By Jan Nekovar

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**Additional info for Algebra 2 [Lecture notes]**

**Example text**

That α is injective). Indeed, τ Hτ −1 Ker(α) = Sn , Ker(α) ⊂ H = An , Sn , τ ∈Sn which implies, by (ii), that Ker(α) = {e}. It follows that n ≤ |X| = (Sn : H). , which implies that (An : H) ≥ n. 18) Symmetric functions and representation theory. Symmetric polynomials in n variables naturally appear as characters of algebraic representations of the matrix group GLn (C). Such a representation on a finite-dimensional complex vector space V is a group homomorphism ρV : GLn (C) −→ GL(V ) given by rational functions (with complex coefficients) of the matrix entries.

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).

Proof. Let S be the set of all ideals of R which do not contain any product of prime ideals. If S is non-empty, then it contains a maximal element I. By definition of S, I is not a prime ideal, hence there exist x, x ∈ R such that x, x ∈ I and xx ∈ I. As I J = I + (x) and I J = I + (x ), there exist prime ideals Pi , Pj such that J ⊃ P1 · · · Pr and J ⊃ P1 · · · Ps , which implies that I ⊃ JJ ⊃ P1 · · · Pr P1 · · · Ps . This contradiction shows that S is empty, as claimed. 12) Proposition. 6(i) is satisfied in any noetherian integral domain A.