Prime Ideals in Group Rings of Polycyclic Groups

1. The statements made in lines −7 to −4 on page 444 are false and should be deleted. Example 26 on page 424 is also incorrect. Not only is there a misprint in the definition of G : x y = x −1 should read x y = x −1 a : but the argument given to show G satisfies (2) is false. In fact G does not satisfy (2). There is therefore some doubt left about the relationship of (2) to (1) and the ◃ 2 ‐condition. In line −2 of page 418 the inclusions O ⩽ δ g ( F ) ⩽ A should be replaced by O ⩽ A. 2. In §8.1 I asserted without proof that certain processes preserved neighbourliness. I now realize that I had no proof. The corrections which follow are designed to show that the validity of Theorems H1 and H2 is not affected, and that Corollary H3 and Theorem H4 are true for з‐groups even though there must remain some doubt about their truth as stated. p. 399, l. –15. Replace the sentence beginning ‘Although this is false’ by ‘This is false for polycyclic groups in general.’ In the sentence following, replace ‘easy’ by ‘obvious’, ‘shall prove’ by ‘state’, and ‘polycyclic’ by ‘in з’. p. 400, l. 1. Replace ‘polycyclic’ by ‘in 3’. p. 431, l. 11. In the sentence beginning ‘In general the processes are reversible enough’ delete what follows ‘neighbourliness’, and replace ‘neighbourliness’ by ‘heights’. p. 431, l. –11. In the sentence beginning ‘The same remarks about’, delete the words ‘and vice versa’. p. 435, l. –9. Replace the sentence beginning ‘Again since F/A is finite’ by ‘We show below that P ∩ kA = ° P 1 and Q ∩ kA = ° Q 1 , with P 1 ⩽ Q 1 neighbouring primes of kA , and proceed as if this had already been shown’. p. 436, l. 4. At the end of §8.3 insert paragraph 3 below. p. 436, l. 5. Delete ‘ Corollary H3 ,’. p. 436, l. 6. In the sentence beginning ‘Theorem H1 asserts’, delete what follows ‘length of G ’. p. 436, l. –12. Delete the sentence beginning ‘If P < P 1 are neighbouring primes’, and also the sentence following it. p. 436, l. –3. Insert ‘with G in з’ after ‘a prime of kG ’. p. 437, l. 9. Delete ‘that the heights of neighbours differ by 1 and’. 3. Since F /Δ. is finite, it is clear (from Lemma 29 for example) that P ∩ k Δ ⩽ Q ∩ k Δ. Since P = ( P ∩ k Δ) kG , it follows from Lemma 8 that any G ‐prime of k Δ between P ∩ k Δ and Q ∩ k Δ is the contraction of a prime of kG between P and Q . Because P and Q are neighbours, we deduce that P ∩ k Δ and Q ∩ k Δ are neighbouring G ‐primes of k Δ. From §8.1 there are neighbouring primes X ⩽ Y of k Δ with P ∩ k Δ = ° X and Q ∩ k Δ = ° Y ; and it is clear, since Δ/Δ is finite, that P 1 = X ∩ kA ⩽ Q 1 = Y ∩ kA . It remains only to prove that P 1 and Q 1 are neighbours. We shall deduce this from the following lemma, and then prove the lemma. LEMMA. Suppose that S is an integrally closed Noetherian domain contained in the centre of a prime polycentral ring R and that R is finitely generated as an S‐module. If V is a prime of S and W a prime of R minimal over VR, then W∩S = V. We take R = kΔ /Δ and let ϕ be the natural projection of k Δ onto R . Let N be the finite radical of Δ. The ideal X ∩ kN is clearly a maximal Δ‐ideal of kN , and therefore ( kN ) ϕ is a ring with no proper non‐trivial Δ‐ideals. Because Δ/ N is Abelian, an argument exactly similar to that of Lemma 7 of [ 23 ] shows that R is a polycentral ring. By Noether's Normalization Lemma, ( kA ) ϕ is integral over some polynomial subring S = k [ X 1 ,…, X r ]. Since Y ϕ has height 1, the Lemma shows that Y ϕ ∩ S also has height 1. Because ( kA ) ϕ is integral over S , it follows that Y ϕ ∩( kA ) ϕ , which is equal toQ 1 φ , also has height 1. We deduce that P 1 and Q 1 are neighbours, as required. Proof of the lemma . Because R is polycentral, the ideal VR has a primary decomposition VR = U 1 ∩…∩ U m , say, with U i primary belonging to a prime W i and with W 1 , W 2 ,…, W m all different. Since W is minimal over VR , we may as well suppose that W = W 1 and that m > 1. Let x be a non‐zero element of W ∩ S , and suppose that ξ = x 8 is in U 1 We write U for U 2 ∩…∩ U m , so that ξ U ⩾ VR . No power of U is contained in W , for otherwise some U i with i ⩾ 2, and so W i would be contained in W . Since R has Max‐ r it follows that there exists an element y of U which has no power in W . We put z for ξ y , so that z is in VR . We now follow the reasoning of § 3 of Chapter V of [ 33 ]. Because z is integral over S , it satisfies an equation f ( z )= z n +υ n −1 z n−1 +…+υ 1 z +υ 0 =0, with the υ i in V . Let K be the quotient field of S . Exactly as is done by Zariski and Samuel, we may suppose that f ( X ) is the minimal polynomial of z over K . Therefore the minimal polynomial of y over K isX n +υ n - 1ξ X n - 1 + ... +υ 1ξ n - 1X +υ 0ξ n= 0. However y is integral over S , so again as in [ 33 ] the coefficients υ n–j /ξ i lie in S . If x is not in V then neither is ξ i for any j . Therefore the υ n–j /ξ i all lie in V . Equation (*) now shows that y n is in V , and hence in W . This is a contradiction which establishes the lemma.