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Additional resources for Algebra 2 [Lecture notes]
This was used Proof. A non-zero element of the domain R/(f ) can be written as g = g (mod f ), where g ∈ R, g ∈ (f ). We must show that g is invertible. The ideal (f, g) = (h) is principal, with h | g and h | f . Irreducibility of f implies that h = uf or h = u, for some u ∈ R∗ . If h = uf , then f | h | g, which contradicts our assumptions; thus h ∈ R∗ and (f, g) = (1). In particular, there exist a, b ∈ R such that af + bg = 1, which implies that b = b (mod f ) is a multiplicative inverse of g in R/(f ).
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 . (iii) The quotient module X/Y is isomorphic to Rn−r ⊕ R/(d1 ) ⊕ · · · ⊕ R/(dr ). [Of course, if d1 , . . ] (iv) The ideals (d1 ), . . , (dr ) depend only on the pair Y ⊂ X. Proof. 2. 4(i) implies that there is another basis e1 , .
Xn ) = f (xσ(1) , . . , xσ(n) ) (σ ∈ Sn , f ∈ R[x1 , . . , xn ]). If R = K is a field, the same formula defines an action of Sn on the field of rational functions K(x1 , . . , xn ) = Frac(K[x1 , . . , xn ]). 3) Definition. A polynomial f ∈ R[x1 , . . , xn ] (resp. a rational function f ∈ K(x1 , . . , xn ), if R = K is a field) is symmetric if ∀σ ∈ Sn σ · f = f . They form a subring R[x1 , . . , xn ]Sn (resp. a subfield K(x1 , . . , xn )Sn ) of R[x1 , . . , xn ] (resp. of K(x1 , . . , xn )).