B2.2: COMMUTATIVE ALGEBRA KONSTANTIN ARDAKOV All rings in this course will be assumed commutative and containing an identity element. For a ring R we denote by R[t ,...,t ] the poly- 1 n nomial ring in indeterminates t with coefficients in R. A subset S of i R is said to be multiplicatively closed if 1 S and whenever x,y S ∈ ∈ then xy S. ∈ 1. Introduction Examples 1.1. We begin by listing a number of examples of commu- tative rings, arising from disparate parts of pure mathematics. (0) Every field F is a ring. (1) Let X be a set and F a field. (a) Fun(X,F) := f : X F is a ring under pointwise addi- { → } tion and multiplication of functions. (b) If X is a topological space and we endow F with the discrete topology, Cont(X,F) := f : X F,f is continuous is a { → } subring of Fun(X,F). (c) IfF = RorCandX isamanifoldoverF, thenSm(X,F) := f : X F : f is smooth is a subring of Fun(X,F). { → } (2) (a) Z Q, (b) Z[i] Q[i], (c) Z[√ 3] Z[ω] Q(√ 3) ⊂ ⊂ − ⊂ ⊂ − where ω := 1+√ 3, and more generally, (d) K for a − 2 − OK ⊂ finite field extension K of Q, where := α K : monic f(X) Z[X] such that f(α) = 0 K O { ∈ ∃ ∈ } is the ring of integers of K. (3) Let F be a field. (a) The rings of polynomials F F[t ] F[t ,t ] F[t ,...,t ]. 1 1 2 1 n ⊂ ⊂ ⊂ ··· ⊂ (b) finitely generated F-algebras; these are the same things as quotients of polynomial rings F[t ,...,t ] by an ideal — see 1 n Definition 2.9 below. Date: Hilary term, 2019. 1 2 KONSTANTIN ARDAKOV Examples (1) come from Topology and Analysis; examples (2) come from Algebraic Number Theory, and examples (3) come from Algebraic Geometry. The main object of study of (Affine) Algebraic geometry are the affinealgebraicvarieties (whichwewillcallalgebraicsets inthiscourse). Let F be a field, n N and let R := F[t ,...,t ] be the polynomial 1 n ∈ ring in n variables t , and let Fn denote the n-dimensional vector space i of row vectors. Definition 1.2. (a) Let S R be a collection of polynomials from R. Define ⊆ (S) := x = (x ) Fn f(x) = 0 f S . i V { ∈ | ∀ ∈ } (b) A set U Fn is an algebraic set if U = (S) for some S R ⊆ V ⊆ (equivalently U = (I) for some ideal I of R). V Thus (S) is just the subset in Fn of common zeroes for all polyno- V mials in S (it may happen of course that this is the empty set). It is easy to see that (S) = (I) where I = S is the ideal generated by V V h i S in R. Here are some examples: Every singleton point a Fn is algebraic, because • { } ⊂ a = x Fn : x = a ,...,x = a = ( t a ,...,t a ). 1 1 n n 1 1 n n { } { ∈ } V { − − } Iff(x,y) = y2 x3+xthen ( f ) = (a,b) F2 : b2 = a3 a • − V { } { ∈ − } is an example of an algebraic curve. We may consider an opposite operation associating an ideal to each subset of Fn. Definition 1.3. Let Z Fn be any subset. Define ⊆ (Z) := f(t ,...,t ) R f(x) = 0 x Z . 1 n I { ∈ | ∀ ∈ } Thus (Z) is the set of polynomials which vanish on all of Z. It is I clear that (Z) is an ideal of R. I Proposition 1.4. Let I I R be ideals and Z Z Fn subsets. ′ ′ ⊆ ⊆ ⊆ ⊆ (1) (I ) (I), ′ V ⊆ V (2) (Z ) (Z). ′ I ⊆ I (3) I ( (I)), ⊆ I V (4) Z ( (Z)), moreover there is equality if Z is an algebraic set. ⊆ V I (cid:3) Proof. Exercise. B2.2: COMMUTATIVE ALGEBRA 3 Proposition1.4showsthat and areorder reversing mapsbetween I V the set of ideals of R and the algebraic subsets of Fn: algebraicsubsets I ideals −→ . (cid:26) Z Fn (cid:27) (cid:26) I F[t ,...,t ] (cid:27) V 1 n ⊂ ⊂ ←− Moreover, is surjective because (S) = ( S ), whereas Proposition V V V h i 1.4(4) shows is injective. Understanding the relationship between I an algebraic set Z and the ideal (Z) is the beginning of algebraic I geometry which we will address in Section 4. In C2.6 Scheme Theory you will see how appropriate generalisations of the constructions in Example 1.1(1) gives meaning to the slogan every commutative ring is a ring of functions on some topological space. The Theory of Schemes, underpinned by the solid foundation of Com- mutative Algebra, allows geometric intuition and techniques to be ap- plied to Algebraic Number Theory, leading to deep results such as Wiles’ proof of Fermat’s Last Theorem. The aim of this course is to study basic structural properties of the class of Noetherian rings which are commonly found in Algebraic Ge- ometry and Algebraic Number Theory: the rings appearing in Exam- ples 1.1(2) and (3) all satisfy the Noetherian condition. 2. Noetherian rings and modules Let R be a ring and let M be an R-module. Recall that M is said to be finitely generated if there exist elements m ,...,m M such that 1 k ∈ M = k Rm . i=1 i P Lemma 2.1. The following three conditions on M are equivalent. (a) Any submodule of M is finitely generated. (b) Any nonempty set of submodules of M has a maximal element under inclusion. (c) Any ascending chain of submodules N N N even- 1 2 3 ≤ ≤ ≤ ··· tually becomes stationary. Proof. (c) implies (b) is easy. (b) implies (a): Let N be a submodule of M and let X be the collection of finitely generated submodules of N. X contains 0 and { } so by (b) there is a maximal element N X. We claim that N = N. 0 0 ∈ Otheriwise there is some x N N and then N + Rx is a finitely 0 0 ∈ \ 4 KONSTANTIN ARDAKOV generated submodule of N which is larger than N, contradiction. So N = N is finitely generated. 0 (a)implies(c): LetN N beanascendingchainofsubmod- 1 2 ≤ ≤ ··· ulesandletN := N . ThenN isasubmoduleofM whichisfinitely ∪∞i=1 i generated by (a). Suppose N is generated by elements x ,...,x . For 1 n each x there is some N such that x N . Take k = max k . We i ki i ∈ ki i{ i} see that all x N and so N = N . Therefore the chain becomes i k k stationary at N∈. (cid:3) k Definition 2.2. An R-module M is said to be Noetherian if it satisfies any of the three equivalent conditions of Lemma 2.1. Proposition 2.3. Let N M be two R-modules. Then M is Noether- ≤ ian if and only if both N and M/N are Noetherian. (cid:3) Proof. Problem sheet 1, Q4. As a consequence we see that Mn := M M M is Noetherian ⊕ ⊕···⊕ for any Noetherian module M. Definition 2.4. A ring R is Noetherian if R is a Noetherian R- module. Examples of Noetherian rings are fields, Z, PIDs and (as we shall see momentarily) polynomial rings over fields. An example of a ring which is not Noetherian is the polynomial ring of infinitely many inde- terminates Z[t ,t ,...]. 1 2 Proposition 2.5. A homomorphic image of a Noetherian ring is Noe- therian. Proof. Let f : A B be a surjective ring homomorphism with → A Noetherian. Then B A/kerf and the ideals of B are in 1 1 ≃ − correspondence with the ideals of A containing kerf. Now A satis- fies the ascending chain condition on its ideals and therefore so does A/kerf B. ≃ Proposition 2.6. Let R be a Noetherian ring. Then an R-module M is Noetherian if and only if M is finitely generated as an R-module. Proof. If M is Noetherian then M is finitely generated as a module. Conversely, soppose that M = k Rm for some m M. Then M i=1 i i ∈ is a homomorphic image of thePfree R-module Rk with basis: Define the module homomorphism f : Rk M by f(r ,...,r ) := r m . Since R and Rk are Noetherian mod→ules so is M1 Rk/kkerf.Pi i (cid:3)i ≃ B2.2: COMMUTATIVE ALGEBRA 5 The main result of this section is Theorem 2.7 (Hilbert’sBasisTheorem). Let R be a Noetherian ring. Then the polynomial ring R[t] is Noetherian. Corollary 2.8. Let F be a field. Then every ideal of F[t ,...,t ] has 1 n a finite generating set. Proof of Theorem 2.7. It is enough to show that any ideal I of R[t] is finitely generated. If I = 0 this is clear. Suppose I is not zero. Let { } M be the ideal of R generated by all leading coefficients of all non-zero polynomials in I. Then M is finitely generated ideal and hence there are some polynomials p ,...,p I such that p has leading coefficient 1 k i ∈ c andM = Rc +Rc + +Rc . LetN = max degp 1 i k and i 1 2 k i ··· { | ≤ ≤ } let K = I R Rt RtN . Note that K is an R-submodule of ∩ ⊕ ⊕···⊕ the Noetheri(cid:0)an R-module RN an(cid:1)d hence K is finitely generated as an R-module, say by elements a ,...,a K I. Let J be the ideal of 1 s ∈ ⊂ R[t] generated by a ,...,a ,p ,...,p . We claim that J = I. Clearly 1 s 1 k J I and it remains to prove the converse. Let f I and argue by ≤ ∈ induction on degf that f J. If degf N then f K = Ra ∈ ≤ ∈ i i and so f J. Suppose that degf > N. Let a M be the Pleading ∈ ∈ coefficient of f. We have a = r c for some r R. Consider the j j j j ∈ polynomial g := f r tdegPf degpjp and note that degg < degf. − j j − j Since g I we can asPsume from the induction hypothesis that g J. ∈ ∈ Therefore f J. Hence I = J is finitely generated ideal of R[t]. Therefore R[t∈] is a Noetherian ring. (cid:3) Definition 2.9. Let A B be two rings. ≤ (1) Given elements b ,...,b B, A[b ,...,b ] denotes the small- 1 k 1 k ∈ est subring of B containing A and all b . i (2) We say that B is finitely generated as an A-algebra, or that B is finitely generated as a ring over A if there exist elements b ,...,b B such that B = A[b ,...,b ]. 1 k 1 k ∈ Thisisequivalenttotheexistenceofasurjectiveringhomomorphism f : A[t ,...,t ] B 1 k → which is the identity on A and f(t ) = b for each i. i i Corollary 2.10. Let R be a Noetherian ring and suppose S R is a ≥ ring which is finitely generated as R-algebra. Then S is a Noetherian ring. Proof. The above discussion shows that S is a homomorphic image of the polynomial ring R[t ,...,t ] and with Theorem 2.7 and induction 1 k 6 KONSTANTIN ARDAKOV on k we deduce that R[t ,...,t ] is a Noetherian ring. Therefore S is 1 k (cid:3) a Noetherian ring. This has the following central application to algebraic geometry. Corollary 2.11. Let X F[t ,...,t ] be any subset. Then there is a 1 k ⊆ finite subset Y X such that (X) = (Y). ⊆ V V Proof. Since F[t ,...,t ] is a Noetherian ring by Corollary 2.8, the set 1 k of ideals of R satisfies the ascending chain condition by Lemma 2.1. So X = Y for some finite subset Y of X. We conclude that h i h i (X) = ( X ) = ( Y ) = (Y). (cid:3) V V h i V h i V 3. The Nilradical Definition 3.1. A prime ideal P of a ring is said to be minimal if P does not contain another prime ideal Q P. ⊂ Theorem 3.2. Let R be a Noetherian ring. Then R has finitely many minimal prime ideals and every prime ideal contains a minimal prime ideal. Proof. Let’s say that an ideal I of R is good if I P P for some 1 k ⊇ ··· prime ideals P , not necessarily distinct. We claim that all ideals of i R are good. Otherwise let be the set of bad ideals and since R is S Noetherian, by Lemma 2.1 there is a maximal element of , call it J. S Clearly J is not prime. So there exist elements x,y outside J such that xy J. Let S = J+Rx,T = J+Ry, we have ST J and both S and ∈ ⊆ T are strictly larger than J and hence must be good ideals. Therefore P P S,P P T for some prime ideals P ,P of R. But then 1··· k ⊆ 1′··· l′ ⊆ i i′ P P P P TS J and so J is good, contradiction. So all 1··· k 1′··· l′ ⊆ ⊆ ideals of R are good an in particular 0 is good and so P P = 0 1 k { } ··· for some prime ideals P . Let Y be the set of minimal ideals from the i set P ,...,P . We claim that Y is the set of all minimal prime ideals 1 k { } of R. Indeed if I is any prime ideal, then P P I and so P I 1 k i ··· ⊆ ⊆ for some i, justifying our claim. This also proves the second statement (cid:3) of the theorem. Definition 3.3. Let R be a ring. (a) Let I be an ideal of R. An ideal P of R is said to be a minimal prime over I if P is prime, I P, and whenever I Q P with ⊆ ⊆ ⊆ Q prime, we must have Q = P. (b) min(I) denotes the set of all minimal primes over I. (c) x R is nilpotent if xn = 0 for some n 1. ∈ ≥ B2.2: COMMUTATIVE ALGEBRA 7 (d) The nilradical of a ring R, denoted by nilrad(R), is the set of all nilpotent elements of R. It follows from Theorem 3.2 that if R is Noetherian then min(I) is finite for every ideal I of R. An easy exercise shows that nilrad(R) is always an ideal of R. Proposition 3.4. Let I be an ideal of a ring R consisting of nilpotent elements (such ideal is called a nil ideal). Suppose that I is finitely generated as an ideal. Then I is nilpotent. Proof. Let x I be such that I = Rx +Rx + +Rx . Let xni = 0 i ∈ 1 2 ··· k i for some integers n N and take n = n + +n . Now i 1 k ∈ ··· In = (Rx +Rx + +Rx )n Rxs1 xsk 1 2 ··· k ⊆ 1 ··· k s1+·X··+sk=n where the sum is over all tuples s subject to k s = n. We must i i=1 i rhiagvhetahtanledasstidoenaebjovsuecihs tthheatzesrjo≥idnejalaannddthsoenIPnxsj=j =0.0. Therefore th(cid:3)e Corollary 3.5. The nilradical of a Noetherian ring is nilpotent. In the absence of the Noetherian hypothesis on the ring, the nil- radical may not be nilpotent: take any field F and consider the ideal generated by t ,t ,... in the ring ∞ F[t ,...,t ]/ t ,t2,...,tk . 1 2 1 k h 1 2 ki kS=1 There is another very useful characterization of the nilradical. Theorem 3.6 (Krull’s Theorem). For any ring R, nilrad(R) is the intersection of all prime ideals of R. The proof of this fact uses Zorn’s Lemma. Recall that a partial order on a set X is a reflexive and transitive relation on X such that a b ≤ ≤ and b a implies a = b. If is a partial order on X, we call the ≤ ≤ pair (X, ) a partially ordered set, or a poset for short. A chain C in ≤ a poset X is a subset C X which is totally ordered: for any a,b C ⊆ ∈ we have a b or b a. If S is any subset of the poset X then an ≤ ≤ element b X is an upper bound for S if s b holds for all s S. ∈ ≤ ∈ The following result is known as Zorn’s Lemma. It is equivalent to the Axiom of Choice and also to the Well-ordering principle. Lemma 3.7 (Zorn’s Lemma). Let (X, ) be a non-empty partially ≤ ordered set such that every chain of elements of X has an upper bound in X. Then X has a maximal element. 8 KONSTANTIN ARDAKOV A typical application of Zorn’s lemma is the existence of maximal ideals in any non-zero unital ring R: recall that an ideal I of R is said to be maximal if I is proper (I = R) and if J is another ideal of R 6 with I J R then either J = I or J = R. Let X be the set of ⊆ ⊆ all proper ideals of R, ordered by inclusion. Note that X is not empty since 0 X. If C is a chain in X we easily check that C X and { } ∈ ∪ ∈ so the condition of Lemma 3.7 is satisfied. Therefore X has maximal elements, i.e. maximal ideals. Proof of Krull’s Theorem. If x is nilpotent and P is a prime ideal then xn = 0 P for some n and so x P. So nilrad(R) J := ∈ ∈ ⊆ P P prime ideal of R . For the converse suppose that x is not ∩{ | } nilpotent. Let S = xn n 0 , then S is a multiplicatively closed { | ≥ } subset of R avoiding 0. By Lemma 3.7, we can find an ideal P of R which is maximal subject to having P S = . By problem sheet 1 ∩ ∅ Q1, this ideal P is prime. So x J. Thus J nilrad(R) and so nilrad(R) = J. 6∈ ⊆ (cid:3) Definition 3.8. Let I be an ideal of R. The radical of I is √I := rad(I) := x R xn I, for some n N . { ∈ | ∈ ∈ } So by definition rad(I)/I = nilrad(R/I). Using Theorem 3.6 and Theorem 3.2, we obtain the following Corollary 3.9. Let I be an ideal of a ring R. Then (a) rad(I) = P P prime ideal of R with I P . ∩{ | ⊆ } (b) If R is Noetherian and min(I) = P ,...,P then 1 k { } rad(I) = P P . 1 k ∩···∩ Connection with algebraic sets. Recall the definitions of the maps and from the Introduction. The following Proposition is an easy V I exercise. Proposition 3.10. Let I , j = 1,2,... be ideals of the polynomial ring j R = F[t ,...,t ]. Then 1 k (1) ( I ) = (I ). V j j ∩jV j (2) (PI I ) = (I I ) = (I ) (I ). 1 2 1 2 1 2 V ∩ V V ∪V (3) rad( (Z)) = (Z) for any subset Z Fk. I I ⊆ When studying algebraic sets it is natural first to express them as union of ’simpler’ algebraic sets. For example the algebraic set W = (t t ) can be written as W = L L , a union of the two lines L = 1 2 1 2 i V ∪ B2.2: COMMUTATIVE ALGEBRA 9 (t ),i = 1,2. This leads us to consider algebraic sets which cannot be i V decomposed further and we make the following definition. Definition 3.11. A non-empty algebraic set W is said to be irreducible if whenever W = W W for some algebraic sets W ,W then W = W 1 2 1 2 1 ∪ or W = W. 2 Proposition 3.12. An algebraic set W is irreducible if and only if (W) is a prime ideal. I Proof. Suppose (W) is a prime ideal and W = W W with each 1 2 I ∪ W = W. Then by Proposition 1.4, (W ) is strictly larger than (W) i i 6 I I and we can find some f (W ) (W) for i = 1,2. Then the polyno- i i ∈ I \I mial f f vanishes on both W and W hence it vanishes on W and so 1 2 1 2 f f (W). Thus (W)isnotaprimeideal,contradiction. Therefore 1 2 ∈ I I W must be irreducible. (cid:3) We leave the converse as an exercise in Problem sheet 2. Theorem 3.13. Every algebraic set is a union of finitely many irre- ducible algebraic sets. (cid:3) Proof. See Problem sheet 2. Lemma 3.14. Let W be a non-empty algebraic set and suppose that W = V V where V are irreducible algebraic sets and n is 1 n i ∪ ··· ∪ minimal possible. Let P := (V ) for each i = 1,...,n. Then i i I min( (W)) = P ,...,P . 1 n I { } Proof. Note that V V for any i = j otherwise we may omit V from i j i 6⊆ 6 the union, and hence P P for any i = j. Now (W) = n (V ). i 6⊆ j 6 I ∩i=1I i If P is a prime ideal containing (W) then P must contain at least one I of the ideals P := (V ). It follows that P ,...,P are precisely the j i 1 n minimal primes of thIe ideal (W). (cid:3) I In the setting of Lemma 3.14, it follows from Proposition 1.4 that the irreducible sets V in the minimal decomposition W = V V i 1 n ∪···∪ are determined uniquely by W, as one can recover the V from the ideal i (W) as the vanishing sets of the minimal primes above (W). I I Definition 3.15. The V are called the irreducible components of the i algebraic set W. It remains to determine the relationship between the algebraic set W = (I) and the ideal (W). This is the topic of the next section. V I 10 KONSTANTIN ARDAKOV 4. The Nullstellensatz Theorem 4.1 (weak Nullstellensatz). Let F E be two fields such ⊆ that E is finitely generated as an algebra over F. Then E/F is a finite extension. We postpone the proof in order to first explore the important con- sequences of this theorem. Corollary 4.2. Let F be a field and let R be a finitely generated F- algebra. Let M be a maximal ideal of R. Then dim R/M is finite. F Proof. R/M is a field which is finitely generated as F-algebra. (cid:3) The next corollary describes the maximal ideals of polynomial rings over algebraically closed fields. First we need some notation. Let F be a field, let R = F[t ,...,t ] be a polynomial ring and let 1 n denote the set of maximal ideals of R. Define a function M µ : Fn → M by n µ(a ,...,a ) := R(t a ) = t a ,...,t a . 1 n i i 1 1 n n − h − − i Xi=1 It is easy to check the following: µ(a ,...,a ) , 1 n • ∈ M the map µ is injective. • Corollary 4.3. Assume that the field F is algebraically closed. Then µ is bijective. Proof. It remains to show that µ is surjective. Let M be a maximal ideal of R. By Corollary 4.2 R/M is a finite field extension of F, and since F is algebraically closed, it follows that R/M F and so ≃ dim R/M = 1. This implies M + F = R. In particular for each t F i there exists a F such that t a M. Then µ(a ,...,a ) M i i i 1 n and hence M =∈µ(a ,...,a ). − ∈ ⊆ (cid:3) 1 n Theorem 4.4 (The Nullstellensatz). Let F be an algebraically closed field and let I be an ideal of the polynomial ring R = F[t ,...,t ]. 1 n Then ( (I)) = rad(I). I V Lemma 4.5. Let R be a polynomial ring over algebraically closed field F and let I be an ideal of R.