Download Algebra 2 [Lecture notes] by Jan Nekovar PDF

Download Algebra 2 [Lecture notes] by Jan Nekovar PDF

By Jan Nekovar

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The domain R = {f ∈ Q[X] | f (0) ∈ Z} is a B´ ezout ring (every finitely generated ideal is principal), but R is not noetherian, nor is it a UFD. 4. 1) Matrices. Let R be a ring. If we write elements of Rn as column vectors with entries in R, then a module homomorphism f : Rm −→ Rn can be identified with a matrix A ∈ Mn×m (R): f (x) = Ax for any x ∈ Rm . , there exists a matrix B ∈ Mn (R) such that AB = BA = In . The latter condition implies that det(A) ∈ R∗ . Conversely, the identity A · adj(A) = adj(A) · A = det(A)In (“Cramer’s rule for solving a system of linear equations”) satisfied by the adjoint matrix adj(A) ∈ Mn (R) (where (−1)i+j adj(A)ij is the determinant of the matrix obtained by removing from A the i-th column and the j-th row) shows that A is invertible if det(A) ∈ R∗ , with inverse B = det(A)−1 adj(A).

2, that each Rn (n ≥ 1) is a noetherian R-module. 2. 7) Proposition. Let f : R −→ R be a surjective ring homomorphism. If R is noetherian, so is R . Proof. For any chain of ideals J1 ⊂ J2 ⊂ · · · ⊂ R of R the chain f −1 (J1 ) ⊂ f −1 (J2 ) ⊂ · · · ⊂ R of ideals of R must stabilise: there exists j such that f −1 (Jk ) = f −1 (Jj ) for all k ≥ j. Surjectivity of f then yields Jk = f (f −1 (Jk )) = f (f −1 (Jj )) = Jj for all k ≥ j, which means that R is noetherian. 8) Theorem (“Hilbert’s basis theorem”).

12) Exercise. (i) Compute the discriminants disc(xn + ax + b) for n = 2, 3, 4. [Hint: relate the discriminant of a cubic (resp. quartic) polynomial to the discriminant of its quadratic (resp. 8) (resp. ] (ii) What happens for general n ≥ 2? 13) Exercise. Let K be a field such that 2 ∈ K ∗ . Show that, for any n ≥ 2, the field of rational functions invariant under An is equal to K(x1 , . . , xn )An = {f + g∆ | f, g ∈ K(σ1 , . . , σn )}. 14) Resolvents revisited. , let x1 , . . , xn be independent variables.

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