Problemas de Algebra Lineal by J. Ikramov

By J. Ikramov

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Proof. That (1) implies (2) follows immediately from part (d) of Princ. 11, together with Princ. 9. , bn ] d for some bi ∈ B. Since dt (bi ) ∈ K[t] for each i, there exists s ∈ S so that d d s dt (B) ⊂ B. Since s ∈ K, s dt is locally nilpotent. If D denotes the restriction d of s dt to B, it follows that D is also locally nilpotent, and ker D = B ∩ d ) = B ∩ K = A. ⊓ ⊔ ker (s dt Principle 14. (Compare to Princ. II of [97]) Suppose B is the graded ring B = ⊕i∈Z Bi , and let D ∈ LND(B) be given. Suppose that, for integers m ≤ n, D admits a decomposition D = m≤i≤n Di , where each Di ∈ Derk (B) is homogeneous of degree i relative to this grading, and where Dm = 0 and Dn = 0.

Conversely, suppose P (t) ∈ ker ( dt ). If deg P ≥ d 1, then since this kernel is algebraically closed, it would follow that t ∈ ker ( dt ), d a contradiction. Therefore, ker ( dt ) = A. For (c), let D ∈ LNDA (B) be given, D = 0. By Prop. 8(c), for any d d p(t) ∈ A[t], D(p(t)) = p′ (t)Dt. Consequently, D = Dt dt . Since both D and dt are locally nilpotent, Princ. 7 implies that Dt ∈ A. Therefore, LNDA (A[t]) ⊆ d A · dt . The reverse inclusion is implied by Princ. 7. ⊓ ⊔ Principle 9. Let S ⊂ B − {0} be a multiplicatively closed set, and let D ∈ Derk (B) be given.

Let A = ker D. If r is a local slice of D and f = Dr, then B ⊂ Bf = Af [r]. We therefore have frac(B) ⊂ frac(A)(r) ⊂ frac(B), which implies frac(B) = frac(A)(r). Now suppose δg = 0 for g ∈ frac(B), and write g = P (r) for the rational function P having coefficients in frac(A). Then 0 = P ′ (r)δr, and since δr = 0, P ′ (r) = 0. It follows that g = P (r) ∈ frac(A), which shows ker δ ⊂ frac(ker D). The reverse containment is obvious. 24. k S = 1 and LND(S) = {0}, then S = K [1] , where K is a field algebraic over k.

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