Commutative Algebra. Instructor: Graham Leuschke & Typist: Caleb McWhorter. Spring 2016

Transcription

1 Commutative Algebra Instructor: Graham Leuschke & Typist: Caleb McWhorter Spring 2016 Contents 1 General Remarks 3 2 Localization Localization The Construction of R S Prime Spectrum The Residue Field & Krull s Theorem Zariski Topology Localization of Modules Primary Decompositions Primary Submodules & Ideals Primary Decomposition Applications of Associated Primes Ideal/Variety Correspondence Affine Algebraic Varieties Hilbert s Nullstellensatz & Krull s Intersection Theorem Krull Dimension Nullstellensatz Krull s Principle Ideal Theorem Dimension Artin-Rees Lemma Completions Inverse Limits I-adic Completion Completing Modules Flatness Fibers Integral Extensions Primes in Integral Extensions Lying Over

2 Commuative Algebra 2 7 Valuation Rings 56 8 Regular Sequences Comparing Depth and Dim Auslander-Buchsbaum Gorenstein Rings Higher Dimensional Gorenstein Rings

3 3 C. McWhorter 1 General Remarks The notes were taken in the Spring of 2016 at Syracuse University in a course taught by Professor Graham Leuschke. Throughout this text, rings will be taken to be commutative with identity. For propriety, any mistakes found in this text are to be attributed to the typist - Caleb McWhorter - and not to any referenced text or on the part of the instructor. 2 Localization 2.1 Localization Recall a few definitions: Definition 2.2 (Multiplicatively Closed). If R is a ring with S R, S is said to be multiplicatively closed if (i) a, b S then ab S (ii) 1 S (this is taken for convenience to avoid having to say it in each example) Example 2.3. (1) If R is an integral domain, then R \ {0} is multiplicatively closed. (2) If I R is an ideal, then S = R \ I is multiplicatively closed if and only if I is a prime ideal. (3) For any a R, S = {1, a, a 2, } is multiplicatively closed. Definition 2.4 (Localization). Let S R be multiplicatively closed. The localization of R at S is a ring, denoted R S, together with a ring homomorphism i : R R S satisfying (i) i(s) is a unit in R S for all s S (ii) For any ring homomorphism φ : R S such that φ(s) is a unit in A for every s S, there is a unique homomorphism ψ : R S A such that R i φ R S A!ψ This is also sometimes denoted S 1 R or R[S 1 ]. Also, if S = R \ P for some prime ideal P, we write R P instead of R R\P or R S. Example 2.5. (1) If S consists entirely of units, R S = R (this should perhaps be = instead of =, but we will not bother with such details).

4 Commuative Algebra 4 (2) If R is an integral domain and S = R \ {0} then R S = Q(R) is the quotient field of R. (3) S = {1, a, a 2, }, then R S = R[x]/ax 1. This is because if we want a n to be invertible for each n N, it is sufficient to invert a. So one need only append to R an element x such that ax = 1. (4) If a S is a zero divisor, say ab = 0 with b 0. Then i(0) = 0 because i is a ring homomorphism. But then i(ab) = i(a)i(b) = 0 and as i(a) is a unit in R S so i(b) = 0. But then R i R S is not injective. Exercise: If no element of S is a zero divisor, then i : R R S is injective. Furthermore, show that if 0 S, then R S is the zero ring. [Note: These are more easily shown using the construction of R S.] 2.6 The Construction of R S We define an equivalence relation on R by (r 1, s 1 ) (r 2, s 2 ) if and only if there is a t S such that t(r 1 s 2 r 2 s 1 ) = 0. We write r/s for the class of (r, s). Then we have r 1 /s 1 = r 2 /s 2 if and only if there is a t S that kills the cross ratio. Define R S = (R S)/ with map i : R R S given by a a/1. Example 2.7. (1) If R = Z, S = Z \ {0}, then R S = Q. (2) If R = Z and S = R \ (S) (observe that S is prime) then R S = {r/s r Z, s Z \ (S)} = {r/s s not divisible by 5} (3) If R = Z and S = {1, 2, 4, 8, 16, }, then R S = {r/s s = 2 k }. (4) If R = k[x], where k is a field, P = (x + 1) and S = R \ P, then { } f(x) R S = f(x) k[x], g(x) k[x] \ (x + 1) g(x) { } f(x) = x + 1 g(x) g(x) { } f(x) = g( 1) 0 g(x) = All rational functions defined at x = 1 In particular, R S k(x), the field of all rational functions. (5) R = k[x, y]/(xy), where k is a field, with S = {1, x, x 2, } (really 1, x, ). In R S, x becomes a unit. We have xy = 0 so y = 0 in R S. That is, x/1 is a unit, xy/1 = 0/1 so y/1 = 0/1. Then R S = k[x, x 1 ].

5 5 C. McWhorter 2.8 Prime Spectrum Definition 2.9 (Prime Spectrum). The (prime) spectrum of R, denoted Spec R, is the set of all prime ideals in R. The spectrum is one of the primary bridges between commutative algebra and algebraic geometry. Remark If φ : A B is a ring homomorphism and P is a prime ideal of B, that is P Spec B, then φ 1 (P ) is a prime ideal in A, i.e. φ 1 (P ) Spec A. Proof: If a, a φ 1 (P ), then φ(aa ) P which implies that φ(a)φ(a ) P. So one of φ(a), φ(a ) are in P so that one of a, a are in φ 1 (P ). We then get a function φ # : Spec B Spec A, where φ # (P ) = φ 1 (P ). We don t in general get a function Spec A Spec B. If Q is a prime ideal of A, then φ(q) is probably not even an ideal of B, nevertheless a prime ideal! Remark If φ is surjective, we get a map. Let I R be an ideal, by Noethers First Isomorphism Theorem, we have a map π : R R/I, the canonical projection. Then Noethers Fourth Isomorphism Theorem, we have an inclusion preserving bijection {ideals I R containing I} {ideals J in R/I} given by J π(j) and J π 1 (J). This bijection preserves prime ideals in both directions. So we get a bijection between P Spec R such that P I and Spec(R/I). Note that this function is injective: if φ $ (p) = φ # (q), then φ 1 (p) = φ 1 (q) so that applying φ yields p = q. We want to create a similar correspondence for localization. If S R is multiplicatively closed, we have a ring homomorphism i : R R S given by a a/1. If I R is an ideal, we get I S = {a/s a I, s S}. Observe that this is (probably) bigger than i(i) as it is the closure of i(i) under multiplication by elements of R S. One should verify that I S is an ideal of R S. Example Let R = k[x, y]/(xy), where k is a field. Let S = {1, x, x 2, }, I = (y), and J = (x). Both ideals I, J are prime in R. Why? To see that I is prime in R, note that R/I = (k[x, y]/(xy))/(y) = (k[x, y]/(y))/(xy) = k[x] as (xy) is already 0 in the right congruence. But this is an integral domain so that I is prime in R. The same holds true mutatis mutandis for J. Note that I S = (0) in R S since y/1 = 0/1 in R S. Furthermore, J S contains a unit - x/1 (as x is invertible) - so J S = R S. This example is an example of bad behavior. But this does not rule out the injectivity of the function I I S for prime ideals. Note that we want the ideal to contain no units. If this is the case, then the ideal under this function will either survive or blow-up to R S. Theorem Let S R be multiplicatively closed. Then there is a bijection {P Spec R P S} {Q Spec(R S )} given by P P S in the forward direction and Q i 1 (Q) = i # (Q) in the reverse direction.

6 Commuative Algebra 6 Proof: Let P Spec R, then P S is prime in R S. Take a/s, b/t R S such that a b s t P S. Then ab st p/w for some p P and w S. Then there is a u S such that u(abw stp) = 0. Therefore, uabw = ustp. Writing the left as uw ab and observing the right has ustp P, this shows that uw P or ab P. But u, w S, which is multiplicatively closed, so that uw S. But this shows that uw / P. Then ab P so that a P or b P. Therefore, a/s P S or b/t P S. We know that q i 1 (Q) = i # (Q) takes primes to primes. We only need check that they satisfy the given condition: i # (Q) S =. If this is not the case, then Q = R S, which is not the case. So to complete the proof, we only need check that these maps are inverses. First, we show P = i 1 (P S ). Let r P, then r/1 P S so r i 1 (P S ). So P i 1 (P S ). Take r i 1 (P S ), then i(r) P S. But i(r) = r/1 so that r/1 = a/s for some a P and s S. So there is a u S such that u(rs a) = 0. But then urs = ua. Writing the left side as us r and observing that ua P, this shows that us P or r P. But as above, us / P so that r P. Finally, we show Q = i 1 (Q S ). Let r Q. Then r/1 Q S so that r i 1 (Q S ), showing that Q i 1 (Q S ). Let r Q S. Then r = p/s, where p P and s S. But then i 1 (r) = i 1 (p/s) = p Q so that i 1 (Q S ) Q. Consider the special case of S = R \ P, for some prime ideal P Spec R. Spec(R P ) {p Spec R p P } In particular, R P has a unique maximal ideal, namely P S. This is why this process is called localization (recalling that a local ring is a ring which contains a unique maximal ideal). Some authors, however, call this quasi-local leaving the term local to mean quasi-local and noetherian. Remark The notation is clunky so that we introduce a new notation. For a ring homomorphism A B and I A, we write IB to be the smallest ideal in B containing the image of I, i.e. the ideal generated by the image of I. Example R P is a local ring with maximal ideal P R P. Corollary If R is noetherian or artinian, so is R S for any multiplicatively closed subset S R. Lemma Localization commutes with quotients, i.e. R S /I S = (R/I)S for any I R and multiplicatively closed S R, where S denotes the image of S in R/I. Proof: Let φ : R S /I S (R/I) S be defined by r/s + I S r/s. We show that this is well defined. Suppose that r/s + I S = r /s + I S so that (r/s r /s ) + I S = I S.

7 7 C. McWhorter Example Let P Spec R. R P /P R P = (R/P )P One the left, we have a field (as we have a quotient by a maximal ideal). On the right, we have R/P is a domain and P = (0). So localizing at P means that every element outside of (0) has been inverted. But then we have a field. This field is the quotient field of R/P. We refer to this as the residue field at P, denoted κ(p ) = R P /P R P = (R/P ) P 2.19 The Residue Field & Krull s Theorem Definition 2.20 (Residue Field). The residue field of R at a prime ideal P, denoted κ(p ), is κ(p ) def = R P /pr P = (R/P )P Example Consider the ring R = k[x, y]. If P = (0), then R P = k(x, y), the field of rational functions. The unique maximal ideal of this is 0 (as it is a field) so that κ(p ) = k(x, y). If we take P = (x, y), then R/P = k, as k is a field its unique maximal ideal is 0 so that κ(p ) = k. Notice also { } f(x, y) R P = k[x, y] (x,y) = f(x, y), k[x, y], g(x, y) / (x, y) g(x, y) But g(x, y) / (x, y) if and only if g(x, y) has nonzero constant term. But this happens if and only if g(0, 0) 0. So R P is simply all rational functions defined at (0, 0). If P = (x), we have Definition For an ideal I R, define κ(p ) = (R/(x)) (x) = k[y] (x) = k[y] 0 = k(y) V (I) = {p Spec R p I} This is in one-to-one correspondence with Spec(R/I). Our goal is to show V (I) defines a topology on Spec R - the Zariski topology. Recall that I = {r R r n I for some n 1} is the radical of I. If I = 0, 0 def = Nil(R), the nilradical, i.e. the set of nilpotent elements. We know that I is an ideal of R. Theorem 2.23 (Krull s Theorem). I = p V (I) i.e., the radical of I is the intersection of all prime ideals p containing I. Proof: If I = (0), we must show Nil(R) = p Spec R p. Suppose r Nil(R). Then r n = 0 for some n. As r n p for all primes p, this shows immediately upon induction upon the power of r using the fact that these ideals are prime that r p for all p so that r p. Now suppose that r p Spec(R) p and assume that r is not nilpotent. Let S = {1, r, r 2, }. Then as r is not nilpotent, 0 / S. So the localization R S is not the zero ring. So R S has a maximal ideal by Zorn s Lemma. In particular, Spec R S is nonempty (as this maximal ideal must also be prime). But Spec R S is in one-to-one p

8 Commuative Algebra 8 correspondence with the set {p Spec R p S }. So there is some prime p Spec R such that r n / p for all n. But r p Spec R p so it must be in all p, this is a contradiction. We can no do the general case. Let π : R R/I be the canonical projection. Then I is π 1 (Nil(R/I)) = π 1 ( p Spec(R/I) p by the particular case. But this is p Spec(R/I) π 1 (p) = pıv (I) p Zariski Topology Proposition V (I) has the following properties: (i) V ((0)) = Spec R (ii) V (R) = (iii) α Λ V (I α ) = V ( α Λ I ) α ( ) (iv) n j=1 V (I j) = V n j=1 I j (v) V (I) = V (J) if and only if I = J Proof: (i),(ii) This is routine. (iii) This follows similarly to the arguments we have made previously. (iv) Suppose that p V (I 1 ) V (I n ). So p V (I k ) for some k if and only if p V (I 1 I n ). Note this uses the fact that p is prime: if p I 1 I n but p I k for any k = 1, 2,, n, then there would be a a k I k \ p for all k. Then a 1 a n I 1 I n p but a i / p for all i, contradicting the fact that p is prime. (v) If V (I) = V (J), then I = p V (I) p = pinv (J) p = J, where the middle equality follows by assumption and the end equalities follow from the Krull Intersection Theorem. On the other hand, if I = J, then I J so that any prime containing J contains I. So V ( J) V (I). The other direction is shown mutatis mutandis so that V (I) = V (J). Notice that the preceding proposition shows that the V (I) are the closed sets in some topology. Definition 2.26 (Zariski Topology). The Zariski topology on Spec R has closed sets V (I) for any I, an ideal of R. Note that this topology is well defined by the proposition. The Zariski Topology also satisfies the T 0 axis: open sets separate points. Why? If p q Spec R, then if p q, we have open set (Spec R \ V (p)) = U with p / U and q U as q q. If p q, then the open set U = Spec R \ V (q) is such that q / U and p U. Furthermore, notice that the Zariski Topology does not satisfy the T 1 axioms: that singleton sets are closed. To see this, let p Spec R. We know that {p} is closed if and only if {p} = V (I) for some ideal I if and only if {p} = V (p). But then p is maximal. So the only closed points of Spec R are maximal ideals.

9 9 C. McWhorter Corollary {p} = V (p) The Zariski topology is also quasi-compact: every open covering has a finite subcovering. Note that Bourbaki defines Hausdorff as points being closed, quasi-compact as being that open covers have finite subcoverings, and compact being Hausdorff and quasi-compact. Note that the Zariski topology is not Hausdorff if R has any non maximal prime ideals so we truly need the distinction between quasi-compact and compact. This distinction essentially only exists for algebraic geometers and algebraic number theorists since they are essentially the only topology they consider is the Zariski topology and it is not Hausdorff. Proposition The Zariski topology is quasi-compact. Proof: Let X = Spec R for convenience. Let {U α } be an open cover of X. For each α, U α = X \V (I α ) for some ideal I α. We have X = α U α. So X = α U α = (X \ V (I α )) α = X \ V (I α ) α ( ) = X \ V I α Therefore, V ( α I α) =. If no proper prime ideal contains α I α, as all maximal ideals are prime, it must be that α I α = R. Then 1 α I α. As the sum representing 1 must be finite, we must have 1 = a a n, where a i I αi. But then I α1 + + I αn = R. Reversing the equalities from before implies that n n X = (X \ V (I αi )) = i=1 α i=1 U αi Definition 2.29 (Principal Open Sets). For a R, let D(a) denote the set D(a) = Spec R \ V ((a)) = {p Spec R a / p} Note that this is an open set in Spec R. If Spec R \ V (I) is an open set and I is generated by {a α } α Λ, then U = Spec R \ V (I) ( ) = Spec R \ V a α = Spec R \ α α V ((a α )) = α Spec R \ V ((a α )) = D(a α ) α

10 Commuative Algebra 10 The family of principal open sets, D(a), form a basis for the Zariski topology. Recall that if ϕ : A B is a ring map, we get ϕ # : Spec B Spec A given by q ϕ 1 (q). Proposition ϕ # is continuous in the Zariski topology. Proof: We need to show that the preimage of open sets are open or the preimage of closed sets are closed. Take a closed set V (I) Spec A. q (ϕ # ) 1 (V (I)) ϕ # (q) V (I) ϕ 1 (q) I q ϕ(i) IB q q V (IB) which shows the primage of the closed set V (I) is the closed set V (IB). Corollary The association φ given by R Spec R and A B ϕ # : Spec B Spec A is a contravariant functor from Commutative Rings to Topological Spaces. Proof: This is routine verification. Remark There are two useful things to note: (i) Topological spaces coming from commutative rings are called affine schemes. A scheme is obtained by gluing together affine schemes. (ii) The corollary explains why we have to have non-closed points in the Zariski topology. We could try to define a functor from Commutative Rings to Topological spaces by R MaxSpec R. But this is not a functor - ring homomorphisms to not necessarily map to continuous maps. It is a good exercise to find a ring map A B that does not induce a continuous map MaxSpec B MaxSpec A Localization of Modules Definition 2.34 (Localization of Modules). Let M be a R-module and S R be a multiplicatively closed set. Put M S = {x/s x M, s S}. Specifically, this is an equivalence class x/s y/t if and only if there is a u S such that u(xt sy) = 0. Then M S is a R S -module via r/s x/t def = (rx)/(st). Note there is a canonical map M M S taking x M to x/1. Remark (i) If S consists of nonzero divisors on M, then x x/1 is injective. (ii) If there is s S such that sm = 0, then M S = 0. Definition 2.36 (Support). The support of M is the set Supp M = {p Spec R M p 0}

11 11 C. McWhorter Notice the second remark above says that if Ann M p, then M p = 0. Supp M V (Ann R M). Proposition If M is finitely generated, then Supp M = V (Ann R M). In other words, Proof: M S = 0 if and only if every x M is annihilated by some s S. If p / Supp M, then M p = 0. So every element of M is killed by some s R \ p. Now suppose x 1,, x n generates M. So every element is of the form r 1 x r n x n for some r i R. We know that x i is killed by some s i R \ p. Then n i=1 s i kills every element of M and n i=1 s i R \ p. So p / V (Ann R M). Corollary 2.38 (Local Zeroness ). The following are equivalent for a R-module M: (i) M = 0 (ii) M p = 0 for all p Spec R (iii) M m = 0 for all maximal ideals m Proof: We first prove this using the previous proposition for finitely generated R-modules M (the (iii) (i) part requires this). (i) (ii): Routine (ii) (iii): Maximal ideals are prime. (iii) (i): We must have Supp M containing no maximal ideals then Ann M is not contained in any maximal ideal. The only such ideal is the whole ring so Ann R M = R. So IM = 0, then M = 0. Now let M be an arbitrary R-module. Let x M. Then we have Rx is exact. Localizing at any maximal ideal m 0 ( M Rx) m m = 0 This implies (Rx) m = 0 for all x and m. So by the finitely generated case, we know that Rx = 0 for all x. That is, M = 0. Proposition M S = M R R S In fact, every element of M R R S can be written as a simple tensor: x r/s. Notice that the proposition states that localization is a tensor product. So localization is a functor R mod R S mod given by M M S and M f f/1 N M S N S where we have f ( x ) def 1 s = f(x) s. Notice that r f ( x ) = r f(x) = rf(x) = f(rx) = f ( r x ) s 1 t s t st st 1 s t so f 1 is R S-linear. Note that there are still things to show to prove this fact but they are routine verifications.

12 Commuative Algebra 12 Proposition Localization is an exact functor, i.e. if is a short exact sequence, then is a short exact sequence. 0 A f B g C 0 0 A S f/1 B S g/1 C S 0 Proof: We will show that for any R-map f : M N that localization preserves images and kernels, i.e. (i) ker f/1 = (ker f) S (ii) im f/1 = (im f) S Suppose that x s ker f/1. Then f ( x ) 1 s = 0 f(x) 1 so that s = 0 1. Then there is a u S such that u(f(x) 0) = 0. But then this implies uf(x) = 0 so that f(ux) = 0. But then ux ker f so that ux us = x s (ker f) S. Now suppose that x s (ker f) S. Then f(x) s = 0 1. But then f ( x ) 1 s = 0 1 so that ker f/1. The proof of (ii) follows similarly. x s Corollary A localization R S is a flat R-algebra. That is, the functor R R S is exact. Not that this functor is not generally faithfully flat since localization can send nonzero modules to 0.

13 13 C. McWhorter 3 Primary Decompositions 3.1 Primary Submodules & Ideals Definition 3.2 (Primary Submodule). Let N M be R-modules. We say that N is a primary submodule of M if and only if for every element a R, multiplication by a on quotient M/N is injective or nilpotent. Equivalently, if rx N for some x M, then either x N or r n : M N. Observe that N M is primary if and only if 0 M/N is primary. Furthermore, I R is a primary ideal, i.e. a prime submodule of a ring if and only if ab I for two elements a, b R then either b I or a n I for some n 0. Consequently for any prime ideal, the power P n is a prime ideal. To see this, suppose ab P n. Then ab P so that a p or b P. But then a n P n or b n P n. Now suppose that N M is primary. Then Ann R (M/N) is a prime ideal. Proposition 3.3. If N, M are R-modules with N a primary submodule of M, then AnnR (M/N) = {r R rm N} = {r R r n M N} Proof: Suppose ab Ann R (M/N) with a / Ann R (M/N). Then for some n, (ab) n Ann R (M/N) for all k. So multiplication by a is not nilpotent on M/N. So this map must be injective since N M is primary. Then we have 0 = a n (b n (M/N)) so that b n (M/N) = 0 so that b Ann R (M/N). Note that if N M is primary by the preceding result, if P = Ann R (M/N), we say that N is P -primary or that P belongs to N. If I R is primary, Ann R (R/I) = I is a prime ideal by the preceding remark. Furthermore, primary ideals have prime radicals but the converse is false. Example 3.4. Let R = k[x, y], where k is a field and let I = (x 2, xy). Intersection Theorem (for the first equality) I = p = p = (x) p V (I) x 2,xy p p = x p Then using Krull s is a prime ideal (note the middle equality follows from the fact that if x 2 p, then x p so that xy p automatically). It only remains to show that this is not primary. Observe xy I and x / I. But now power of y I. So that I is not primary. Proposition 3.5. If I is a maximal ideal, then I is primary. In fact, if m = I then I is m-primary. Proof: Suppose ab I and b n / I for all n 0. We want to show that a I. Since b n / I, then b / I = m. So the ideal (I, b) is not contained in m. Then it must be that (I, b) = R so 1 (I, b); that is, 1 = x + rb for some x I and r R. But then 1 = x + rb so that a = ax + rab. But ax I and rab I so that a I.

14 Commuative Algebra 14 Example 3.6. Let R = k[x, y], where k is a field and let I = (x 3, x 2 y 4, y 5 ) is primary to a maximal ideal generated by (x, y) since P I contains both x, y so that by Krull s Intersection Theorem, I = (x, y), which is prime. Example 3.7. In Z, an ideal I is primary if and only if I = (0) or I = (p n ) for a prime P and some n 0. Consequently by the Fundamental Theorem of Arithmetic, every ideal in Z can be written uniquely as an intersection of primary ideals with distinct radicals: if (m) Z, where p a 1 1 pa 2 2 par r, where the p i are distinct prime integers, then (m) = (p a 1 1 (par r ) and (p a i i ) = (p i). It is our goal to show that this is the case for any noetherian ring. That is, given any submodule N of a noetherian module M has a primary decomposition N = Q 1 Q n where each Q i is primary. The Q i are not unique but minimal primary decompositions are unique and prime ideals belonging to them are. 3.8 Primary Decomposition Definition 3.9 (Irreducible Submodule). We say that a submodule N M is irreducible if N can be written as N = N 1 N 2 for N i M, then N 1 = N and N 2 = N. It is our goal to use this to show the existence of primary decompositions. But we do this in an easier manner by showing existence of irreducible decompositions, which we will show are primary. We then worry about the uniqueness. Lemma Assume that M is a noetherian module. Then irreducible submodules are primary. Proof: Let N M be irreducible. Let r R and assume that multiplication by r is not nilpotent on M/N. Since M is noetherian, M/N is noetherian. We have a chain ker r ker r 2 ker r 3 ker r n This must stabilizer as M/N is noetherian, say at n. Then by Fitting s Lemma, ker r n im r n = 0 in M/N as r n x = 0 and x = r n y. Then r 2n y = 0 so that r n y = 0 by stabilization showing that x = 0. Since N is irreducible in M, 0 is irreducible in M/N (by the Correspondence Theorem) showing ker r n = 0 or im r n = 0. But then ker r n = 0 showing that multiplication by r is injective via the chain of kernels. Theorem 3.11 (Noether, 1921). Let M be a noetherian module and N a submodule, then we can write N = Q 1 Q n of irreducible submodule, hence primary submodules.

15 15 C. McWhorter Proof: Let Γ = {N M N has no such decomposition}. We want to show that Γ =. Suppose that Γ. By noetherianness, Γ has a maximal element, say N. It must be that N is reducible otherwise it is its own decomposition, so that N / Γ. So N = N 1 N 2 with N 1, N 2 N. By maximality, these have irreducible decompositions so that N has an irreducible decomposition, a contradiction. Therefore, Γ =. Corollary Primary decompositions exist for submodules of noetherian modules. Now we need to discuss what we mean by saying we can make this decomposition unique as aforementioned. Definition 3.13 (Irredundant). Let N = Q 1 Q n be a primary decomposition. Let P i def = AnnR (M/Q i ) be the set of prime ideals belonging to Q i (note that if N = I is an ideal, the Q i are prime ideals so that P i = Q i ). We say that such a decomposition is irredundant or reduced if no Q i can be removed from the intersection with equality still holding and that all the P i need be distinct. It is clear that such a decomposition satisfying the first condition is possible: if you are given a primary decomposition and one of the Q i is not needed for equality in the intersection, simply leave it out. We need to demonstrate that the second condition can be made to hold. Lemma If Q 1,, Q n are p-primary, then so too is Q 1 Q k (so if several of the Q i have the same P i, simply clump them together to form the smallest Q i ). Proof: We need to show (i) the intersection Q 1 Q k is primary (ii) P = Ann R (M/Q 1 Q k ) To prove (ii), observe Ann R (M/Q 1 Q k ) = k i=1 Ann R(M/Q i ) ( knocking M into all the Q i s must knock M into each one). Since I J = I J, we have AnnR (M/Q 1 Q k ) = k i=1 AnnR (M/Q i ) = k i=1p = p To prove (i), let r R. Assume that multiplication by r is not nilpotent on M/Q 1 Q k. So r / Ann R (M/Q 1 Q k ) = p = Ann R (M/Q i ) for i = 1, 2,, k. So multiplication by r is not nilpotent on any M/Q i. Therefore, multiplication by r is injective on each M/Q i so that multiplication is injective on M/Q 1 Q k. Corollary Irredundant primary decompositions exist for noetherian modules. Theorem Let N = Q 1 Q n be an irredundant primary decomposition for an R-submodule N, where R is noetherian. Let P i be the prime ideal belonging to Q i : (i) the set {p 1,, p n } is determined uniquely by N (ii) the Q i s corresponding to isolated primes P i (does not contain the other P j s) are also determined by N

16 Commuative Algebra 16 Definition 3.17 (Colon Ideals). If N M is a submodule, we define (N : R M) def = {r R rm N} If x M, we define (N : R x) def = {r R rx N} Lemma Let Q M be a p-primary submodule, i.e. AnnR (M/Q) = p Let x M. If x Q, then (Q : R x) = R and if x / Q then (Q : R x) = p. This shows we can write (Q : R x) to obtain p instead of using Ann R (M/Q). Proof: If x Q, then (Q : R x) = R. To prove the second equality, let r (Q : R x) then r n x Q for some n. So multiplication by r n : M/Q M/Q is not injective so that it must be nilpotent. So r m Ann R (M/Q) for some m. Hence, r p. For the other inclusion, let r p so that r n annihilates M/Q for some n. In particular, r n x Q. Theorem 3.19 (First Uniqueness Theorem). Let R be a noetherian ring and N M be R- modules. Suppose that N = Q 1 Q s is an irredundant primary decomposition of N in M. Let p i = Ann R (M/Q i ) be the primes belonging to the Q i. For any p Spec R, we have p {p i } s i=1 if and only if p = (N : R x) for some x M \ N. Hence, the p i are uniquely determined by N and do not depend on the choice {Q i } s i=1. These primes are called the associated primes of N and we write {p i } s i=1 = Ass R(M/N). In the special cases of I R, we write Ass R (I) def = Ass R (R/I) and if I = (0), we write Ass R (R) def = Ass R ((0)). Proof: To prove the reverse direction, suppose that p = (N : R x) R. Then we have p = (N : x) = (Q 1 Q s : x) = (Q 1 : x) (Q s : x) (Q 1 : x)(q 2 : x) (Q s : x) But as p is prime, we must have p (Q i : x) for some i. As p R, we must have x / Q i. Furthermore as p is prime, p (Q i : x) = p i by the preceding lemma. On the other hand, p = (Q j : x) (Q i : x) (Q i : x). Therefore, p = (Q i : x) = p i. For the forward direction, without loss of generality, we show that p 1 = (N : x) for some x M \ N. As the primary decomposition is irredundant, N Q 2 Q s. Then the set Γ = {I R I = (N : x), x (Q 2 Q s ) \ N} is nonempty. As R is noetherian, we can find a maximal element, say (N : x). We claim that Ann R (M/Q 1 ) (N : x) p 1 = Ann R (M/Q 1 ). To see that, Ann R (M/Q 1 ) (N : x), suppose that r Ann R (M/Q 1 ). Then rx Q 1 and we also have x Q 2 Q s. Then it must be that rx Q 1 Q s = N. To see that (N : x) p 1, suppose that s (N : x). Then sx N Q 1. However, x / Q 1 so that s is not injective on M/Q 1 so that s is nilpotent. Therefore, s Ann R (M/Q 1 ) = p 1.

17 17 C. McWhorter The fact that we have Ann R (M/Q 1 ) (N : x) p 1 = Ann R (M/Q 1 ) implies that p 1 = (N : x). We want to show p1 = (N 1 : x); that is, we want to show that the radical is redundant. It is enough to show that (N : x) is prime. Let a, b R such that ab (N : x) and a / (N : x). Then abx N so that b (N : ax). We show that (N : ax) = (N : x) via maximality. Note that (N : x) (N : ax). To it is sufficient to show (N : ax) Γ. For this to be so, we need ax (Q 2 Q s ) \ N. But x Q 2 Q s so that ax Q 2 Q s also. But a / (N : x) so that ax / N. But then (N : ax) Γ so that (N : ax) = (N : x) by maximality. But then b (N : x), as desired. Remark Note that if we say that p is an associated prime to the R-module M if N = 0 in the First Uniqueness Theorem. This is equivalent to saying that there is a m M such that p = Ann R (m). Example Let R = k[x, y], where k is a field, and I = (x 2, xy). Notice that I = (x) (x 2, y) and I = (x) (x 2, xy, y 2 ). These are both irredundant primary decompositions. Then (x) is prime showing that it is primary. Furthermore, (x 2, y) = (x 2, xy, y 2 ) = (x, y) is maximal, so it is primary also. But then Ass R (R/I) = Ass R (I) = {(x), (x, y)}. We should be able to write these in the form (I : f) for some f / I. Notice that xy I so that x (I : y). In fact, if g (I : y), then gy I (x) so that g (x). Hence, (x) = (I : y). For the other, notice that x 2, xy I so that (x, y) (I : x). But as (x, y) is maximal and (I : x) R (as x / I), we have (x, y) = (I : x). Definition 3.22 (Isolated/Embedded Primary Component). Suppose that N = Q 1 Q s is an irredundant primary decomposition and p i = Ann R (M/Q i ). If p i is a minimal element of Ass R (M/N), we say that Q i is an isolated primary component. Otherwise, we say that Q i is an embedded primary component Example Given the notation of the previous example, we have MinAss(R/I) = {(x)} is isolated and (x, y) is embedded. Theorem 3.24 (Second Uniqueness Theorem). Let N M by R-modules in a noetherian ring and let N = Q 1 Q s be an irredundant primary decomposition for N in M. Suppose that Q i is isolated (so p i does not contain any p j ), then Q i = {x M x } 1 N p i = ϕ 1 (N pi ) where ϕ : M M pi. In particular, Q i is in every primary decomposition for N. Proof: For notational ease, let p = p i. Suppose that x M. Then x 1 N p and x 1 = n s for some n N, s R \ p. So there is a t / p with t(sx n) = 0, i.e. tsx = tn N. Set r = ts so that rx N Q. Notice that r / p and p = Ann R (M/Q) so multiplication by r is not nilpotent on M/Q. It follows that multiplication by r must be injective. But then x Q. Now assume that x Q. Since p is minimal in Ass R (M/N), j i p j p (if it were, then p 1 p 2 ˆp i p n p 1 so that p j p i, a contradiction). So choose a j i p j \ p. Then a Ann R (M/Q j ) for all j i. But then multiplication by a is nilpotent on M/Q j for all j i. Choosing k sufficiently large such that a k M j i Q j, then a k x j i Q j and also Q i.

18 Commuative Algebra 18 But then a k x N so that x 1 = ak x N a k p since a k / p. To see that Q i appears in every primary decomposition for N, note that we have Q i = ϕ 1 (N pi ) so that Q i is uniquely determined by N. By the First Uniqueness Theorem, we know that p i depends only on N. But Q i is determined by p i so that Q i is determined by N Applications of Associated Primes Definition 3.26 (Zerodivisor). We say that r R is a zero divisor on M if there is a x M \ {0} such that rx = 0. We denote by Z R (M) the set of zero divisors in M. Proposition Let R be noetherian and N M be finitely generated. Then Z(M/N) = p Ass R (M/N) Proof: If p Ass(M/N), then p = (N : R x) for some x M \ N. So every r p satisfies rx = 0 in M/N and so every element of p is a zero divisor. Now suppose that rx = 0 for some x M/N with x 0. Let N = Q 1 Q s be an irreducible primary decomposition for N in M. Since x / N then x / Q i for some i. However, rx Q i so that multiplication by r is not injective on M/Q i. Hence, multiplication by r must be nilpotent. This shows that r Ann R (M/Q i ) = p i. p Example Let R = k[x, y]/(x 2, y), where k is a field. Then Z(R) = p Ass(R) p. In k[x, y], (x 2, xy) = (x) (x 2, y) so that in R, (0) = (x) (x 2, xy). So Ass R = {(x), (x, y)} and Z(R) = (x, y). But then R is local (recalling that a ring is local if and only if the non units form an ideal). Proposition Let R be noetherian and I R. Then I = p Ass R (R/I) Proof: If I = Q 1 Q s, then I = Q i = Q i = p. p Definition 3.30 (nth-symbolic power). Let p Spec R. The nth symbolic power of p is p (n) = p n R p R = ϕ 1 (p n R p ) where ϕ : R R p. Remark We have the following properties for symbolic powers: (i) p n p (n) (ii) If ϕ : A B and I B is primary, then ϕ 1 (I) is primary. (iii) p n R p is a power of the maximal ideal of R p so that it is primary.

19 19 C. McWhorter (iv) p (n) is p-primary. [It is enough to show that p (n) p. If x p (n) then x 1 pn R p pr p so that x 1 = r s for some r p, s / p. But then there is a t / p with stx = tr p. However, s, t / p so that x p.] (v) p (n) is the p-primary component of p n R. Because p is isolated, any other associated prime of p n must contain p. Lemma 3.32 (Prime Avoidance). Let R be a ring and I, P 1, P 2,, P n be ideals of R with n 2 of the P i primes. If I n i=1 P i, then I is contained in P i for some i = 1, 2,, n. Proof: We proceed by induction on n. For the case of n = 1, there is nothing to show. If n = 2, then we have p 1 p 2 I. Assume to the contrary that I p 1, p 2. Choose x I such that x / I \ p 1 and y I \ p 2. Now we have x + y I but x + y / p 1, p 2. But then y = (x + y) x p 2, a contradiction. Notice this did not even use the fact that p i was prime. Now assume that n 3 and I p i but I p j for j = 1, 2,, n. If I p 1 ˆp k p n, then by induction I must be in one of them and we are done. So if I p j with j k for all k. That is, I j k p j Then at least one of the p j s are prime, say p 1. For each k = 1, 2,, n, choose a k I \ j i p j. Note that a k p k for each k. But then y def = a 1 + (a 2 a 3 a n ) I. If y p 1, then as a 1 p 1, we have a 2 a n p 1 so that a k p 1 for some k 1, a contradiction. Then it must be that y / p 1 so that y = p 1 for some j 1 so that a 1 p j, a contradiction. Note that this Lemma gets its name from its contrapositive: if I is not in any p j, then I is not contained in the union of the p j s. So there is some element of I avoiding all of the p j s. Example If (R, m) is a noetherian local ring with maximal ideal m and m is not an associated prime of R, then m p for all p Ass(R) (by maximality). Then there is a r M such that r is not an element of any associated primes. But then r is a nonunit and nonzerodivisor. Remark If a ring contains an infinite field, then more than one of the p i have to be prime. This is observed using the fact that there is no vector space with dimension greater than 1 is an infinite field is a union of finitely many subfields Ideal/Variety Correspondence Theorem 3.36 (Hilbert s Basis Theorem). If R is noetherian then R[x] is noetherian. Proof (Sketch): Let I be an ideal of R[x]. We will show that I is finitely generated. Set I d to be the set of leading coefficients of polynomials f(x) I with degree d along with 0. Then I d is an ideal of R and we have a chain of ideals as I d I d+1. As R is noetherian, this must stabilize at some n, say N. Then for d = 0,, N, write I d = (c d,1,, c d,nd ). Then the c i,j are the leading coefficients of some f i,j I so that I = (f i,j ) N,ni by choosing elements of minimal degree of the left hand side but not the right, a contradiction. Corollary If R is noetherian, then so is any finitely generated R-algebra.

20 Commuative Algebra 20 Proof: If S = R[u 1,, u n ], where u i S, then S = R[x 1,, x n ]/ ker φ, where φ : R[x 1,, x n ] S given by x i u i. But the quotient of a noetherian module is noetherian. Remark Using roughly the same argument as in the Hilbert Basis Theorem, one can show that if R is noetherian then R[[x]] is noetherian except one uses I to be coefficients of least degree Affine Algebraic Varieties Throughout this section, let k be a field. Definition 3.40 (Affine Space). An affine n-space over k is A n k = kn. The elements of an affine space are called points. Definition 3.41 (Zero Set). Let S be any set of polynomials in k[x 1,, x n ]. The zero set of S is Z(S) def = {p = (a 1,, a n ) A n k f(p) = 0 for all f S}. Such a set is called an (affine algebraic) variety. Example We can represent common graphs as affine algebraic sets (giving them an algebraic definition): (i) If k = R and S = {x 2 + y 2 1} k[x, y]. Then Z(S) is the unit circle in R 2. (ii) If k = R and S = {(x y)(y x 2 )}, then Z(S) is the union of the parabola y = x 2 and the line y = x. (iii) If k = R and S = {x, y(y 1)}, then Z(S) consists of the points (0, 0) and (0, 1). (iv) If k = R then Z(y 2x, z 3x, 3y 2z) is the line spanned by the vector 1, 2, 3. Remark Observe that if S S then Z(S) Z(S ) as anything that kills everything in S certainly kills everything in S, so zero sets are inclusion reversing. Furthermore, we have Z(S) = Z( S ). To see this, note the forward inclusion is immediate as S S. To see the reverse inclusion, let p Z(S) and f S. Then there is a f i S and g i k[x 1,, x n ] such that f = g i f i. But then f(p) = g i (p)f i (p) = 0. Remark Since Z(S) depends only on ideals by the Hilbert Basis Theorem for any S k[x 1,, x n ], there is a finite generating set for the ideal S, say f 1,, f m. Then Z(S) = Z( S ) = Z(f 1,, f m ). So an affinity algebraic varieties is the common zero set of a finite number of polynomials. Notice that we can actually choose the generators f i to be in S. To see this, we use the noetherian property. Let Γ be the set of ideals in k[x 1,, x n ] that are finitely generated by elements of S. Note that we require S to be nonempty. Then Γ has a maximal element, say J. We want J = S. Note that if J S and J S, we could add more generators, contradicting the maximality of J. Definition 3.45 (Hypersurface). A hypersurface is a zero set in A n k fo a single polynomial: Z(f). Notice that if S = f 1,, f m, then Z(S) = Z(f 1,, f m ) = Z(f 1 ) Z(f n ). So any affine algebraic variety is an intersection of finitely many hypersurfaces.

21 21 C. McWhorter Example 3.46 (Macaulay s Curve). Let C A 4 k be the locus of points of the form (s4, s 3 t, st 3, t 4 ) for s, t k, i.e. the image of the map φ : A 2 k A4 k given by (s, t) (s4, s 3 t, st 3, t 4 ). Then C is a 2-dimensional variety in A 4 k. Using (x, y, z, w) for A4 k, then C is cut out by the polynomials xw yz, y 4 x 3 w, z 4 xw 3. One can show that these generate C. Notice that codim C = dim A 4 k dim C = 4 2 = 2. Does there exist two polynomials alone cutting out C? If char k = p, then the answer is yes (this is due to Hartshorne in the 1960s). However, if char k = 0, then this is an open problem. Remark We have Z( S) = Z(S) as any points killing any one of the S s must kill them all. Furthermore, we have Z(I 1 ) Z(I m ) = Z(I 1 I m ). So see this, if p is an elements of the left side, then p Z(I j ) for some j. The preceding remarks, as before, show that affine algebraic sets form the closed sets for a topology, called the Zariski topology. Definition 3.48 (Zariski Topology). The topological space over A n k algebraic sets is called the Zariski topology. whose closed sets are the affine The Zariski topology is T 1 (that is, points are closed). To see this, let p = (a 1,, a n ) A n k and set m p (x 1 a 1,, x n a n ) k[x 1,, x n ]. Then we have Z(m p ) = {p}. Notice that this was not the case for the Zariski topology on Spec R. However, the Zariski topology is still not Hausdorff if k is infinite; that is, the Zariski topology is not T 2. If Z(I) for all ideals I k[x 1,, x n ], then the Zariski topology is pseudo-compact (quasi-compact). Remark Quasi-Compact means that any open covering has a finite subcovering. Often, one meets this as the definition of compact. However, nearly all spaces one typically deals with in Topology, all those in Analysis, et cetera are Hausdorff. However, the only topology an algebraic geometer would care to work with - the Zariski topology - is not. There needs to be a distinct. Quasi-compact is compact as one typically knows it while compact is reserved for compact and Hausdorff. This tedious (though necessary) distinction exists only for algebraic geometers. However, Z(I) does not have to hold. For example, take k = R and observe Z(x 2 + 1) A 2 k is empty. The moral of this story is this: algebraic geometry works best (perhaps only at all) over an algebraically closed field. It is also interesting to note that the Zariski topological space on an affine R-space over the reals is coarser than the Euclidean topology. Now we go the other direction: Definition 3.50 (Vanishing Ideal). For Y A n k, the vanishing ideal of Y is I(Y ) def = {f k[x 1,, x n ] f(p) = 0 for all p I} Observe that if Y Y then I(Y ) I(Y ). Furthermore, we know that Z(I(Y )) Y. In fact, Z(I(Y )) = Y Y. Lemma Z(I(Y )) = Y Proof: We know that Z(I(Y )) contains Y and is closed. So assume that Y is contained in some closed set Z(S). We want to show that Z(I(Y )) Z(S). Let p Z(I(Y )). Then we have f(p) = 0 for all f I(Y ). Let g S so that g S I(Z(S)) I(Y ), as Y Z(S). Therefore, g(p) = 0 so

22 that p Z(S). Commuative Algebra 22 Then we know that I(Y ) = I(Y ). Furthermore, we have I( Y α ) = α I(Y α ). Finally, we have I(Y 1 Y 2 ) I(Y 1 ) + I(Y 2 ) but equality need not hold. To see this, let f n I(Y ) so that f n (p) = 0 for all p Y. But we are in a field so that f(p) = 0 so that f I(Y ). Now I( ) is always a radical ideal. The sum of radical ideals need not be radical. Example Let Y 1 = Z(y) and Y 2 = Z(y x 2 ). Then I(Y 1 ) is all the multiples of y and I(Y 2 ) = (y x 2 ). We have I(Y 1 ) + I(Y 2 ) = (y, x 2 ). However, we have Y 1 Y 2 = (0, 0) so that I(Y 1 Y 2 ) = (x, y). If I is a proper ideal of the polynomial ring k[x 1,, x n ], then Z(I) (this follows from the generalized Fundamental Theorem of Algebra). Finally, we have I(Z(I)) = I for all ideals I of k[x 1,, x n ]. It is our goal to move forwards toward and prove Hilbert s Nullstellensatz: Theorem 3.53 (Hilbert s Nullstellensatz). Let k be an algebraically closed field. (i) There are many good results which follow from this theorem: Proposition The maps I Z(I) and Y I(Y ) is a inclusion reversing bijection between the set of radical ideals in k[x 1,, x n ] and varieties in A n k. Proof: We want to show that I(Z(I)) = I if I is radical and Z(I(Y )) = Y if Y is a variety. For the first direction, observe that I = I be the Nullstellensatz. For the reverse direction, if Y = Z(I) for some ideal I, we may assume that I is radical as Z(I) = Z( I). But then we have Z(I(Y )) = Z(I(Z(I))) = Z( I) = Y. (ii) Note that this proposition is an example of a Galois correspondence. Proposition A system of polynomial equations f 1 (x 1,, x m ) = 0, f 2 (x 1,, x n ) = 0,, f m (x 1,, x n ) = 0 has a simultaneous solution if and only if I(f 1,, f n ) is a proper ideal of the polynomial ring k[x 1,, x n ]. Equivalently, the above system of polynomial equations has no solutions if and only if 1 = m i=1 p if i for some polynomials p 1,, p m. Proof: If 1 is a linear combination of the f s, then the system cannot have a solution for then the right side of the equality would vanish at that point while the left side would not. Conversely, if Z(f 1,, f m ) =, then the ideal is not proper by the Nullstellensatz. Definition 3.56 (Coordinate Ring). For a variety, X A n k, the coordinate ring of X, k[x], is k[x 1,, x n ]/I(X). Observe that k[x] is a finitely generated k-algebra and is reduced since I(X) is a radical ideal. If a k-algebra is finitely generated and reduced, then it is an affine k-algebra. Conversely, if k = k, then any affine k-algebra is a coordinate ring. If S is an affine k-algebra is generated

23 23 C. McWhorter by u 1,, u n then we get a surjection π : k[x 1,, x n ] S via x i u i. So we have S = k[x 1,, x n ]/I, where I = ker π is a radical ideal since S is reduced. So S is in the coordinate ring of X = Z(I). Then we get an equivalence of categories between varieties in affine n-space and affine k-algebras. (iii) The maximal ideals of k[x 1,, x n ] are all of the form m p = (x 1 a 1,, x n a n ) for p = (a 1,, a n ) A n k. Note that this fails if k k. As an example, (x2 + 1) R[x] is maximal as R[x]/x = C, a field. So the map A n k MaxSpec k[x 1,, x n ] given by p m p is a bijection. (iv) If X A n k is a variety, then X MaxSpec k[x] given by p m p is a bijection. [p X if and only if I(X) I(p) = m p ] In order to prove the Nullstellensatz, we will need to divert and discuss dimension theory and some advanced field theory. 4 Hilbert s Nullstellensatz & Krull s Intersection Theorem 4.1 Krull Dimension Definition 4.2 (Transcendence Degree). Let X be a topological space. A subset Y X is called irreducible if Y is not a union of two proper closed subsets (note the difference between this and connected is these sets need not be disjoint). Otherwise, the subset is called reducible. The (combinatorial) dimension of X is the sup of the longest chain of irreducible subsets. In particular, if X is an affine algebraic variety then the subset Y X is irreducible. If not, it is a union of proper subvarieties. The dimension of X is the sup of lengths of chains of irreducible subsets. Example 4.3. Points are irreducible as dim({p}) = 0. In fact, dim({p 1,, p n }) = 0. Example 4.4. The set Z(y x 2 ) is clearly irreducible in A 1 k. We know dim Z(y x2 ) 1. The question is, is this dimension actually 1? The answer is yes - as one would expect - but we need more theory to prove it. This will follow from the next proposition as k[x, y]/(y x 2 ) = k[x], a domain. Proposition 4.5. A variety X A n k is irreducible if and only if I(X) is prime. Proof: Suppose that X is irreducible and let f, g k[x 1,, x n ] such that f, g I(X). Let Y 1 = Z(I(X) + (f)) and Y 2 = Z(I(X) + (g)). We know that Y 1, Y 2 are subvarieties of X. We claim that X = Y 1 Y 2. To prove, let p X. Then we know that (fg)(p) = 0 so that f(p) = 0 and g(p) = 0. So p Y 1 or p Y 2. We need also show that Y 1, Y 2 X. As X is an irreducible union of subvarieties, we know X = Y 1 or X = Y 2. If X = Y 1, then f I(X) (as X is among the zeros of f). Mutatis mutandis, if X = Y 2 then g I(X). To prove, assume that X is reducible. We want to show that I(X) is not prime so that we have proper subvarieties Y 1, Y 2 X such that X = Y 1 Y 2. Since Y 1 X, we have I(X) I(Y i ) for i = 1, 2. If I(X) = I(Y i ), then Y i = Z(I(Y i )) = Z(I(X)) = X and Y i, X are closed so that this gives equality. Then I(X) I(Y i ) for i = 1, 2. Take f i I(Y i ) \ I(X). Then f 1 f 2 I(Y 1 )I(Y 2 ) I(Y 1 ) I(Y 2 ) = I(Y 1 Y 2 ) = I(X) so that I(X) is not prime.

24 Commuative Algebra 24 Definition 4.6 (Krull Dimension). The Krull dimension of a commutative unital ring R is dim R = sup{n there is a chain of prime ideals of R, p 0 p 1 p n } Corollary 4.7. dim X = dim k[x] for a variety X A n k. Proof: Y X is an irreducible subvariety if and only if I(Y ) I(X) is a prime ideal. Definition 4.8 (Height). The height of a prime ideal p Spec R is ht p def = sup{n there is a chain of prime ideals in R, p 0 p 1 p n = p} We know that Spec(R p ) = {q Spec R q p}. That is, Spec(R p ) = {q Spec R q (R/p) = }. This is the same as ht p = dim R p. We know also that Spec(R/p) = {q Spec R q p}. So ht p + dim R/p dim R. Notice the sum on the left is the sup of the lengths of chains containing p. Strict inequality can occur if in the inclusion diagram looks like the following: Then notice we have ht p = 1, dim R/p = 1 but dim R = 3. Definition 4.9 (Catenary). We say that R is catenary if all saturated chains of primes between any two primes have the same length. Recall a saturated chain is one which cannot be refined. Theorem 4.10 (Ratliff). If R is catenary then ht p + dim R/p = dim R for all primes p Spec R. Furthermore, any affine algebraic variety over a field is catenary. Proof: If (f) is nonzero prime, then (0) (f) is a chain of length 1 so dim R 1. To show that dim R 1, we need to show that if (f) (g) for nonzero primes then we have equality, so we cannot extend the chain. Since (f) (g), then f = rg for some r so that rg (f). As (f) is prime, r (f) or g (f). We show that r / (f): if r = sf, then f = rg = sfg so that (1 sg)f = 0. But then f = 0, a contradiction. So 1 sg = 0 so that 1 = sg, showing that (g) is not prime. Note that the converse is false as dim Z[ 5] = 1 but this is not a UFD so that it is trivially not a PID. The first example of a non-catenary ring is due to Nagata in the 1950s and is not trivial. One should note that height was once called codimension and is still sometimes referred to as this - most notably in algebraic geometry. Example To see some examples of this theorem: