Solutions of Some Exercises
Solutions of Some Exercises 1.3. n ∈ Let Specmax(R) and consider the homomorphism ϕ: R[x] → R/n,f→ f(0) + n. m ∩m n n ∈ The kernel of ϕ is a maximal ideal of R[x], and R = ,so Specrab(R). 2.5. (a)ThatS generates A means that for every element f ∈ A there exist finitely many elements f1,...,fm ∈ S and a polynomial F ∈ K[T1,...,Tm]inm indeterminates such that f = F (f1,...,fm). Let n P1,P2 ∈ K be points with f(P1) = f(P2). Then F (f1(P1),...,fm(P1)) = F (f1(P2),...,fn(P2)), so fi(P1) = fi(P2) for at least one i. This yields part (a). (b) Consider the polynomial ring B := K[x1,...,xn,y1,...,yn]in2n inde- terminates. Polynomials from B define functions Kn × Kn → K.For f ∈ K[x1,...,xn], define Δf := f(x1,...,xn) − f(y1,...,yn) ∈ B. n So for P1,P2 ∈ K we have Δf(P1,P2)=f(P1) − f(P2). Consider the ideal | ∈ ⊆ I := (Δf f A)B B. G. Kemper, A Course in Commutative Algebra, Graduate Texts 217 in Mathematics 256, DOI 10.1007/978-3-642-03545-6, c Springer-Verlag Berlin Heidelberg 2011 218 Solutions of exercises for Chapter 5 By Hilbert’s basis theorem (Corollary 2.13), B is Noetherian, so by Theorem 2.9 there exist f1,...,fm ∈ A such that I =(Δf1,...,Δfm)B . We claim that S := {f1,...,fm} is A-separating. For showing this, take n two points P1 and P2 in K and assume that there exists f ∈ A with f(P1) = f(P2).
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