩ N and pr(X + N ) = pr(Y + N ), then X = Y . Proof. Let y ∈ Y ; we must show that y ∈ X. The condition pr(X + N ) = pr(Y + N ) implies that there exist x ∈ X and n ∈ N such that y = x+n. It follows that n = y −x ∈ N ∩Y = N ∩X, hence y = x+(y −x) ∈ X. 4) Definition. , if the following equivalent conditions are satisfied. (i) Every ideal I of R is finitely generated. (ii) Ascending chain condition. Every ascending chain of ideals I1 ⊂ I2 ⊂ · · · ⊂ R of R stabilises: there 31 is an index j such that Ik = Ij for all k ≥ j.

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