Algebra 2 [Lecture notes] by Jan Nekovar PDF

By Jan Nekovar

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Extra resources for Algebra 2 [Lecture notes]

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7) Theorem on symmetric functions. We have R[σ1 , . . , σn ] = R[x1 , . . , xn ]Sn and there is no polynomial relation between the elements σ1 , . . , σn . In other words, every symmetric polynomial over R can be written in a unique way as a polynomial over R in σ1 , . . , σn . Proof. 5), as follows. 6(i) it is enough to show that (∀I) sI ∈ R[σ1 , . . , σn ]. ,0 = 1, we can assume, by induction, that I = (i1 = · · · ik > ik+1 ≥ · · · ≥ in ≥ 0 (1 ≤ k ≤ n) and that we already know that sK ∈ R[σ1 , .

4) For any α, β ∈ K the subfield K(α, β) ⊂ L is a finite extension of K, by (2), hence is algebraic over K, by (1). In particular, the elements α ± β, αβ and αβ −1 (if β = 0) also lie in K . 16) Corollary. Let K → L → M be field extensions. The extension M/K is algebraic ⇐⇒ both extensions L/K and M/L are algebraic. Proof. The implication “=⇒” is automatic. Conversely, if both L/K and M/L are algebraic, then each β is a root of a suitable polynomial f = X n + a1 X n−1 + · · · + an ∈ L[X] (n ≥ 1).

Xn ]Sn (resp. a subfield K(x1 , . . , xn )Sn ) of R[x1 , . . , xn ] (resp. of K(x1 , . . , 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 , ...

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Algebra 2 [Lecture notes] by Jan Nekovar

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