SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU CİHAN BAHRAN I will collect my olution to ome of the exercie in thi book in thi document. Section 2.1 1. Let A = k[[t ]] be the ring of formal power erie with coefficient in a field k. Determine Spec A. Note that every nonzero f A can be written a f = T n g where n 0 and g i a power erie with nonzero contant term, in other word a unit in A. Thu if p i a nonzero prime ideal in A, then picking 0 f p with f = T n g a above, primene force T n p. Here n mut be greater than zero, o uing the primene again we conclude T p. Thu (T ) p, but (T ) i a maximal ideal o p = (T ). Therefore Spec A = {0, (T )} whoe cloed ubet are, Spec A and {(T )}. 2. Let ϕ : A B be a homomorphim of finitely generated algebra over a field. Show that the image of a cloed point under Spec ϕ i a cloed point. Write k for the underlying field. Let pare the tatement. A cloed point in Spec B mean a maximal ideal n of B. And Spec(ϕ)(n) = ϕ 1 (n). So we want to how that p := ϕ 1 (n) i a maximal ideal in A. Firt of all, p i definitely a prime ideal of A and ϕ decend to an injective k-algebra homomorphim ψ : A/p B/n. But the map k B/n define a finite field extenion of k by Corollary 1.12. So the integral domain A/p i trapped between a finite field extenion. Such domain are necearily field, thu p i maximal in A. 3. Let k = R be the field of real number. Let A = k[x, Y ]/(X 2 + Y 2 + 1). We wih to decribe Spec A. Let x, y be the repective image of X, Y in A. (a) Let m be a maximal ideal of A. Show that there exit a, b, c, d k uch that x 2 + ax + b, y 2 + cy + d m. Uing the relation x 2 + y 2 + 1 = 0, how that m contain an element f = αx + βy + γ with (α, β) (0, 0). Deduce from thi that m = fa. Let firt how the lat claim. So aume that f = αx+βy +γ uch that (α, β) (0, 0) lie in m. WLOG we may aume β 0 and then by caling f we may aume β = 1. 1
Then SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU 2 A/fA = R[X, Y ]/(X 2 + Y 2 + 1, Y + αx + γ) = R[X]/(X 2 + ( αx γ) 2 + 1) = R[X]/((α 2 + 1)X 2 2αγX + γ 2 + 1) Note that the dicriminant of the quadratic polynomial we are factoring out from R[X] at the lat tep i 4α 2 γ 2 4(α 2 + 1)(γ 2 + 1) = 4(γ 2 + α 2 + 1) < 0 therefore the quadratic i irreducible hence it generate a maximal ideal in R[X] and the correponding quotient i a field. Thu A/fA i a field and hence fa i a maximal ideal of A. But we have fa m, therefore fa = m. To how uch an f exit, firt note that by Corollary 1.12 A/m i a finite extenion of R, hence i iomorphic to R or C, hence [A/m : R] 2. If x R, then x + γ m for ome γ R and if y R then y + γ m for ome γ R. So we can take f to be thi linear polynomial. Aume neither x nor y lie in R. Then their minimal polynomial over R have degree 2, o there exit irreducible polynomial g(x) = X 2 + ax + b R[X] and h(y ) = Y 2 + cy + d R[Y ] uch that g(x) m and h(y) m. Now ince x 2 + y 2 = 1, h(x) + g(y) = ax + cy + (b + d 1) m. If (a, c) 0 we can take f = ax + cy + (b + d 1). So aume (a, c) = (0, 0). In that cae m contain the calar b + d 1 o necearily b + d = 1. Alo x 2 + b, y 2 + d m. Let n be the maximal ideal of k[x, Y ] which i the invere image of m under the projection k[x, Y ] A. So we have X 2 + b, Y 2 + d n. If b 0, then X 2 + b ha a linear factor X + b which lie in n (n i prime!), hence x + b m and we can take f = x + b. Similarly for d 0. So aume both b and d are poitive. Then d(x 2 + b) b(y 2 + d) = dx 2 by 2 = ( dx by )( dx + by ) lie in n, therefore one of it factor, which i of the form αx + βy with (α, β) (0, 0) lie in n. So αx + βy m. 4. Let A be a ring. (a) Let p be a minimal prime ideal of A. Show that pa p i the nil radical of pa p. Deduce from thi that every element of p i a zero divior in A. We know that in general for a multiplicative ubet S of A the prime ideal of S 1 A are in one-to-one correpondence with the prime ideal of A that do not interect S. When S = A p thi mean that the prime ideal of A p are in one-to-one correpondence with the prime ideal of A contained in p. Since p i minimal, thi mean that pa p i the only prime ideal of A p, hence equal to the nilradical of A p. Nilpotent element are clearly zero divior. (b) Show that if A i reduced, then any zero divior in A i an element of a minimal prime ideal. Show with an example that thi i fale if A i not reduced (ue Lemma 1.6). See alo Corollary 7.1.3(a).
SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU 3 Let a A be a zero divior. So there exit 0 b A uch that ab = 0. Since A i reduced, the multiplicative ubet S = {1, b, b 2, } doe not contain 0 and hence S 1 A i nonzero. Note that a map to 0 under the ring homomorphim A S 1 A becaue b 1 a 1 = 0 and b 1 i a unit in S 1 A. Now let q be a minimal prime ideal of S 1 A (nonzero ring alway have minimal prime by Zorn lemma). Then q = S 1 p for ome prime ideal p of A which doe not interect S and i minimal among prime ideal with thi property. Clearly it follow that p i (unconditionally) a minimal prime ideal of A. And ince 0 q, a mut lie in p. For a counterexample let k be any field and let A = k[x, Y ]/((Y ) (X, Y ) 2 ). Since XY (Y ) (X, Y ) 2 and X, Y / (X, Y ) 2, the element X A i a zero divior. We claim that X i not in any minimal prime ideal of A. For, uppoe p Spec A contain X. We can write p = q ((Y ) (X, Y ) 2 for ome q Spec k[x, Y ] uch that ) (Y ) (X, Y ) 2 q. Note that Y 2 q, o Y q a q i prime. And ince p contain X, q contain X. Thu q contain the ideal (X, Y ), which i already maximal! So q = (X, Y ) and p = (X, Y ). But there i a chain of trict incluion (Y ) (X, Y ) 2 (Y ) (X, Y ). So (Y ) i a prime ideal in A (becaue (Y ) i prime in k[x, Y ]) which i trictly contained in p = (X, Y ) thu p i not prime. 5. Let k be a field. Let m be a maximal ideal of k[t 1,..., T n ]. (a) Let P 1 (T 1 ) be a generator of m k[t 1 ] and k 1 = k[t 1 ]/(P 1 ). Show that we have an exact equence 0 P 1 (T 1 )k[t 1,..., T n ] k[t 1,..., T n ] k 1 [T 2,..., T n ] 0. In general if A i a ring and I i an ideal of A and if X i a et of variable, the ring homomorphim between the polynomial ring A[X] (A/I)[X] ha kernel I[X] = I A[X]. In the quetion ituation we may take A = k[t 1 ], I = (P 1 ) and X = {T 2,..., T n }. (b) Show that there exit n polynomial P 1 (T 1 ), P 2 (T 1, T 2 ),..., P n (T 1,..., T n ) uch that k[t 1,... T i ] m = (P 1,..., P i ) for all i n. In particular, m i generated by n element. Let proceed by induction. The cae n = 1 follow from the fact that k[t 1 ] i a PID. Now aume the claim hold for n 1. By (a), there exit a polynomial P 1 (T 1 ) k[t 1 ] uch that k[t 1 ] m = (P 1 ) and writing k 1 = k[t 1 ]/(P 1 ), we have a urjective ring homomorphim ϕ : k[t 1,..., T n ] k 1 [T 2,..., T n ] with kernel P 1 (T 1 )k[t 1,..., T n ]. Firt, note that the ideal (P 1 ) of k[t 1 ] i the invere image of m under the incluion k[t 1 ] k[t 1,..., T n ] o i a prime ideal. But k[t 1 ] i a PID, o (P 1 ) i maximal. Thu k 1 i a field. Second, ince P 1 m, the ideal P 1 (T 1 )k[t 1,..., T n ] i contained in m. So m 1 = ϕ(m) i a maximal ideal of k 1 [T 2,..., T n ].
SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU 4 Now by the induction hypothei there exit n 1 polynomial Q 2 (T 2 ), Q 3 (T 2, T 3 ),..., Q n (T 2,... T n ) uch that k 1 [T 2,..., T i ] m 1 = (Q 2,..., Q i ) for all 2 i n. By definition, for every 2 i n, the ring homomorphim ϕ map k[t 1,..., T i ] onto k 1 [T 2,..., T i ] with kernel P 1 (T 1 )k[t 1,..., T i ]. Therefore there exit P i (T 1,..., T i ) uch that ϕ(p i ) = Q i. Conider the retricted ring homomorphim ϕ i : k[t 1,..., T i ] k 1 [T 2,..., T i ]. Chaing the commutative quare formed by retricting ϕ to ϕ i, we ee that k[t 1,..., T i ] m = ϕ i 1 (k 1 [T 2,..., T i ] m 1 ) = ϕ i 1 (Q 2,..., Q i ) = (P 1, P 2,..., P i ). 8. Let ϕ : A B be an integral ring homomorphim. (a) Show that Spec ϕ : Spec B Spec A map a cloed point to a cloed point, and that any preimage of a cloed point i a cloed point. Let n be a maximal ideal of B. Then ϕ induce an injective ring homomorphim ψ : A/ϕ 1 (n) B/n. Since ϕ i integral, o i ψ. But B/n i a field and by the following lemma A/ϕ 1 (n) i a field. Hence ϕ 1 (n) i maximal in A. Lemma 1. If B i a integral domain with a ubring A uch that the incluion A B i integral, then A i a field if and only if B i a field. Proof. Suppoe B i a field and let a A {0}. Since a 1 B i integral over A, there exit b 0,..., b n 1 A uch that o multiplying both ide by a n 1 yield 0 = b 0 + b 1 a 1 + b n 1 (a 1 ) n 1 + (a 1 ) n = b 0 + b 1 a 1 + b n 1 a n+1 + a n a n = (b 0 + b 1 a 1 + + b n 1 a n+1 ) a 1 = (b 0 a n 1 + b 1 a n 2 + + b n 1 ) A. Suppoe A i a field and let b B {0}. Let A[b] be the ubring of B generated by A and b. Then there i a urjection ɛ : A[X] A[b] that end X to b. Since b i integral over A, ker ɛ 0. And ince A[b] i a domain ker ɛ i a prime ideal in A[X]. But A[X] i a PID, o 0 ker ɛ mut be a maximal ideal. Thu A[b] = A[X]/ ker ɛ i a field. Thu b ha an invere b 1 in A[b] and hence in B. The econd part require howing that given a prime ideal q in B uch that ϕ 1 (q) i maximal in A, then q i maximal. Again uing the injection A/ϕ 1 (q) B/q and the lemma (in the other direction thi time) we get that B/q i a field hence q i maximal.
SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU 5 (b) Let p Spec A. Show that the canonical homomorphim A p B A A p i integral. We will how that in general for a multiplicative et S, the ring homomorphim S 1 A S 1 B i integral, which i enough ince B A S 1 A = S 1 B in a natural way. And indeed given b B and, ince b i integral over A we have a 0 + a 1 b + a 2 b 2 + a n 1 b n 1 + b n = 0 for ome a 0,..., a n 1 A. Therefore in S 1 B, we have 0 = a 0 + a 1 b + a 2 b 2 + a n 1 b n 1 + b n n = a 0 n + a 1 n 1 b + a 2 n 2 which how that b/ i integral over S 1 A. Ç å b 2 + + a n 1 Ç å b n 1 + Ç å b n (c) Let T = ϕ(a p). Let u uppoe that ϕ i injective. Show that T i a multiplicative ubet of B, and that B A A p = T 1 B 0. Deduce from thi that Spec ϕ i urjective if ϕ i integral and injective. Being a ring homomorphim, ϕ map the multiplicative ubet A p in A to a multiplicative ubet of B, which i called T here. Since the functor A A p i exact, A A A p = A p inject in B A A p becaue A inject in B. A A p 0, we deduce that B A A p 0. We want to how Spec ϕ i urjective. So let p be a prime ideal in A and write T = p) a above. Since T 1 B = B A A p 0, T 1 B ha a maximal ideal which i necearily of the form T 1 q for a prime ideal q of B uch that q T =. Now ϕ p : A p B A A p i integral by (b), o by (a) f 1 (T 1 q) i a maximal ideal in A p. But there i only one maximal ideal in A p, o ϕ 1 p (T 1 q) = pa p. Taking the invere image of T 1 q in two different way given by the commutative diagram ϕ(a A ϕ B yield Spec(q) = ϕ 1 (q) = p. A p ϕ p B A A p 9. Let A be a finitely generated algebra over a field k. (a) Let u uppoe that A i finite over k. Show that Spec A i a finite et, of cardinality bounded from above by the dimenion dim k A of A a a vector pace (Contruct a trictly decending chain of ideal by taking interection of maximal ideal). Show that every prime ideal of A i maximal. For a given A-module M, denote it iomorphim cla by [M]. There i a bijection {maximal ideal of A} {iomorphim clae of imple A-module} m [A/m] ann A (S) [S]
SOLUTIONS TO ALGEBRAIC GEOMETRY AND ARITHMETIC CURVES BY QING LIU 6 (commutativity of A i crucial here). The aumption on A ay that A i a finitedimenional k-algebra hence an artinian ring. Therefore A ha finitely many iomorphim clae of imple module, hence finitely many maximal ideal. Let S 1,..., S r be a et of repreentative for imple A-module. Since A/ Rad A i emiimple, A/ Rad A = S n 1 1 Sr nr for ome n 1,... n r. In particular r dim k A = dim k Rad A + n j dim k S j r. Thu we howed that the number of maximal ideal in A i le than or equal to dim k A. To finih, we how that every prime i maximal. Let p Spec A. Then A/p i an artinian integral domain. But artinian domain are field (conider the decending chain (a) (a 2 ) for any nonzero a), hence p i maximal. j=1 (b) Show that Spec k[t 1,... T d ] i infinite if d 1. Note that B := k[t 1,..., T d ] i a UFD. So irreducible element are prime, that i, they generate prime ideal. So it i enough to how that there are finitely many irreducible element. The exitence of uch element i guaranteed by the condition d 1, a it tell u that B i a noetherian ring which i not a field and hence the collection of proper principal ideal ha a maximal element which i not zero (recall that p A irreducible (p) i maximal among proper principal ideal). The old and beautiful Euclidean proof for the infinitude of the prime doe the ret. (c) Show that Spec A i finite if and only if A i finite over k. One direction wa etablihed in (a). Suppoe Spec A i finite. By the Noether normalization lemma, there exit an integer d 0 uch that there i an injective finite k-algebra homomorphim ϕ : k[t 1,..., T d ] A. Being finite, ϕ i integral. Therefore by part (c) of the previou quetion, Spec(ϕ) : Spec A Spec k[t 1,..., T d ] i urjective. Thu Spec k[t 1,..., T d ] i finite. But then part (b) force d = 0. Hence A i finite over k via ϕ.