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Home , Krull dimension, Regular sequence

KRULL

E. BALLICO

1. Krull : the main theorems Let R be a with unit 1, R 6= 0. We recall that the dim R of R is the supremum of all integers n ≥ 0 such that there is an increasing chain

P0 ( P1 ( ··· ( Pn (1) of prime ideals of R. There are rings R such that dim R = +∞. A chain in (1) is said to have length n. Note that in a any strictly increasing chain of ideals is finite, but the supremum may be +∞. As an exercise prove that in Z the supremum of the lengths of chains of ideals (NOT of PRIME ideals) is +∞. We will see in the next sections large classes of Noetherian rings (e.g. finitely generated algebras over a field or over Z) which have finite dimension and many other nice properties. For the moment we do not assume noetherianity. Fix P ∈ Spec(R). There are two interesting rings associated to P : the domain R/P and the RP . The dimension of RP is called the height ht(P ) ([8, 9]) or the codimension codim of P ([5]). Exercise 1.1. Prove that dim RP + dim R/P ≤ dim R (2) and discuss the informations you get if you are told that equality holds for all prime ideals of R. In 1928 W. Krull proved the following key theorem, usually called with a German name whose English translation is Theorem.

Theorem 1.2. (PIT) Let R be a Noetherian ring. Let I ( R be an ideal generated by c elements. Then ht(P ) ≤ c for any minimal prime containing I. We present the proof of PIT and its consequences from [5, pp. 232-233] (see [7, Theorem 152] for another proof). Theorem 1.7 is also [7, Theorem 153]. You need noetherianity and to know the to make sense of the words “ minimal prime ”, because in the definition of noetherianity never you met minimality (maximality, yes, minimality, no). The minimal primes containing I are the minimal, i.e. the non-embedded√ ones; more elementary, they are the prime ideals appearing in a minimal description of I = P1∩· · ·∩Ps as a finite intersection of prime ideals. Sometimes one call ht(I) the height of I, sometimes it is called the little height of I. Theorem 1.2 implies that ht(I) ≤ c, but in general (for certain rings) it is stronger, because minimal primes of I may have different heights. Proof of the case c > 1 of PIT, assuming that it is true when c = 1: By induction on c we may as- sume that the theorem√ is true (for any noetherian ring) for ideals generated by at most c − 1 elements. Write I = P ∩ P1 ∩ · · · ∩ Ps, s ≥ 0 with P,P1,...,Ps the minimal primes of I. Note that the height of P is equal to the height of the maximal√ ideal of RP . Thus taking RP instead of R (which is still Noetherian) we may assume that I = P with P maximal. If ht(P ) = 0, then the theorem is true. Thus we may assume that P strictly contains another prime. Let P1 be a prime such that P ) P1 and there is no prime Q with P ) Q ) P1 (to prove he existence of such P1 use the ascending condition property). To prove that ht(P ) ≤ c it is sufficient to prove that ht(P1) ≤ c − 1. By the inductive assumption it is sufficient to prove that P1 is a minimal prime containing an ideal generated by c − 1 elements. Write I = (x1, . . . , xc) with xi ∈ R. Since P is a minimal prime containing I and P1 ( P , there is i ∈ {1, . . . , c} such that xi ∈/ I. With no loss of generality we may assume i = 1, i.e. x1 ∈/ P1. Since there is no prime strictly between P1 and P , P is a minimal prime of the ideal (P1, x1) := P1 + Rx1. Thus (since we assuming that P is p the of R), (P1, x1) = P . Since the ring is noetherian, P is finitely generated, say

1 2 E. BALLICO

p ki P = (a1, . . . , am). Since (P1, x1) = P , there is an integer ki > 0 such that ai ∈ (P1, x1). Prove n the existence of an integer n > 0 such that P ⊆ (P1, x1) (you may take n = k1 + ··· + km − 1 if m ≥ 2 and n = k1 if m = 1). Write n xi = aix1 + yi, i = 2, . . . , c with ai ∈ R and yi ∈ P1. We claim that P1 is minimal among the primes containing y2, . . . , yc (which would be sufficient to prove he theorem by the inductive assumption). Assume the existence of a prime P2 such that P1 ) P2 ⊇ (y2, . . . , yc). By assumption P is nilpotent modulo (x1, y2, . . . yc). Thus a power of P/(y2, . . . , yc) is contained in (x1, y2, . . . , yc)/(y2, . . . , yc). Thus P/(y2, . . . , yc) is a minimal prime containing the principal ideal (x1, y2, . . . , yc)/(y2, . . . , yc) of the noetherian ring R/(y2, . . . , yc). By the case c = 1 of PIT, P/(y2, . . . , yc) has at most eight 1. Thus either P2/((y2, . . . , yc) = P or P2/((y2, . . . , yc) = P1. Hence either P2 = P or P2 = P1, a contradiction. 

Remark 1.3. Let R be a noetherian ring and P a . Let j : R → RP be the localization map. The ideal µ := RP P is the maximal ideal of RP . For any positive integer k we have k k k µ = RP P . We know (notes on the primary decomposition) that µ is a µ-primary ideal. Hence P (k) := j−1(µk) is a P -primary ideal (same notes). The P -primary ideal P (k) is called the k-th symbolic power of P . We have P k ⊆ P (k) and equality holds if and only if P k is a primary ideal. By construction in a primary decomposition of P k the ideal P (k) is the P -primary component of P k. Proof of the case c = 1 of the PIT (Theorem 1.2) assuming a part of Theorem 2.3: It is sufficient to prove that if Q ( P is a prime ideal, then Q is a minimal prime of R. Replacing R with RP we may assume that R is local with P as its maximal ideal. Since P is minimal over x, the noetherian ring R/(x) has dimension 0. Thus it is artinian (Theorem 2.3). Thus the descending chain (x) + Q(n), n > 0, stabilizes, say with Q(n) ⊆ (x) + Q(n+1). Thus for each f ∈ Q(n) there are a ∈ R and g ∈ Q(n+1) such that f = ax + g. Thus ax ∈ Q(n). Since P is a minimal prime over (x), we have x∈ / Q and hence a ∈ Q(n). This implies that Q(n) = (x)Q(n) + Q(n+1). Since (n) (n+1) x ∈ P , Nakayama’s lemma implies Q = Q . Applying again Nakayama’s lemma in RQ we n get (RQQ) = 0. Thus dim RQ = 0, i.e. Q has height = codimension 0.  The following lemma is often called the prime avoidance lemma ([5, Lemma 3,3], [8, Lemma 1.B at p. 2]).

Lemma 1.4. Fix an integer r ≥ 2 and take r + 1 ideals I1,...,Ir,J. Suppose that at least r − 2 of the ideals I1,...,Ir are prime and that J ⊆ I1 ∪ · · · ∪ Ir then there is j ∈ {1, . . . , r} such that J ⊆ Ij Proof. Since the lemma is obvious for r = 1, we may assume r ≥ 2 and use induction on r. In particular we may assume that for any S ⊂ {1, . . . , r} such that #S = r − 1 we have J * ∪j∈SIj. Since J ⊆ I1 ∪ · · · ∪ Ir, for each S = {1, . . . , r}\{j} there is xj ∈ Ij ∩ J such that xj ∈/ ∪h6=jIj. If r = 2 use that x1 + x2 ∈ J, but x1 + x2 is neither in I1 nor in I2. If r > 2 and, say I1 is a prime, Q x1 + j>1 xj cannot be in any Ij, but it is in J, a contradiction 

Exercise 1.5. Lemma 1.4 is obvious without assuming that at least r − 2 of the ideals Ij are primes if R contains an infinite field K, because (since K is infinite) no K-vector space V 6= 0 is a finite union of proper subspaces. Check that PIT (see the proof of Theorem 1.7 below for some help) that in a noetherian ring the prime ideals satisfies the descending chain condition . Corollary 1.6. ([5, Corollary 10.3]) The prime ideals in a noetherian ring satisfy the descending chain condition, with the length of a chain descending from a prime P upper bounded by the number of generators of P . Theorem 1.7. (Converse of PIT). Take R noetherian. Then any prime P of R has finite codi- mension (= height) and the codimension, c, is the minimal number of generators of an ideal I such that P is minimal over I. 3

Proof. Since R is noetherian, P is finite generated, say P = (a1, . . . , am). By PIT we have ht(P ) ≤ m. By PIT we also have that for any ideal J such that P is minimal over J, ht(P ) is at most the number of generators of J. Take an ideal I such that P is minimal over I and that I has the minimal number of generators for any such ideals, say I = (y1, . . . , yc) with c ≥ 0. If c = 0, i.e. if √I = (0), then P is a minimal prime of R, i.e. one prime appearing in the writing of the nilradical 0 as a minimal finite intersection of prime ideals and hence ht(P ) = 0. Now assume c > 0 and that the theorem is proved for all prime ideals of all noetherian ring which are minimal over an ideal with at most c − 1 generators. Assume to have found for some integer t with 0 ≤ t < c some element u1, . . . ut ∈ P such that (u1, . . . , ut) has a minimal prime of height t. By Lemma 1.4 there is ut+1 ∈ P not contained in any of the finitely many primes minimal over (u1, . . . , ut). We get that (u1, . . . , ut+1) has a minimal prime of height t + 1.  Even the fact that in a noetherian ring any prime ideal has finite height is NOT obvious or easy.

Corollary 1.8. Let (R, µ) be a noetherian local√ ring. Then dim R < +∞ and dim R is the minimal number of generators of an ideal I such that I = µ. √ Proof. Recall (notes on the primary decomposition) that any ideal I such√ that I is a maximal ideal is primary. Thus µ is a minimal prime of an ideal J if and only if J = µ. Apply Theorem 1.7.  1.1. Krull dimension and polynomial rings. Let A 6= {0} be a ring. We consider the relations between the Krull dimension of A and the Krull dimension of the A[t] in one variable. Lemma 1.9. We have dim A[t] ≥ dim A + 1. Proof. Fix any strictly increasing chain with lenth n ≥ 0:

P0 ( P1 ( ··· ( Pn of prime ideals of A. Consider the chain

P0[t] ⊆ P1[t] ⊆ · · · ⊆ Pn[t] ⊆ (Pn, t). of ideals of A[t]. We claim that each of these n + 2 ideal is prime and that each of these inclusions ∼ is strict (which would be sufficient to prove the lemma). We have A[t]/Pi[t] = (A/Pi)[t], which is a domain, because A/Pi is a domain (since Pi is a prime). Thus Pi[t] is a prime ideal. We also see that ∼ Pi[t] ( Pi+1[t], because the A/Pi+1 is a proper quotient of A/Pi. We have A[t]/(Pn, t) = A/Pn, which is a domain (hence (Pn, t) is a prime ideal) and the strict quotient of A[t]/Pn[t] obtaining taking the quotient by t.  The lemma implies that if dim A = +∞, then dim A[t] = +∞. In [1, Ch. 11, Ex. 6] it is proved that dim A[t] ≤ 2 dim A + 1. Iterating this result we get that if dim A < +∞, then each polynomial ring A[t1, . . . , tr] in finitely many variables has finite dimension. In [2, VIII.1, §2, Proposition 3] there is a result that gives (when r ≥ 2) a better upper bound for dim A[t1, . . . , tr]; it is proved that dim A[t1, . . . , tr] ≤ (r + 1) dim A − 1. In [1, Ch 11, Ex. 7] you find the following theorem. Theorem 1.10. Let A 6= 0 be a noetherian ring. Then dim A[t] = dim A + 1. For very different proofs (they use the notion of flatness) see [8, Theorem 22, section, 14.A, p. 83 ] or [9, Th. 15.4] or [5, Th. 10.10].

2. Artinian rings and rings of dimension 0 By the definition of Krull dimension a commutative ring A has dim A = 0 if and only if every prime ideal of A is maximal. Thus a domain has dimension 0 if and only if it is a field. Let M be an A-. A strictly increasing chain

N0 ( N1 ( ··· ( Nn of submodules of M is said to have length n, i.e. n counts the number of strict inclusions. A module M 6= 0 is said to have length n if each chain of its submodules has length at most n and there is a chain of lengthy n. 4 E. BALLICO

Definition 2.1. A module M is said to have finite length if it has length n for some n ∈ N.A module M is said to be Artinian if it satisfied the descending chain conditions for its submodules, i.e. if for each descending chain of submodules

M0 := M ⊇ M1 ⊇ M2 ⊇ · · · ⊇ Mn ⊇ Mn+1 ⊇ · · · then there in no such that Mt = Mt+1 for all t ≥ n0. A commutative ring is said to be Artinian if it is Artinian as an A-module, i.e. if its ideals satisfy the descending chain condition. Exercise 2.2. Check that A is artinian if and only if any non-empty family of ideals of A has a minimal elements (use Zorn’s lemma). Theorem 2.3. Let A 6= 0 be a commutative ring. The following conditions are equivalent: (1) A is Artinian; (2) A is noetherian and dim A = 0; (3) the A-module A has finite length. Remark 2.4. A module with finite length is artinian and noetherian, because there is an upper bound on the lengths of its chains (increasing or decreasing) of submodules. Let R 6= 0 be a ring and J the intersection of all maximal ideals of A. J is a proper ideal and it is usually called the Jacobson radical of R. We recall the following form of Nakayama’s lemma ([1, Proposition 2.6]). Lemma 2.5. Let R 6= 0 be any ring, M a finitely generated R-module, J the Jacobson radical of R and I ⊆ J be any ideal. If M = IM, then M = 0. Lemma 2.6. A ring A is artinian if and only if the A-module A has finite length. Proof. Remark 2.3 gives the “ if ” part. Now assume that A is artinian. Claim 1: A has only a finite number of maximal ideals. Proof of Claim 1: Suppose that A has infinitely many maximal ideals. Hence it has maximal ideals mi, i ≥ 1 with mi 6= mj for all i 6= j. Since each mi is primes and mi * mj for any i 6= j, we saw in the proof of the primary decomposition that the sequence of ideals

m1 ) m1m2 ) m1m2m2 ) ··· is strictly decreasing, contradicting the assumption that A ia an Artin ring. Let p1, . . . , pr be all distinct maximal ideals (we have r > 0, since we may assume A 6= 0). Set I := p1 ··· pr. By the Chinese Remainder Theorem I is the intersection of all maximal ideals of A, i.e. the Jacobson radical of A. Since A is an , the descending chain I ⊇ I2 ⊇ · · · is stationary, i.e. there is an integer s > 0 such that Is = It for all t > s. Set J := ((0) : Is). We have (J : I) = (((0) : Is): I) = ((0) : Is+1) = J. Claim 2: J = A and Is = 0. Proof of Claim 2: Since J = ((0) : I) we have J = A if and only if Is = 0. Suppose J ( A. Since A is artinian, there is a minimal member J 0 of the set of ideals (maybe improper) strictly containing J (Exercise 2.2). Thus J 0 = Ax + J for any x ∈ J 0 \ J. Thus J 0/J is finitely generated (it is even generated by a single element). Since I is the Jacobson ideal of A Nakayama’s lemma in the form of Lemma 2.5 gives Ix + J 6= J 0. By the minimality of J 0 we get Ix + J = J and hence Ix ⊆ J. Thus x ∈ (J : I) = J, a contradiction. Claim 2 was difficult to prove, because we are not assuming that A is noetherian. By Claim 2 we have a chain f rs inclusions s−1 s A ⊇ p1 ⊇ p1p2 ⊇ · · · ⊃ p1 ··· pr−1 ⊇ I ⊇ Ip1 ⊇ Ip1p2 ⊇ · · · ⊇ I p1 . . . pr−1 ⊇ I = (0). and heac subquotient of this chain is isomorphic to A/pi for some i. Since each mi is maximal, each module A/mi is isomorphic to a field and hence it has length 1. PProve that length is additive for short exact sequences.  Proof of Theorem 2.3: By Remark 2.3 if A has finite length, then it is artinian and noetherian. By Lemma 2.4 artinian is equivalent to finite length and hence it implies that A is noetherian. Now we check that ifA is artin, then dim A = 0. Let P be a prime of A. In the notation of Lemma 2.4 s s let p1, . . . , pr be the maximal ideals of A. We proves that there is s > 0 such that p1 ··· pr = 0. 5

s s Thus P ⊇ p1 ··· pr. Since P is prime, we proved in the file on the primary decomposition that P ⊇ pi for some i. Since pi is maximal, then P = pi. Now assume that A is noetherian and that dim A = 0. Let p1, . . . , pr be the minimal primes of p A, i.e. the minimal primes of a primary decomposition of (0). We know that (0) = p1 ∩ · · · ∩ pr. k k Since each pi is finitely generated, there is an integer k such that p1 ··· pr = 0. From this we minic the proof of Lemma 2.4 to conclude that A has finite length.  3. Constructible sets and a theorem of Chevalley Let (X, τ) be a topological space. We recall that (X, τ) is called noetherian if the descending chain condition for its closed sets holds. A closed subset F of X is called irreducible if it is not the union of two proper closed subsets. Check that if F is closed, then it is not the union of finitely many proper closed subsets. We proved in the Triennale that if (X, τ) is noetherian then each closed subsets F 6= ∅ is a finite union of closed irreducible sets, say F = F1 ∪ · · · ∪ Fn and that the sets Fi are uniquely determined, up to the ordering, if we also assume that Fi * Fj for all i 6= j. If this condition is satisfied the sets F1,...,Fn are called the irreducible components of F . If A is a noetherian ring, then Spec(A) with the is a noetherian topological space; we checked in the triennale (see my notes or do it an exercise, that the irreducible closed subsets of Spec(A) (any A 6= 0, no noetherianity assumption) are the closed subsets V (P ), P ∈ Spec(A), i.e the sets {Q ∈ Spec(A) | Q ⊇ P }; (see Lemma 3.1 for a proof with some terminology from ). In this way we got in the Triennale in a few lines the primary decomposition for radical ideals. We recall that a locally closed subset of a topological space is just the intersection of an open set with a closed set. Take a noetherian topological space (X, τ). A set S ⊆ X is said to be constructible if it is a finite union of locally closed subsets. Check that the set C of all constructible subsets of X is closed under finite union, finite intersection and taking the complements and that C is the smallest family of subsets of X containing the closed one and closed under these operations. The definition of constructible set and the theorem of Chevalley are given as exercises in Hartshorne’s book ([6, Cap II, Ex 3.22, part (b)]). Lemma 3.1. Let A 6= 0 be a commutative ring. A closed set S ⊆ Spec(A) is irreducible if and only if S = V (P ) for some P ∈ Spec(A). If S = V (P ) is irreducible, the prime P is unique and it is called the generic point of S. √ Proof. Assume that S is irreducible. Since√S is assumed to be closed, S = V (√ I) for some ideal I. Recall (my Triennale notes) that V ( I) = ∩P ∈Spec(A),P ⊇I P . Check that I is uniquely determined by S (it is ∩P ∈SP ). This gives the uniqueness part of the lemma,√ which also follows from the fact that a prime P is the√ unique minimal√ element√ of the set V (P ). If I is not a prime, there are a, b ∈ A such that a∈ / I and b∈ / I, but ab ∈ I. Thus S ⊆ V (a) ∪ V (ab) (check the last equality, it uses the definition of prime ideals). Thus S = (S ∩ V (a)) ∪ (√S ∩ V (b). Since√ S is irreducible either S = V (a) or S = V (b), contradicting the assumptions a∈ / I and b∈ / I. For the converse note that any closed set of Spec(A) containing {P } contains V (P ) and hence V (P ) is the unique closed set S1 ⊆ V (P ) containing {P }.  Remark 3.2. Let F be an irreducible closed subset of a topological space (X, τ). Each non-empty open subset of F (with the topology of F induced by τ) is dense in F . The following lemma is [8, Proposition (6.C)] with more details on the second part as in [10, §2.3]. Lemma 3.3. Let (X, τ) be a noetherian topological space. Let Z ⊆ X be any subset. Then Z is constructible if and only if the following condition £ is satisfied: Condition £: For each irreducible closed subset X0 of X either X0 ∩ Z = ∅ or X0 ∩ Z contains a non-empty open subset of X0.

Proof. We first check that if Z is constructible, then £ holds. Since F is constructible, X0 ∩ Z is constructible and hence we may write m X0 ∩ Z = ∪i=1(Ui ∩ Fi) (3) with Ui open in X and Fi closed in X. Writing each Fi as the union of its irreducible components we may also find a decomposition (3) in which each Fi is irreducible and Ui ∩ Fi 6= ∅ for all i. By 6 E. BALLICO

m Remark 3.2 we have Ui ∩ Fi = Fi. Thus X0 = X0 = ∪i=1X0 ∩ Fi = F1 ∪ · · · ∪ Fm. Since X0 is irreducible, there is i with X0 = Fi. Thus Ui ∩ X0 = Ui ∩ Fi is a non-empty open subset of X0 contained in X0 ∩ Z. Thus £ holds for X0. Now we prove the other implication. We assume that Z satisfies £. Let S the set of all closure Z of sets satisfying £, but not constructible. Assume S= 6 ∅. Since (X, τ) is noetherian, there is some Z satisfying £, but not constructible, such that Z is a minimal element of S. Since ∅ is constructible, we have Z 6= ∅. Thus Z 6= ∅ and every Z0 ( Z with £ and with Z0 ( Z is constructible. Let Z = F1 ∪ · · · ∪ Fr be a decomposition of Z into irreducible components. Claim 1: F1 ∩ Z is dense in F1. Proof of Claim 1: Since Z = (Z ∩ F1) ∪ · · · ∪ (Z ∩ Fr), we have F1 = F1 ∩ Z ∩ (Z ∩ F1 ∪ · · · ∪ Z ∩ Fr). Since F1 is irreducible and F1 * Fi for any i 6= 1, we have F1 = Z ∩ F1, i.e., F1 ∩ Z is dense in F1. 0 0 00 0 By £ and Claim 1 there is a closed set F ⊆ F1 such that F1 \F ⊆ Z. Set F := F ∪F2 ∪· · ·∪Fr. 0 00 00 We have Z = (F1 \ F ) ∪ (Z ∩ F ). The set F1 \ F is locally closed (and hence constructible) in X. Claim 2: The set Z ∩ F 00 satisfies £. 00 Proof of Claim 2: Fix any closed and irreducible set X0 ⊆ X such that Z ∩ F ∩ X0 = X0. 00 0 The closed se F must contain X0. Thus Z ∩ F ∩ X0 = Z ∩ X0. We have Z ∩ F 00 ⊆ F 00 ( Z. Thus we may apply the inductive assumption to Z ∩ F 00 and get 00 00 00 that Z ∩ F is constructible. Thus Z = (F1 \ F ) ∪ (Z ∩ F ) is constructible.  The following lemma is useful and shows why Chevalley’s theorem below is very good Lemma 3.4. Let B be a noetherian ring. Set Y := Spec(B) with its Zariski topology. Let Y 0 ⊆ Y be a constructible subset of Y . Then there is a finitely generated B-algebra B0such that the morphism f : Spec(B0) → Y associated to the B-algebra structure j : B → B0 is exactly Y 0. Proof. First assume Y 0 = U ∩ F , where F = V (I) is closed and U is a very particular open subset of Y usually called an elementary open subset of Y , i.e. assume the existence of b ∈ B such that n U = {P ∈ Y | b∈ / P }. Put S := {b }n∈N. S is a multiplicative subset of B and it may be seen as a multiplicative subset of B/I. Check that U ∩ F is exactly the image of the natural map Spec(B0) → Y . Now we consider the general case. Claim 1: Since B is a noetherian, every open subset U of Y is a FINITE union of elementary open subsets. Proof of Claim 1: Write U = Y \ V (I) for some ideal I. Since B is noetherian, I = m (b1, . . . , bm) for some bi ∈ A. Note that U + ∪i=1Ubi , where Ubi = {P ∈ Y | bi ∈/ P }. 0 Since Y is constructible, Claim 1 implies that Y is a finite union of sets Ui ∩ Fi with Fi closed and Ui a principal open subset of Y . By the first part there is a finitely generated B-algebra Bi such that the natural map Spec(Bi) → Y has Ui ∩ Fi as its image. Check that the B-algebra 0 m B : ⊕i=1 ⊕ Bi satisfies the thesis of the lemma.  Lemma 3.5. Let j : A → B be a homomorphism of rings A 6= 0 and B 6= 0. Let f : Spec(B) → Spec(A) be the continuos map induces by j. The set f(Spec(B)) is dense in Spec(A) if and only if ker(j) is contained in the nilradical p(0) of A. In particular if p(0) = 0, then f(Spec(B)) is dense in Spec(A) if and only if j is injective.

−1 Proof. Set I := ∩P ∈Spec(B)j (P ). By the definition of the map f the closed set V (I) ⊆ Spec(A) is the closure of f(Spec(B)). Since the radical of an ideal J is the intersection of all prime ideals −1 p containing J, we have I = j ( (0B)). Suppose that f(Spec(B)) is dense in Spec(BA). Then p p V (I) = Spec(B), i.e. (since (0A)) is the intersection of all prime ideals of A, we get I = (0A) p p −1 p and hence ker(j) ⊆ (0A). Conversely, suppose ker(j) ⊆ (0A). We get I = j ( (0B)) = p (0A), i.e. f(Spec(B)) = V (I) = Spec(A).  Theorem 3.6. (Claude Chevalley, 1955). Let A be a noetherian ring and B a finitely generated algebra. Set X := Spec(A) and Y := Spec(B). Let ϕ : A → B be the morphism implicit in the definition of A-algebra and f : Y → B the continuos map induced by ϕ, i.e. the map P 7→ ϕ−1(P ). Let Y 0 ⊆ Y be a constructible set. Then f(Y 0) is constructible. 7

Proof. By Lemma 3.4 there is a finitely generated B algebra B0 such that Y 0 is the image of the map Spec(B0) → Y . Taking B0 instead of B we reduce to prove the theorem in the case Y 0 = Y . By Lemma 3.1 to conclude the proof it is sufficient to prove that f(Y ) satisfies £. Let X0 be an 0 irreducible closed subset of X. Write X0 = V (P ) with P ∈ X (Lemma 3.1). Set A := A/P and B0 := B/P B. B is a finitely generated A0-algebra and Spec(A0) is naturally homeomorphic to V (P ). By Lemma 3.5 the map ϕ0 : A0 → B0 induced by ϕ is injective and that A0 is a domain. Recall that to prove the theorem it is sufficient to prove that X0 ∩f(Y ) contains a non-empty open 0 0 0 subset of X0. By replacing A, B and ϕ by A , B and ϕ respectively, we see that to conclude the proof it is sufficient to prove the following Assertion (†): (†): Let R be a noetherian domain and E a ring containing R and which is a finitely generated R-algebra. Let u : Spec(E) → Spec(R) be the continuous map induced by the inclusion R ⊂ E. Then there is a ∈ R \{0} such that the elementary open set D(a) := {P ∈ Spec(A) | a∈ / P } is contained in u(Spec(E)). Proof of (†): Write E = R[a1, . . . an] and suppose that a1, . . . , ar, r ∈ {0, . . . , n}, are algebraically independent over R, while each aj, r < j ≤ n, is algebraic over R[a1, . . . , ar]. Set 0 R := R[a1, . . . , ar]. For each j ∈ {r + 1, . . . , n} choose a relation

dj dj −1 gjoxj + gj1(x)xj + ··· = 0, 0 Qn with gjh(x) ∈ R for all h and gj0 6= 0. Thus a := j=r+1 gj0(a1, . . . ar) is a non-zero polynomial in the algebraically independent (over R) elements a1, . . . , ar. We claim that D(a) ⊆ u(Spec(E)). Indeed, fix P ∈ Spec(R) such that a∈ / P and set P 0 := PR0. Since R0 is a polynomial ring over R, 0 0 ∼ 0 0 we have R /P = (R/P )[t1, . . . , tr]. Since (R/P )[t1, . . . , tr) is a domain, P is a prime ideal of R . 0 0 Since a∈ / P , EP 0 is integral over RP 0 . By the Lying Over theorem discussed in section 4 there is 0 0 a primes Q of EP 0 lying over AP 0 . We have Q ∩ R = Q ∩ R ∩ R = P [a1, . . . , ar] ∩ R = P . Thus P = Q ∩ R = (P ∩ E) ∩ A ∈ u(Spec(E)), concluding the proof of (†).  Definition 3.7. Let A be a ring. Fix P,P 0 ∈ Spec(A). We say that P 0 is a specialization of P and that P is a generalization of P 0 if P ⊆ P 0. A subset Z ⊆ Spec(B) is said to be stable under specialization (resp. stable under generalization) if for each Q ∈ ZZ contains all Q1 ∈ Spec(A) wich are specializations (resp. generalizations) of Q. Remark 3.8. Take any S ⊆ Spec(A). (1) S is stable under generalization if and only if Spec(A) \ S is stable under specialization. (2) Any open subset of Spec(A) is stable under generalization. Any closed subset of Spec(A) is stable under specialization. Proposition 3.9. Let A be a noetherian ring. Set X := Spec(A) with its Zariski topology. Let S ⊆ X be a constructible subset. S is closed (resp. open) if and only if it is stable under specialization (resp. generalization). Proof. The second part of Remark 3.8 gives the “ only if ” part. By the first part of Remark 3.8 it is sufficient to prove that S is closed if it is constructible and stable under specialization. Let W be an irreducible component of Z. Write W = V (P ) with P ∈ X (Lemma 3.1). W ∩ Z is dense in W . By Lemma 3.1 W ∩ Z contains a non-empty open subset U of W . Since P ∈ U and Z is stable under specialization, we get W ⊆ Z. Thus Z = Z.  Proposition 3.9 gives a geometrical interpretation of the Going Up and Going Down theorems discussed in another section.

4. Integral extensions I assume the basis definitions and properties of integral extensions. More precisely, I assume my course in the triennale on this topic, i.e., [12, Ch. IV] and give many theorems not stated or proved in [12], but which require very little work after that chapter. I follows [1, Ch. 5], which start with all the definitions and so you can get the needed background starting with the first pages of [1, Ch. 5]. Lemma 4.1. Let A ⊆ B rings with B integral over A. (a) If Q is an ideal of B and P := Q ∩ A, then A/P ⊆ B/Q and B/Q is integral over A/P . 8 E. BALLICO

(b) Let S be multiplicatively closed subset of A. Then S−1B is integral over S−1B and the natural map S−1A → S−1B is injective. Proof. The composition of the inclusion A ⊆ B with the surjection B/Q has P as its kernel and hence the inclusion A ⊆ B induces an injective homomorphism A/P → B/Q. Take any y ∈ B/Q and a representative x of y in B. Any integral relation for x over A gives (modulo P ) an integral relation of y over A/P . Now we prove part (b). Composing the inclusion A ⊆ B with the map B → S−1B we get a map A → S−1A in which each s ∈ S is sent to s/1 with 1/s ∈ S−1B. By the universal property of S−1 there is a natural map S−1A → S−1B. This map is injective, because we proved (in the Triennale) that S−1 sends exact sequences of A-modules to exact sequence of S−1A-module. Fix b/s ∈ S−1B and take a relation

n n−1 b + a1b + ··· + an = 0 with ai ∈ A. We get a relation n −1 n (b/s) + (a1/s)(b/s) + ··· + an/s = 0

−1 i −1 in S B with ai/s ∈ S A.  You arrive at [1, Proposition 5.2], which I state, but not prove, because it is [12, Proposition at pp. 66–67] and I proved it at the Triennale. Proposition 4.2. Let A ⊆ B be integral domains such that B is integral over A. Then B is a field if and only if A is a field. Corollary 4.3. Let A ⊆ B rings with B integral over A. Fix Q ∈ Spec(B) and set P := A ∩ Q ∈ Spec(A). Then Q is maximal if and only if P is maximal. Proof. By Lemma 4.1 we have an inclusion A/P ⊆ B/Q with B/Q integral over A/P . By Propo- sition 4.2 B/Q is a field (i.e., Q is maximal) if and only if A/P is a field (i.e. P is maximal).  The next corollary is called incomparability. Corollary 4.4. Let A ⊆ B rings with B integral over A. Let Q, Q0 be prime ideals such that Q ⊆ Q0 and Q ∩ A = Q0 ∩ A. Then Q = Q0.

Proof. Set P := Q ∩ A. By part (b) of Lemma 4.1 BP is integral over AP . Call µ the ideal AP P 0 0 0 (it is the maximal ideal of the local ring AP ) and β = BP Q, and β = BP Q . We have β ⊆ β and 0 0 0 0 β ∩ Ap = β ∩ AP = µ. By Corollary 4.3 β and β are maximal. Since β ⊆ β , we get β = β . By the injectivity of the map Spec(B) → Spec(BP ) (which in [1] is [1, Proposition 3.10 (iv)]) we have 0 Q = Q .  The following theorem is often called the Lying Over theorem or LO for short. Theorem 4.5. Let A ⊆ B be rings with B integral over A. Then the map Spec(B) → Spec(A) is surjective.

Proof. Fix P ∈ Spec(A). By part (b) of Lemma 4.1 BP is integral over its subring AP . Let µ be a maximal ideal of BP (it exists by the Krull-Zorn’s theorem). By Proposition 4.2 µ ∩ Ap is a maximal ideal of AP . Since AP is local with AP P as its maximal ideal, we get AP P = µ ∩ A. We know (Triennale) that there is a bijection between the prime ideals of B not intersecting P (resp. the primes ideals of A not intersecting P ) and the prime ideals of BP (resp. AP ). Let Q be the prime ideals of B such that BqQ = µ. Q ∩ A is the unique prime ideal of A such that AP (Q ∩ A) = AP P . Thus Q ∩ A = P . 

Theorem 4.6. (Going up theorem). Let A ⊆ B be rings with B integral over A. Let p1 ⊆ p2 ⊆ pn be a chain of prime ideals of A and suppose you have a chain q1 ⊆ q2 ⊆ · · · ⊆ qm, m < n, of prime ideals of B with pi = qi ∩A for all i ≤ m. Then this chain can be extended to a chain q1 ⊆ · · · ⊆ qn of prime ideals of B with pi = qi ∩ A for all i. 9

Proof. Check that using induction on the integer n − m it is sufficient to prove the theorem for (n, m) = (2, 1), i.e. it is sufficient to find a prime ideal q2 of B containing q1 and with q2 ∩ A = p2. 0 0 0 0 0 Set A := A/p1 and B := B/q1. By part (a) of Lemma 4.1 we have A ⊆ B and B is integral over 0 0 0 A . By Theorem 4.5 there is a prime ideal G of B such that G ∩ A = p2/p1. We have G = q2/q1 for some prime ideal q2 ⊇ q1. We have q2 ∩ A = p2, because p2 is the only ideal J of A containing p1 and with J/p1 = p2.  In [1, pp. 63-64] there is an easy (but longer) proof of the following theorem, called the Going- down theorem. Theorem 4.7. Let A ⊆ B integral domains with A integrally closed in its quotient field and B integral over A. Let p1 ⊇ p2 ⊇ · · · ⊇ pn be a chain of prime ideals of A and suppose you have a chain q1 ⊇ q2 ⊇ · · · ⊇ qm, m < n, of prime ideals of B with pi = qi ∩ A for all i ≤ m. Then this chain can be extended to a chain q1 ⊇ · · · ⊇ qn of prime ideals of B with pi = qi ∩ A for all i. I do not prove the following result ([1, Proposition 5.17], whose proof uses Galois theory. Proposition 4.8. Let A be an integrally closed domain in its field of fraction K. Let L be a field which is a finite and separable extension of K. Let B be the integral closure of A in L. Then B is a finite A-module. In particular if A is noetherian, then B is noetherian Remark 4.9. The section 3 allows us to see the geometric content of the theorems of this section in terms of topological properties of the continuous map f : Spec(B) → Spec(A) if we also assume (as alway in this Remark) that A is noetherian and that B is a finitely generated A-algebra. By a theorem of Chevalley (Theorem 3.6) for any constructible set S ⊆ Spec(B) the set f(S) is constructible. By Proposition 3.9 we know that f(S) is open if and only if it is stable for generalization and that f(S) is closed if and only if it is stable by specialization. (a) Assume A ⊆ B and that B is integral over A. By the Going up theorem (Theorem 4.6) f is closed, i.e. f(F ) is closed for each closed set F ⊆ Spec(B). (b) Assume A ⊆ B with A and B domain, A integrally closed in its field of fraction and B integral over A. By the Going-down theorem (Theorem 4.7) the map f is open, i.e. f(U) is open for each open U ⊆ Spec(B).

5. Finitely generated algebras over a In this section we consider rings that are finitely generated over a field K. By Hilbert’s basis theorem any such ring is noetherian, but we will see that these rings have properties not true for all noetherian rings. Even more important: most proofs are easier than the ones in other sections. I’ll use ’s Normalization Lemma ([12, Ch. IV]) proved in the Triennale. Theorem 5.1. ([5, Corollary 13.13] Let k be a field and A a finitely generated k-algebra. Assume that A is a domain and call K its field of fraction. Let L be any finite extension of K. Let B be the integral closure of A in L. Then B is a finite A-module and in particular it is noetherian and a finitely generated k-algebra. Taking L = K we get the noetherianity of the integral closure of A, a result in general false for noetherian domains (Akizuki). The proof uses E. Noether’s normalization lemma, a important result which she proved to prove Theorem 5.1. Another result is that finitely generated k-algebras A which are domains have finite Krull di- mension (equal to the trascendence degree of the field of fraction of A with respect to k by [5, Ch 13] and that they are catenary, i.e.

dim A = dim A/P + dim AP for each P ∈ Spec(A) ([5, Corollary 13.4], [8, §(14.H), Corollary 3 at p. 93].

6. missing parts

Let A be a commutative ring, A 6= 0, and M an A-module. An element x1 ∈ A is said to be a non-zero divisor of M or an M-regular element if x1m = 0 implies m = 0. Thus 0 is not M- regular, unless M = 0. An ordered set x1, . . . , xr is said to be a weak M- if x1 is M-regular and (by induction on r)(x2, . . . , xr) is a weak M/(x1)M-regular sequence. The ordered 10 E. BALLICO

sequence x1, . . . , xr is said to be an M-regular sequence if they are a weak M-regular sequence and M/(x1, . . . , xr) 6= 0. Regular sequences (introduced by Hilbert in his 1890 key paper which proved his Syzygy Theorem) are best studied using homological methods ([3], [5], [9], [7, Ch 4]). For polynomial rings there are effective computational tools (like Groebner basis) to study them.

References [1] M. F. Atiyah and I. G. Macdonald, Introduction to , Addison-Wesley, Reading MA, 1969. [2] N. Bourbaki, Algebr`eCommutative, Ch. 8, Masson, Paris, 1983. [3] W. Burns and H. Herzog, Cohen-Macaulay Rings, Cambridge Univ. Press, [4] A. Caruth, A short proof of the principal ideal theorem, Quart. J. Math. Oxford Ser. (2) 31 (1980), no. 124, 401. [5] D. Eisenbud, Commutative Algebra with a view toward Algebraic Geometry, Springer-Verlag, Berlin, 1995. [6] R. Hartshorne, Algebraic Geometry, New York: Springer-Verlag, (1977). [7] I. Kaplansky, Cummutative Rings, Revised Ed., The University of Chicago Press, London, 1974. [8] H. Matsumura, Commutative Algebra, first ed. , W. A. Benjamin, Inc., New Yoork, 1970; there is a second edition 1980. [9] H. Matsumura, Commutative , Cambridge Univ. Press, Cambrige UK, 1995. [10] D. Murfet, Matsumura: Commutative Algebra, webnotes, version of October 5, 2006. [11] G. Northcott, Lecture on rings, modules and multiplicities, Cambridge UK, 1968.==Ch. 4.8 Lemma 9 [12] M. Reid, Undergraduate Commutative Algebra, Cambridge University Press, 1995.

Dept. of Mathematics, University of Trento, 38123 Povo (TN), Italy E-mail address: [email protected]