This is a folder for courses lectured in Part III Maths at University of Cambridge in year 2022/23. Enjoy!
| Front | Proposition 9.8. Let A be a finitely generated k-algebra and an integral domain. Then dim A = trdegk A . |
| Back | Proof. (1) By Noether’s Normalziation Theorem 3.8, A is integral over a polynomial subalgebra B = k[t1, . . . , tn] (t1, . . . , tn ∈ A algebraically independent over k). (2) By Proposition 9.7, dim A = dim B and trdegk A = trdegk B (3) Thus it suffices to prove that dim B = trdegk B. (4) By Proposition 9.6, dim B ≤ trdegk B. (5) The chain (9.1) (t1, . . . , tn) ) (t1, . . . , tn−1) ) . . . ) (t1) ) (0) of distinct prime ideals of k[t1, . . . , tn] shows that dim B ≥ n (and clearly n = trdegk B). |
| Front | Lemma 16.3. With the above notation we have that 2^rank E(K) = |
| Back | 1/4 |Im αE||Im αE'|. Proof. Since φφˆ = [2]E0 and φφˆ = [2]E we have the exact sequence 0 → E(K)[φ] → E(K)[2] φ→ E'(K)[φˆ] → E'(K)/φE(K) φ→ E(K)/2E(K) → E(K)/φˆE' (K) → 0. As E'(K)/φE(K) ∼= Im αE' and E(K)/φEˆ 0 (K) ∼= Im αE, we deduce that|E(K)/2E(K)|/|E(K)[2]| = 1/4 |Im αE||Im αE'|. By MordellWeil, we can write E(K) = ∆ × Z r with ∆ finite. Thus E(K)/2E(K) = ∆/2∆ × (Z/2Z)^r and E(K)[2] ∼= ∆[2]. As ∆ is nite, |∆/2∆| = |∆[2]|. Hence |E(K)/2E(K)| |E(K)[2]| = 2r . |
| Front | Proposition 12.17. Let a be an ideal of a noetherian ring A. Then (1) Ga(A) is a noetherian ring. |
| Back | (2) If M is a finitely generated A-module and (Mn) a stable a-filtration of M, then G(M) is a finitely generated graded Ga(A)-module. Proof. (i) Since A is noetherian, a is finitely generated by some x1, . . . , xs. Let xi be the image of xi if [$]a/a^2[/$] . Then [$]G_a(A) = (A/a) \osum_{n=1}^{\infty} a^n/a^{n+1}[/$] is generated as an A/a-algebra by x1, . . . , xn. But A/a is a noetherian ring, so Ga(A) is noetherian by Hilbert’s basis theorem. (ii) Take n0 such that [$]M_{n_0+r} = a^r M_{n_0}[/$] for all r ≥ 0. Then G(M) is generated by [$]\osum_{n\leq n_0} M_n/M_{n+1}[/$] as a Ga(A)-module. Each Mn/Mn+1 is a noetherian A-module (M is noetherian since it is finitely generated over the noetherian ring A, and so Mn also is), and is annihilated by a. So it is a finitely generated A/a-module, and thus [$]\osum_{n\leq n_0} M_n/M_{n+1}[/$] is a finitely generated A/a-module. Thus G(M) is finitely generated as a Ga(A)-module. |