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). Since Δf ∈ I,thereexistg1,...,gm ∈ B with
m Δf = giΔfi, i=1 so m gi(P1,P2)Δfi(P1,P2)=Δf(P1,P2) =0 . i=1
Therefore we must have Δfi(P1,P2) =0forsome i,sofi(P1) = fi(P2). (c) S = {x, xy} is R-separating.
3.6. Define a partial ordering “≤”onsetM := {P ∈ Spec(R) | P ⊆ Q} by
P ≤ P ⇐⇒ P ⊆ P