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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). 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 for P, P ∈M.LetC⊆M be a chain (=totally ordered subset) in M.Set C C∪{ } ⊆ := Q and P := P ∈C P . Clearly P is an ideal of R,andP Q. For showing that P is a prime ideal, take a, b ∈ R with ab ∈ P but b/∈ P . There exists P0 ∈C with b/∈ P0.LetP ∈C.SinceC is a chain, we have P ⊆ P0 or P0 ⊆ P . In the first case, b/∈ P but ab ∈ P ,soa ∈ P .In particular, a ∈ P0.Fromthis,a ∈ P follows in the case that P0 ⊆ P .We have shown that a ∈ P ,soP ∈M. By the definition of the ordering, P is an upper bound for C. Now Zorn’s lemma yields a maximal element of M, which is a minimal prime ideal contained in Q. If R = {0}, there exists a maximal ideal m of R (by Zorn’s lemma applied to {I R | I ideal} with the usual ordering), and by the above, m contains a minimal prime ideal. 5.3. As in the proof of Theorem 5.9 and Proposition 5.10,weonlyhave to show that trdeg(A) ≤ dim(A). By hypothesis, A ⊆ B with B an affine K-algebra. By induction on n, we will show the following, stronger claim: Solutions of exercises for Chapter 6 219 Claim. If trdeg(A) ≥ n, then there exists a chain Q0 ⊆ Q1 ⊆···⊆Qn in Spec(B) such that with Pi := A ∩ Qi ∈ Spec(A) there are strict inclusions Pi−1 Pi for i =1,...,n. The claim is correct for n =0.Toproveitforn>0, let a1,...,an ∈ A be algebraically independent. As in the proof of Theorem 5.9,weseethat there exists a minimal prime ideal Mi of B (not A!) such that the ai are algebraically independent modulo Mi. Replacing B by B/Mi and A by A/A∩ Mi, we may assume that B is an affine domain. Set L := Quot(K[a1]), A := L · A and B := L · B, which are all contained in Quot(B). A has transcendence degree at least n − 1overL. By induction, there is a chain ⊆ ⊆···⊆ Q0 Q1 Qn−1 ∩ ∈ in Spec(B ) such that with Pi := A Qi Spec(A ) there are strict inclu- − ∩ ∈ sions Pi−1 Pi for i =1,...,n 1. Set Qi := B Qi Spec(B)and ∩ ∩ ∈ − Pi := A Qi = A Pi Spec(A). For i =1,...,n 1, we have Pi−1 Pi, since Pi−1 = Pi would imply ⊆ · ∩ ⊆ · · ⊆ · ⊆ Pi (L A) Pi L Pi = L Pi−1 L Pi−1 = Pi−1 Pi . As in the proof of Theorem 5.9,weseethatA/Pn−1 is not algebraic over K.SinceA/Pn−1 is contained in B/Qn−1, it follows from Lemma 1.1(b) that A/Pn−1 is not a field. Choose a maximal ideal Qn ⊂ B which contains Qn−1.ByProposition1.2, Pn := A ∩ Qn is a maximal ideal of A. Clearly Pn−1 ⊆ Pn.SinceA/Pn−1 is not a field, the inclusion is strict. So we have shown the claim, and the result follows. 6.8. (a) We prove that the negations of both statements are equivalent. First, if a ∈ P ,thenUa ∩ P = ∅ since a ∈ Ua.IfP +(a)R = R,then1=b + xa with b ∈ P and x ∈ R,sob =1− xa ∈ Ua ∩ P .Conversely,ifUa ∩ P = ∅, m then a (1 + xa) ∈ P with m ∈ N0 and x ∈ R. This implies a ∈ P or 1+xa ∈ P . In the second case we obtain P +(a)R = R. (b) Assume that dim(R) ≤ n,andletQ0 ··· Qk be a chain of prime −1 ∈ −1 ideals in Ua R,witha R.ByTheorem6.5, setting Pi := ε (Qi) → −1 (with ε: R Ua R the canonical map) yields a chain of length k in Spec(R), and we have Ua ∩ Pi = ∅.Bypart(a), this implies that Pi is not a maximal ideal (otherwise, Pi +(a)R would be R), so we can append a maximal ideal to this chain. Therefore k +1 ≤ dim(R) ≤ n,andwe −1 ≤ − conclude dim Ua R n 1. 220 Solutions of exercises for Chapter 7 −1 ≤ − ∈ ··· Conversely, assume dim Ua R n 1 for all a R.LetP0 Pk be a chain in Spec(R)oflengthk>0. Choose a ∈ Pk \ Pk−1.Then Pk−1 +(a)R = R (both ideals are contained in Pk), so Ua ∩ Pk−1 = ∅ by −1 − part (a). By Theorem 6.5, setting Qi:= Ua Pi (i =0,...,k 1) yields a − −1 − ≤ −1 ≤ chain of length k 1inSpec Ua R . Therefore k 1 dim Ua R n − 1. We conclude dim(R) ≤ n if n>0. If n = 0, the above argument shows that there cannot exist a chain of prime ideals in R of positive length, so dim(R) ≤ 0. (c) We use induction on n, starting with the case n =0.Bypart(b), ≤ −1 { } ∈ dim(R) 0isequivalenttoUa R = 0 for all a R. This condi- tion is equivalent to 0 ∈ Ua, which means that there exist m ∈ N0 and m m m+1 x ∈ R with a (1 − xa) = 0. This is equivalent to a ∈ (a )R,which is (6.5)forn =0. ≤ −1 Now assume n>0. By part (b), dim(R) n is equivalent to dim Ua R ≤ n − 1 for all a ∈ R. By induction, this is equivalent to the follow- ing: For all a0,...,an−1 ∈ R and all u0,...,un−1 ∈ Ua,thereexist m0,...,mn−1 ∈ N0 such that n−1 mi j mi ai aj ai ∈ · j =0,...,n− 1 . ui uj ui −1 i=0 i=0 Ua R Multiplying generators of an ideal by invertible ring elements does not −1 change the ideal. Since the ε(ui) are invertible in Ua R, it follows that the above condition is independent of the ui. In particular, the condition is equivalent to − j n1 mi ∈ · mi − ε(ai) ε(aj) ε(ai) j =0,...,n 1 −1 . Ua R i=0 i=0 By the definition of localization, this is equivalent to the existence of m ∈ N0 and x ∈ R with n−1 j m · mi ∈ · mi − a (1 + xa) ai aj ai j =0,...,n 1 . R i=0 i=0 Writing an and mn instead of a and m, we see that this condition is equivalent to (6.5). 7.4. (a)LetP ∈ Spec(R) be a prime ideal containing I := (x)R. For all nonneg- ative integers i we have (xyi)2 = x · xy2i ∈ I,soxyi ∈ P . Therefore P 2 contains the ideal (x, xy, xy ,...)R, which is maximal. So Solutions of exercises for Chapter 7 221 2 P =(x, xy, xy ,...)R. (b)Theideal 2 3 Q := (xy, xy ,xy ,...)R ∼ is properly contained in P ,andR/Q = K[x], so Q is a prime ideal. The chain {0} Q P shows that ht(P ) ≥ 2. But dim(R) ≤ trdeg(R)=2byTheorem5.5,so ht(P )=2. (c)LetSn = K[x, y1,...,yn−1] be a polynomial ring in n indeterminates (countably many for n = ∞), and set Rn := K + Sn · x.Asin(a), we · see that P = Sn x is the unique prime ideal of Rn containing (x)Rn .For 0 ≤ k<n, we have prime ideals · ∩ ∈ Qk := x (y1,...,yk)Sn = R (y1,...,yk)Sn Spec(R) forming a strictly ascending chain. Since all Qk are properly contained in P ,weobtainht(P ) ≥ n, and equality follows by Theorem 5.5. 7.7. We first show that S is infinite-dimensional. For i ∈ N0, we have strictly ascending chains of prime ideals 2 2 ⊂ ≤ ≤ Qi,j = xi +1,...,xi +j R R (1 j 2i +1) with Qi,j ∩ U = ∅.ByTheorem6.5, this corresponds to a chain of length 2i in Spec(S).
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