Bachelorarbeit. Zur Erlangung des akademischen Grades. Bachelor of Science. Dimension theory for commutative rings. Universität Regensburg

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1 Bachelorarbeit Zur Erlangung des akademischen Grades Bachelor of Science Dimension theory for commutative rings Eingereicht von: Eingereicht bei: Daniel Heiß Prof. Dr. Niko Naumann Universität Regensburg Fakultät für Mathematik Ausgabetermin: (Semester 5) Abgabetermin:

2 Contents Contents I Introduction 1 II Primary Decomposition Primary ideals The Lasker-Noether decomposition theorem III Krull s principal ideal theorem The n-th symbolic powers Krull s principal ideal theorem IV A first boundary for dim(r[x]) 11 V Dimension theory of K[X 1,..., X n ] 14 VI Dimension theory of Noetherian rings Dimension under polynomial extensions Non-transcendental extensions Summary of non-transcendental extensions VII Prüfer domains and a note on Seidenberg s F -rings Dimension theory of Prüfer domains A note on F -rings Summary of polynomial extensions VIII A journey to the power series ring 27 Summary and comparison with polynomial extensions References iii ii

3 I Introduction I Introduction In mathematics there are several concepts of dimension. Some of them are quite intuitive, like the Hamel dimension: The Hamel dimension of a vector space is the minimal number of vectors that are needed to generate the vector space. For the euclidean space R n this coincides with the natural term of dimension which is the minimal number of coordinates that are necessary to specify each point of the space. On the other hand there are less intuitive concepts of dimension: Studying fractals one usually comes across the Hausdorff dimension which associates to any metric space a certain non-negative real number, which (for fractals) can be non-integer. For example the Sierpinski triangle (a famous fractal) has the Hausdorff dimension log(3) log(2) / Q. In algebraic geometry one of the basic objects of study are affine algebraic varieties. The dimension of an affine algebraic variety Y is defined as the supremum of all natural numbers n N such that there exists a chain Z 0 Z 1... Z n of irreducible closed subsets of Y. This idea of dimension is clearly the generalization of the intuitive Hamel dimension; and it can be translated into notion of commutative algebra: For this let k be an algebraically closed field and m 1 an integer. For an affine algebraic subvariety Y of k m the chain of irreducible closed subsets of Y correspond by Hilbert s Nullstellensatz to a chain of prime ideals p 0 p 1... p n of k[x 1,..., X m ] containing the vanishing ideal I(Y ) of Y. This chain in turn corresponds to a chain of prime ideals p 0... p n in the ring k[x 1,..., X m ]/I(Y ). So the dimension of a variety Y can be interpreted as the supremum of the lengths of chains of prime ideals in the ring k[x 1,..., X m ]/I(Y ). So we translated the concept of geometric dimension into notion of pure algebra. There this concept of dimension is called the Krull dimension of a ring and throughout this thesis we will call it just the dimension. Clearly the Krull dimension of any field k is zero and any principal ideal domain that is not a field is one-dimensional. The main interest of this thesis will be the behavior of the dimension of a commutative unitary ring R under certain extensions R S. Mainly we study polynomial extensions and see that in several cases the dimension rises by the number of adjoint indeterminates. This thesis is organized as follows: We start with a brief introduction to primary decomposition in order to proof Krull s principal ideal theorem and similar results in section III. Equipped with this we are able to give a lower and an upper bound for the dimension dim(r[x]) of the polynomial ring R[X] as it is provided by [Sei53]. Then we follow principally [BMRH73] and we will give an elementary proof that dim ( k[x 1,..., X n ] ) = n for any field k or more generally any Artinian ring. Next we pass from Artinian rings to Noetherian rings and we will see still following [BMRH73] that we have a similar behavior: For any Noetherian ring we have dim(r[x]) = dim(r) + 1 fitting the previous result for zero-dimensional Noetherian rings. Still in the Noetherian case and now following [Cla65] we will further study the Krull dimension under ring extensions R R[x] where x is not an indeterminate. It will turn out that the dimension either decreases by one or does not 1

4 II Primary Decomposition change at all and we will give some necessary and sufficient conditions for the dimension to decrease. In section VII we will find out that the beautiful property of the dimension under polynomial extensions in the case of Noetherian rings is also true for semi-hereditary rings, especially it is true for Prüfer domains. Further we will discuss rings that do not behave like Noetherian rings or Prüfer domains under polynomial extentions. Especially we will have a look at F -rings provided by Seidenberg in [Sei53, Sei54]. Eventually we will make a short visit to the formal power series rings in one and several indeterminates observing that in the Noetherian case the behavior is just like in the polynomial case while in the non-noetherian case various weird things can happen. In this last section we only want to give a rough overview of Arnold s work in [Arn73a, Arn73b, Arn82] concerning the large theory about the Krull dimension under power series extensions. Notation. Throughout the whole thesis R denotes a commutative unitary non-zero ring of finite Krull dimension. Further the set of natural numbers N is supposed to contain zero. If a ring does not contain any proper zero-divisors we call it a domain. Moreover if not otherwise specified X, X 1,..., X n denote indeterminates and p R is assumed to be a prime ideal. The set of all prime ideals of R is denoted by Spec(R). When passing to a residue class ring R/a for some ideal a we denote with x the residue class of x for any element x R and similar by b we mean the image of the ideal b. II Primary Decomposition In this section we will briefly develop the theory of primary decomposition which is a basic tool studying ideals. Mainly we follow [ZS75], but also [AM69]. 2.1 Primary ideals We start with some basic facts about the radical of an ideal. Definition 2.1. Let a R be an ideal. Then we call a := the radical of a. { x R } n N: x n a Lemma 2.2. (i) If a b for some ideals a, b R, then a b. (ii) For any prime ideal p Spec(R) the equation p n = p holds for all n 1. (iii) Let I be a finite set and for all i I let q i R be an ideal. Then we have i I q i = i I qi. (iv) Let a R be an ideal. Then a is the intersection of all prime ideals p Spec(R) containing a. Proof. Those are well-known properties. See [AM69, 1.13f.] among others. 2

5 II Primary Decomposition Example 2.3. (i) For the nilradical of R we have N(R) := (0) = p Spec(R) p. (ii) For a prime number p Z we see n N (pn ) = (0) = (0) (p) = n N (p n ). This shows that we cannot omit the finiteness of I in Lemma 2.2 (iii). In some sence the concept of prime ideals corresponds to those of prime elements. Now we introduce ideals that in a similar way correspond to powers of prime elements. Definition 2.4. Let q R be an ideal such that for x, y R with xy q and x / q we have y q. Then q is called primary ideal of R. In this case one says that q is p-primary if p = q. Remark: In Proposition 2.8 we will see that for a primary ideal q the radical q R is a prime ideal. Remark 2.5. An ideal q R is primary if and only if every zero-divisor of R/q is nilpotent. Example 2.6. (i) Every prime ideal p is p-primary. (ii) The primary ideals of Z are (0) and (p n ) where p is a prime number and n 1. (iii) Consider the ideal q := (X, Y 2 ) R := K[X, Y ] where K denotes a field. Since we have R/q = K[Y ]/(Y 2 ) all zero-divisors of R/q are multiples of Y hence nilpotent. So q R is primary. But we have p := q = (X, Y ) and p 2 q p. This shows that a primary ideal need not be a power of a prime ideal. (iv) Conversely not every power of a prime ideal is primary: Let R := K[X, Y, Z]/(XY Z 2 ) and p := ( X, Z). Then p is prime since R/p = K[Y ], but X Ȳ = Z 2 p 2 and X / p 2, Ȳ / p = p. So p 2 is not primary although it is a power of a prime ideal. We have seen that in general the powers of prime ideals need not be primary. However: Lemma 2.7. If p := q MaxSpec(R) then q is p-primary. maximal ideals are primary. In particular the powers of Proof. The second assertion follows from the first since for any n 1 and any maximal ideal m MaxSpec(R) we have m n 2.2 = m. For the first assertion observe that p is the intersection of all prime ideals containing q, so the image of p in R/q is the intersection of all primes of R/q, so it is the nilradical N := N(R/q). As p R is maximal, the nilradical N = p R/q is maximal. Now for any prime ideal a R/q we have N a and since N is maximal it follows that N = a yielding N is the only maximal ideal. An arbitrary zero-divisor x R/q is contained in some maximal ideal hence in N. Therefore it is nilpotent and the proof is complete. 3

6 II Primary Decomposition In Example 2.6 we have seen that for an ideal q R to be primary it is not sufficient that q R is a prime ideal, but the following proposition states that it is necessary. Proposition 2.8. Let q be p-primary. Then p is the smallest prime ideal containing q. Proof. By the fact that p = q we only have to show that p is prime. Let x, y R such that xy p, x / p. So there is a some n 1 such that (xy) n = x n y n q. Since x / p we have x n / q so since q is primary there is m 1 such that (y n ) m = y nm q. Thus y q = p. 2.2 The Lasker-Noether decomposition theorem Definition 2.9. (i) An ideal a R is said to be irreducible if a = b c for some ideals b, c R implies a = b or a = c. (ii) For two ideals a, b R the ideal (a : b) := { x R } xb a R is called the ideal quotient. Remark: It is clear that the ideal quotient is indeed an ideal of R. Example Every prime ideal p R is irreducible while a primary ideal need not be. Proof. The first assertion is clear (see [AM69, 1.11]). Now for a field K let m := (X, Y ) MaxSpec(K[X, Y ]). Then by Lemma 2.7 the ideal m 2 = (X 2, XY, Y 2 ) is primary, but m 2 = m 2 + (X) m 2 + (Y ). Definition A representation a = i q i of an ideal a R as a finite intersection of primary ideals q i R is called a primary decomposition. It is said to be irredundant or reduced if the following conditions hold: (i) q i q j i j j. (ii) q j q i i j. The main purpose of this section is to prove that for any ideal of a Noetherian ring there is an irredundant primary decomposition. First we notice: Lemma Let R be Noetherian and a R an ideal. Then there exist finitely many irreducible ideals q i R such that a = i q i. Proof. Suppose the set F of all ideals failing the condition is non-empty. Therefore F contains a maximal element m. In particular m is not irreducible, so there are ideals b, c R such that m = b c and m b, c. But since m is maximal in F it follows that b, c are finite intersections of irreducible ideals thus m is. 4

7 II Primary Decomposition Lemma In a Noetherian ring R every irreducible ideal is primary. Proof. Let q be non-primary. Then there exist elements a, b R such that ab q, a / q and b / p := q. Since R is Noetherian the increasing sequence {( q: (b s ) )} s N becomes stationary. So there is some n 1 such that ( ) ( q: (b n ) ) = ( q: (b n+1 ) ). Claim: q = ( q + (b n ) ) ( q + (a) ) So the lemma follows since b n, a / q, hence q is not irreducible. Proof of the claim: Since is clear, it suffices to prove the reverse: Let x ( q + (b n ) ) ( q + (a) ). Then there are u, v q, y, z R such that ( ) x = u + yb n = v + za. So we have: bx ( ) = bv + zab q = q bx ( ) = bu + yb n+1. = yb n+1 q = y ( q: (b n+1 ) ) ( ) = ( q: (b n ) ) = yb n q. = x = u + yb n q and the claim follows. Now we are able to get to the main result of this section: Theorem 2.14 (Lasker-Noether decomposition theorem). Let R be Noetherian. Then every ideal a R admits an irredundant finite primary decomposition. Proof. By Lemma 2.12 and Lemma 2.13 we have a = i q i for finitely many primary ideals q i. So we just have to make sure that this representation is irredundant or that we can make it irredundant. First if there is some ideal q j such that i j q i q j we omit q j in the representation. It is clear that we still have a = i j q i. If on the other hand there are i j such that qi = q j then we replace q i and q j by q ij := q i q j. So we just have to make sure that q ij is primary. More generally we show: If q 1,..., q n are p-primary then q := n i=1 q n is p-primary. For this we first observe that n i=1 q i 2.2 = n i=1 qi = i p = p. Now let x, y R such that xy q, x / q. So there is some 1 j n such that x / q j but q j is p-primary and xy q j, so it follows y q j = p. Definition Let a R be an ideal admitting a primary decomposition a = i q i. The radicals p i := q i are called the associated primes or the prime ideals belonging to a or simply the primes of a. If p i is minimal in the set of associated primes, then p i is said to be an isolated prime of a. Remark. In general the primary decomposition is not unique. However throughout this thesis we are only interested in the isolated primes of some ideals and it can be shown that these are uniquely determined. See [ZS75, Cap. IV, 5] for a proof. 5

8 III Krull s principal ideal theorem The following theorem will be used repeatedly: Theorem Let {p i } i be the collection of associated primes of an ideal a R. Then a prime ideal p Spec(R) contains a if and only if it contains some p i. In particular the isolated primes of a are precisely the prime ideals that are minimal among all prime ideals that contain a. Proof. ( ) Clear since a p i p. ( ) We have p a = i q i. Then by [AM69, 1.11] there exists i such that q i p and it follows that p i = q i p = p by Lemma 2.2. Corollary Let a R be an ideal and p Spec(R). (i) Let furthermore b a be an ideal and denote by ā, p the image of a, p respectively in R/b. Then p is an isolated prime of ā iff p is an isolated prime of a. (ii) Let S R be a multiplicatively closed subset of R such that p does not meet S. Then S 1 p is an isolated prime of S 1 a if and only if p is an isolated prime of a. Proof. Follows straight from the definitions, Theorem 2.16 and the ideal theory of the localization and residue class rings (see [AM69, 1.1, 3.11]). III Krull s principal ideal theorem Equipped with the theory of primary decomposition and isolated primes we now formulate Krull s principal ideal theorem together with some generalizations. But first we need a bit preparatory work. 3.1 The n-th symbolic powers Notation. Let S R be multiplicatively closed and ϕ: R S 1 R. (i) For an ideal a R let S 1 a := ϕ (a) denote the image of a in S 1 R. (ii) For an ideal b S 1 R we write b R := ϕ (b) for the pre-image of b in R. (iii) For a prime ideal p Spec(R) we write ar p := (R \ p) 1 a. Proposition 3.1. Let (R, m) be a Noetherian local ring and a R an ideal admitting m as an isolated prime. Then a is m-primary and m is the only prime ideal containing a. Proof. Let p Spec(R) be any prime ideal containing a. Since R is local we have p m. But m is an isolated prime of a yielding p = m (by Theorem 2.16). Now the radical of a is the intersection of all prime ideals containing a, hence m. So by Lemma 2.7 a is m-primary and we are through. 6

9 III Krull s principal ideal theorem We have seen in Example 2.6 that in general powers of prime ideals are not necessarily primary. However in the following we construct a certain primary ideal which we can associate with the power of a given prime ideal. Definition 3.2. For a prime ideal p Spec(R) we define p (n) := p n R p R to be the n-th symbolic power of p. Lemma 3.3. The ideal p (n) R is p-primary. Proof. First pr p is maximal in R p, so by Lemma 2.7 we have p n R p = (pr p ) n is pr p - primary. Then more generally: For any ring homomorphism ϕ: R S the pre-image ϕ (q) R is primary for any primary ideal q S. This is clear since R/ϕ (q) is isomorphic to some subring of B/q (now see Remark 2.5). Finally one immediately verifies p (n) = p. Corollary 3.4. If R is a Noetherian domain and p (0) then p (n+1) p (n) n 1. Proof. Ideal theory provides p (j) R p = p j R p for all j 1, so if p (n+1) = p (n) for some n 1, then p n+1 R p = p n R p in the local Noetherian ring (R p, pr p ) yielding that R p is an Artinian ([AM69, 8.6]) domain, hence it is a field. Contradiction to p (0). Example 3.5. In the situation of 3.4 all conditions are necessary: (i) Let R := Z/4Z and (0) (2) =: p. Then clearly p 2 = (0) = p n for all n 2. (ii) Consider the integral ring extension Z R := Z C and some (0) p Spec(R) (then p is necessarily maximal). Observe the local ring (R p, pr p ): For any α pr p we evidently have α R p. Now α is not a unit in R p otherwise α would be. This states α pr p and it follows pr p α = α α (pr p ) 2 hence pr p = (pr p ) 2. As R p is a domain, this additionally shows that R p is not Noetherian (nor is R). 3.2 Krull s principal ideal theorem Definition 3.6. Let p Spec(R). By ht(p) := dim(r p ) we denote the height of p and dp(p) := dim(r/p) is called the depth of p. Remark 3.7. (i) We have that ht(p) is the supremum of all n N such that there exists a chain of prime ideals p 0 p 1... p n = p ending with p and analogously dp(p) is the supremum of the lengths of chains of prime ideals starting with p. (ii) We have ht(p) + dp(p) dim(r), but in general there is no equation: Take any ring having two maximal ideals of different height. An example is given in

10 III Krull s principal ideal theorem (iii) Whenever we want to show an assertion about the height of a prime ideal p R then we may assume that R is local and p its unique maximal ideal. This follows from the ideal theory of the localization R R p. Now we can prove the first version of Krull s principal ideal theorem: Theorem 3.8 (Krull s principal ideal theorem). Let R be a Noetherian domain and (0) (a) R a principal ideal. Then every isolated prime of (a) is of height one. Proof. Let p be an isolated prime of (a). By Remark 3.7 we may assume p is the only maximal ideal. Now suppose there is some prime ideal (0) v p. Set q n := v (n) and obtain by Lemma 3.3 and Corollary 3.4 an infinitely decreasing chain of primary ideals ( ) q 1... q n q n+1... So we get a decreasing chain ( q n + (a) ) of ideals containing (a). n Since p is maximal and an isolated prime of (a) Proposition 3.1 provides that (a) is p-primary and p is the only prime ideal containing (a). ( ) It follows that R/(a) has only one prime ideal namely the image of p in R/(a). Therefore we have dim ( R/(a) ) = 0, so R/(a) is Artinian and thus fulfills the descending chain condition. So the sequence q n + (a) and therefore also q n + (a) become stationary. So there is n 1 such that q n + (a) = q n+1 + (a) yielding q n q n+1 + (a). Thus for an arbitrary chosen x q n there exist y q n+1 q n, z R such that x = y + za = za = x y q n (+). Evidently a / v, otherwise we get (a) v p in contradiction to ( ). So since q n is primary and a / q n (+) implies z q n. We get: q n q n+1 + aq n and since the other inclusion is obvious, it follows ( ) q n = q n+1 + aq n. Now pass to R := R/q n+1 in which we denote by q the image of q n. The equation ( ) implies: q = ā q = q = (ā) q. Now q is finitely generated as R is Noetherian. Furthermore we have Jac(R ) = p since p is the only prime ideal of R. This gives (ā) p = Jac(R ), so Nakayama s lemma [AM69, 2.6] supplies q = ( 0). So q n = q n+1. to ( ). Having proved the original principal ideal theorem we proceed with some generalizations. Proposition 3.9. A prime ideal p R of a Noetherian ring R has height ht(p) = 0 iff it is an isolated prime of (0). Proof. Clear by Theorem Corollary Let R be a Noetherian ring and p an isolated prime of a principal ideal (a) R. Then ht(p) 1. Note that R needs not be a domain. Proof. Let p p 1 p 2 be a chain of prime ideals of R. Then pass to the Noetherian 8

11 III Krull s principal ideal theorem domain R/p 2 and get p p 1 ( 0). Corollary 2.17 gives that p is an isolated prime of (ā). If (ā) = ( 0) Proposition 3.9 provides ht( p) = 0 and thus p = ( 0) (so p = p 2 ). If (ā) ( 0) it follows by Theorem 3.8 that ht( p) = 1 yielding p 1 = ( 0) or p 1 = p. A further generalization can be made by varying the number of generators of the principal ideal. For this we need the following preparatory lemma: Lemma Let R be Noetherian, p 0... p m a chain of prime ideals and let q i (for i I) be finitely many prime ideals not containing p 0. Then there are prime ideals p 0 p 1... p m 1 p m such that none of the p j are contained in any of the q i. Proof. It suffices to prove the claim for m = 2. Since p 0 q i i I we have: p 0 i q i p 2 (see [AM69, 1.11]). So there exists an element x p 0 : x / i q i p 2. Set a := p 2 + (x) and denote by F := { p i } the set of associated primes of a (existence by Theorem 2.14). Since a p 0 there is by Theorem 2.16 an isolated prime p 1 F such that p 1 p 0. Further as x a p 1 we have by construction: p 1 q i i I and p 1 p 2. So the only thing left to show is that p 0 p 1 : Observe the Noetherian domain R/p 2 in which ā is obviously a principal ideal, generated by x. By Corollary 2.17 p 1 is an isolated prime of ā. Thus if we had p 0 = p 1 it would follow that p 0 is an isolated prime of ā hence by Corollary 3.10 we have ht( p 0 ) 1, but we have by assumption p 0 p 1 p 2 yielding p 0 p 1 p 2 = ( 0). Eventually we obtain the final version of Krull s principal ideal theorem: Theorem 3.12 (Krull s principal ideal theorem, generalized). Let R be a Noetherian ring, a R an ideal generated by r elements and p an isolated prime of a. Then ht(p) r. Proof. Proof by induction on r. (r = 0) The only ideal generated by is (0) and thus ht(p) = 0 by Proposition 3.9. (r 1 r) Let a = (x 1,..., x r ) and b := (x 1,..., x r 1 ). If p is among the isolated primes {v i } of b we are through by induction hypothesis. So assume p is not contained in any v i. (In fact: b a p and v i are the minimal primes containing b.) Now let m := ht(p) and assume m 0. So by definition there is a chain of prime ideals ( ) p =: p 0 p 1... p m 1 p m. Since p 0 is not contained in any of the v i we can by Lemma 3.11 assume that p m 1 v i Now pass to the ring R/b in which ā is a principal ideal, generated by the residue class of x r. By Corollary 2.17 p 0 is an isolated prime of ā hence Corollary 3.10 supplies ht( p 0 ) 1. Recall that p m 1 v i i. Thus p m 1 + b v i while the latter are the isolated primes of ( 0) ( ). i. 9

12 III Krull s principal ideal theorem Claim: p 0 is an isolated prime of p m 1 + b. Proof : Obviously we have p m 1 + b p 0. Then let q be an isolated prime of p m 1 + b contained in p 0 (Theorem 2.16). It follows ht( q) 1 since otherwise q would be an isolated prime ideal of ( 0) (Proposition 3.9) contradicting ( ). So we have q p 0 while ht( q) 1 and ht( p 0 ) 1, so q = p 0. This proves the claim. It follows that p 0 is an isolated prime of p m 1 + b (Corollary 2.17) and therefore p 0 /p m 1 is an isolated prime of (p m 1 + b)/p m 1 which is generated by r 1 elements in R/p m 1. The induction hypothesis implies that ht(p 0 /p m 1 ) r 1. ( ) Moreover the order-preserving one-to-one correspondence of ideals (see [AM69, 1.1]) supplies ht(p 0 /p m 1 ) = m 1 (see ( )). Therefore by ( ) m 1 r 1 hence m r. Corollary (i) Let (R, m) be a local Noetherian ring and n := dim(r). Then m cannot be generated by less than n elements. Remark: If it can be generated by exactly n elements the ring is called regular. (ii) We have ht(p) < for any prime ideal p Spec(R) and thus Noetherian rings satisfies the descending chain condition for prime ideals. Note that dp(p) may be infinite, even in the Noetherian case! An example is provided in [Nag62, Appendix]. The following theorem is in some sence the converse of the principal ideal theorem 3.12: Theorem Let R be Noetherian and p Spec(R) with ht(p) = h. Then there is an ideal generated by h elements admitting p as an isolated prime. Proof. Proof by induction on i < h: Let i < h and a 1,..., a i p such that every isolated prime p j of (a 1,..., a i ) is of height i. Since ht(p j ) = i < h = ht(p) we have p p j j. It follows that there is an element a i+1 p\ j p j. Now an arbitrary chosen isolated prime v of (a 1,..., a i+1 ) contains some p j by Theorem 2.16 and this inclusion is proper since a i+1 v \ p j. Remember that ht(p j ) = i, so we have ht(v) i + 1, but v is an isolated prime of (a 1,..., a i+1 ) hence by Theorem 3.12 we get ht(v) = i + 1. So successively this procedure supplies elements a 1,..., a h p such that the isolated primes v j of (a 1,..., a h ) are of height h. Again Theorem 2.16 yields that p contains some v j and since both ideals are of height h we have equality and the claim follows. Example Consider R := C[X, Y ]/(Y 2 X 3 ) = C[t 2, t 3 ] and m := (t 2, t 3 ). Theorem 3.8 ht ( (Y 2 X 3 ) ) = 1 and as we will see in section V dim ( C[X, Y ] ) = 2, so dim(r) = 1 yielding ht(m) = 1. Thus ht(mr p ) = 1, but mr m cannot be generated by less than two elements. Especially R m is not regular. Cf. Corollary Furthermore we see: In order that a local Noetherian ring (R, m) is at most one-dimensional, it is sufficient but not necessary that m is principal. By 10

13 IV A first boundary for dim(r[x]) IV A first boundary for dim(r[x]) Now the preparatory work is done and we can start studying the Krull dimension under certain ring extensions. We start with R R[X] where X denotes an indeterminate. The purpose of this section is to establish a first boundary for dim(r[x]) in terms of dim(r) as it is provided by [Sei53]. But first we need to take a look on the prime spectrum of R[X]. Notation. Throughout this section let ι: R R[X], p Spec(R) and P Spec(R[X]). Further we write pr[x] := ι (p) for the image of p under ι and analogously P R := ι (P) denotes the pre-image of P under ι. Finally we define { p[x] := a ix i R[X] i } a i p i R[X]. Lemma 4.1. (i) We have p[x] = pr[x] and there is an isomorphism of rings R[X]/pR[X] (R/p)[X]. (ii) For a multiplicative system S R we get (S 1 R)[X] = S 1 (R[X]). Proof. (i) In the first assertion p[x] pr[x] is clear and so we just have to show : Let α pr[x]. Then α = i α iβ i with α i p, β i R[X], thus β i = j γ ijx j, so α = i,j α iγ ij X j p[x]. }{{} p For the second assertion observe η : R[X] (R/p)[X] a i X i ā i X i i i Apply the fundamental theorem on homomorphisms observing ker(η) = p[x] = pr[x]. (ii) Observe R S 1 R η 1 η 2 R[X] with η 1 : S 1 R S 1 (R[X]), η 2 : R[X] (S 1 R)[X], α α a i X i i i a i 1 Xi (S 1 R)[X] g f S 1 (R[X]) We set f(x) := X 1 and by universal properties the existences are clear and it is easily verified that g f = id, f g = id. An immediate consequence of Lemma 4.1 (ii) is: 11

14 IV A first boundary for dim(r[x]) Corollary 4.2. Let R be a domain. Then Lemma 4.1 (ii) in particular implies that ( R \ (0) ) 1 ( R[X] ) = ( (R \ (0)) 1 R ) [X] = Quot(R)[X] Therefore there is an order-preserving one-to-one correspondence Spec ( Quot(R)[X] ) { p Spec ( R[X] ) } p R = (0). In section III we used that for the ring homomorphism ϕ: R S 1 R we have ϕ (ϕ (q)) = q and ϕ (ϕ (p)) p. To the contrary we now have: Corollary 4.3. The following assertions are valid: (i) ι (ι (p)) = p and ι (ι (P)) P. (ii) ι (Spec(R[X])) = Spec(R) and ι (Spec(R)) Spec(R[X]). (iii) p[x] + (X) R[X] is a prime ideal which properly contains p[x]. Proof. (i) follows that in general there is no equality in the latter assertion in (i). In particular it The first assertion is clear since ι (p) = p[x] and the second assertion is true for any ring homomorphism and any ideal. (ii) In the first statement is true for any ring extension and for equality use the first assertion in (i). For the second statement observe that for a prime ideal p Spec(R) the ring (R/p)[X] is a domain. Now use Lemma 4.1 (i). The remaining follows from (iii). (iii) X / p[x] since 1 / p. { Thus p[x] p[x] + (X). } Claim: P := p[x] + (X) = f R[X] ev 0 (f) p R[X]. Then P indeed is a prime ideal: Let a, b R[X] such that ab P, a / P. This gives p ev 0 (ab) = ev 0 (a) ev 0 (b) while ev 0 (a) / p, but p is prime yielding ev 0 (b) p, hence b P. Proof of the claim: For an element x P there are f p[x], g R[X] such that x = f + gx. Then ev 0 (x) = ev 0 (f + gx) = ev 0 (f) + 0 ev 0 (g) = ev 0 (f) p. Conversely let x = i a ix i R[X] such that ev 0 (x) p. Then x = a 0 +X ( n i=1 a ix i 1) p[x] + (X). Finally for the last assertion observe that ( p[x] + (X) ) R = p. For an integral ring extension R S and prime ideals q 1 q 2 in S the equation q 1 R = q 2 R implies q 1 = q 2 (see [AM69, 5.9]). This assertion is wrong in the situation R R[X] because we have seen in Corollary 4.3 that p[x] p[x] + (X) but evidently both ideals are lying over p in R. So there may exist a chain of prime ideals in R[X] contracting to the same prime ideal in R. However the length of such chains is restricted as the following theorem states: 12

15 IV A first boundary for dim(r[x]) Theorem 4.4. Let R be a ring and P 1 P 2 P 3 be a chain of three distinct prime ideals in R[X]. Then P 1 R P 3 R. Proof. Let p := P 1 R. Then we have p[x] P 1. Assume that P 3 R = p. Factoring out p of R and p[x] of R[X] we obtain a chain of prime ideals P 1 P 2 P 3 in the domain R[X]/p[X] = (R/p)[X] with P 1 (R/p) = P 3 (R/p) = ( 0). 4.1 So Corollary 4.2 provides a chain of prime ideals Q 1 Q 2 Q 3 in Quot(R/p)[X] yielding dim ( Quot(R/p)[X] ) 2. (Note that Quot(R/p)[X] is a PID) Corollary 4.5. Let p R be a prime ideal of height ht(p) = 0. Then the lifted prime ideal p[x] has height ht(p[x]) = 0. Proof. Assume there is a prime ideal P Spec(R[X]) such that P p[x]. Then we have P R p[x] R = p yielding P R = p since ht(p) = 0. On the other hand the prime ideal p[x]+(x) also contracts to p contradicting Theorem 4.4. Finally we are able to establish a lower and an upper bound to the Krull dimension of R[X] using Corollary 4.3 and Theorem 4.4: Theorem 4.6. Let n := dim(r). Then n + 1 dim(r[x]) 2n + 1. Proof. By definition there is a chain of prime ideals p 0 p 1... p n in R. Now consider the chain p 0 [X]... p n [X] p n [X] + (X). That this is a strictly ascending chain of prime ideals in R[X] follows from Corollary 4.3. Thus we obtain dim(r[x]) n + 1 and the only thing left to prove is dim(r[x]) 2n + 1: So let P 0 P 1... P m be a chain of prime ideals in R[X]. Then an immediate consequence of Theorem 4.4 is that P 0 R P 2 R P 4 R... P 2 m 2 R is a chain of length m 2 of distinct prime ideals in R. This gives m 2 n hence m 2n + 1 and since we started with an arbitrary chain of prime ideals, it follows dim(r[x]) 2n + 1 as desired. Remark 4.7. Since dim(q[x]) = 1 = dim(q) + 1 is is clear that the lower bound is optimal. Moreover the upper bound cannot be strengthened either: Seidenberg showed in [Sei54, Theorem 3] that for any natural numbers n and m with n + 1 m 2n + 1 there exists a ring R such that dim(r) = n and dim(r[x]) = m. We will discuss this a bit more detailed in section VII. Using induction we can generalize Theorem 4.6: Corollary 4.8. dim(r) + m dim ( R[X 1,..., X m ] ) 2 m( dim(r) + 1 ) 1 m N. 13

16 V Dimension theory of K[X 1,..., X n ] V Dimension theory of K[X 1,..., X n ] In the previous section we have seen that for an arbitrary ring R with dimension dim(r) = n we have n+1 dim(r[x]) 2n+1. Moreover we have mentioned that in general this result cannot be improved. However there are classes of rings for which there can be made stronger statements. In this section we start with a field K: We know that dim(k) = 0 and since K[X] is a principal ideal domain we also know dim(k[x]) = 1. So the dimension rises from 0 to 1 under polynomial extension. This raises the question whether dim(k[x 1,..., X n ]) = n. In this section we will give an elementary proof to verify this. As we will see this nice behavior is true for any zero-dimensional ring like Artinian rings. Lemma 5.1. Let P Spec(R[X]) and p := P R Spec(R). If p[x] P then we have ht(p) = ht(p[x]) + 1. Proof. Proof by induction on ht(p). (ht(p) = 0) First by Corollary 4.5 we have ht(p[x]) = 0, so we need to show that ht(p) = 1. Assume ht(p) > 1 then there are prime ideals ( ) P P 1 P 2. Now clearly p 2 := P 2 R P R = p, but ht(p) = 0 yielding p 2 = p and so all prime ideals in ( ) have the same contraction in R contradicting Theorem 4.4. (m := ht(p) 0) initial step Since ht(p) ht(p[x]) + 1 is clear, it suffices to prove that ht(q) ht(p[x]) for any prime ideal Q P. So let Q P be any prime ideal and q := Q R. As above we have q p. If in the first case q = p then p[x] = q[x] 4.3 Q P yielding p[x] = Q by Theorem 4.4 and therefore clearly ht(q) ht(p[x]). If on the other hand q p, then q[x] p[x] and thus ht(q[x]) + 1 ht(p[x]). Also ht(q) < ht(p) implies by induction hypothesis that ht(q) = ht(q[x]) + 1 ht(p[x]). Remark. One might get the impression that Lemma 5.1 can be proved easier by just assuming ht(p) ht(p[x])+2 and argue that then there is a chain of prime ideals p[x] P P all contracting to the same prime ideal in R. But it is not yet clear that all maximal chains of primes ending with P have the same length. In fact there are (even Noetherian) rings with prime ideals like this: P 2 P 3 P 4... P 1 P 2 P 3 P 4 where both chains cannot be refined ( [Kun80, p. 55]). In this situation there is no prime ideal between P 3 and P 4, but nevertheless we could have ht(p 4 ) ht(p 3 ) + 2 (if the pictured chain is the longest chain of prime ideals ending with P 3 ). Corollary 5.2. Let P R[X 1 ] be any prime ideal such that P (p := P R)[X 1 ]. Then 14

17 V Dimension theory of K[X 1,..., X n ] in the ring R[X 1,..., X n ] we have ht(p[x 2,..., X n ]) = ht(p[x 1,..., X n ]) + 1. Proof. First we have P[X 2,..., X n ] R[X 2,..., X n ] = (P R)[X 2,..., X n ] = p[x 2,..., X n ]. Now observe that P p[x 1 ] implies P[X 2,..., X n ] p[x 1,..., X n ] and we are through by Lemma 5.1 applied to the ring R[X 2,..., X n ][X 1 ]. Theorem 5.3. Let P R[X 1,..., X n ] be some prime ideal and p := P R Spec(R). Then ht(p) ht(p[x 1,..., X n ]) + n. Proof. Proof by induction on n. (n = 1) This is Lemma 5.1. (n 1 n) Let q := P R[X 1 ]. If we have q = p[x 1 ] then we are through by induction hypothesis applied to the ring R[X 1 ][X 2,..., X n ]. If in the other case we have p[x 1 ] q then by Corollary 5.2 we get ht(q[x 2,..., X n ]) = ht(p[x 1,..., X n ]) + 1. Again the induction hypothesis on R[X 1 ][X 2,..., X n ] gives ht(p) (n 1) + ht(q[x 2,..., X n ]). Together we have thus shown ht(p) ht(q[x 2,..., X n ]) + (n 1) = ht(p[x 1,..., X n ]) + n. This leads us to the main theorem of this section: Theorem 5.4. Let K be a field. Then dim(k[x 1,..., X n ]) = n. Proof. Consider the chain of prime ideals of K[X 1,..., X n ]: (0) (X 1 ) (X 1, X 2 )... (X 1,..., X n ). First it is clear that all these ideals are prime since for each 1 k n we have K[X 1,..., X n ]/(X 1,..., X k ) = K[X k+1,..., X n ]. This yields dim(k[x 1,..., X n ]) n. Conversely let P Spec(K[X 1,..., X n ]) be some prime ideal. Then P K Spec(K) = {(0)}. Theorem 5.3 implies ht(p) ht ( (0) ) + n = n. Hence dim(r[x 1,..., X n ]) n. Remark. In the proof of Theorem 5.4 we see that the result of Theorem 5.3 cannot be improved: Choose P := (X 1,..., X n ) and note ht(p) = n while p = (0). To show dim(k[x 1,..., X n ]) n we could alternatively refer to Corollary 4.8, but the way we proved the theorem above is the way [BMRH73] provided the proof and they point out that this proof is somehow special in respect of its elementary facts that are used. The authors emphasize that the proof is no deeper than the basic fact that there cannot be three distinct prime ideals in R[X] all of which contracting to the same prime ideal in R. The proof of this in turn bases upon nothing more than the simple fact that K[X] is a PID if K is a field. 15

18 VI Dimension theory of Noetherian rings As announced in the introduction to this section this fact of dimension theory extends to every zero-dimensional ring (in particular to Artinian rings): Theorem 5.5. Let R be zero-dimensional. Then dim(r[x 1,..., X n ]) = n. Proof. By Corollary 4.8 we have dim(r[x 1,..., X n ]) n. To see the reverse inequality let P Spec(R[X 1,..., X n ]) be an arbitrary prime ideal. It follows that P R Spec(R), but since dim(r) = 0 we have ht(p := P R) = 0. Corollary 4.5 gives ht(p[x 1,..., X n ]) = 0 as well, so like in the proof of Theorem 5.4 we conclude dim(r[x 1,..., X n ]) 0 + n = n. Remark 5.6. Note that this is (in the case n > 1) a real improvement of the general boundary given in Corollary 4.8 where we stated that dim ( R[X 1,..., X n ] ) 2 n 1 (if dim(r) = 0). In particular Seidenberg s surprising result mentioned in Remark 4.7 does not extend to multivariate polynomial rings and the boundary in Corollary 4.8. VI Dimension theory of Noetherian rings 6.1 Dimension under polynomial extensions In section V we have seen that dim(r[x 1,..., X n ]) = n = dim(r) + n for any Noetherian zero-dimensional ring R (those are Artinian). So it seems reasonable to ask whether this behavior extends to all Noetherian rings. Indeed in this subsection we will verify this. Throughout the whole section R denotes a Noetherian ring. Recall that we then always have a primary decomposition. Theorem 6.1. Let R be Noetherian, then dim ( R[X 1,..., X n ] ) = dim(r) + n. Before we prove this theorem we need the following lemma: Lemma 6.2. Let a R be some ideal and p Spec(R) be an isolated prime of a. Then p[x] is an isolated prime of a[x]. Remark. Since the proof of p[x] = pr[x] in Lemma 4.1 does not use the fact that p is prime, it is also true for any ideal a R. Proof (6.2). Clearly a p implies a[x] p[x]. Now by Theorem 2.16 p[x] contains some isolated prime P of a[x]. Then as a[x] P p[x] we have a[x] R }{{} =a P R }{{} =:q p[x] R. }{{} =p But q R is prime while p is an isolated prime of a. This yields q = p and therefore we get P p[x] = q[x] P hence p[x] = P. Now we can proof the main result of this section: 16

19 VI Dimension theory of Noetherian rings Proof (Theorem 6.1). By Hilbert s basis theorem (see [AM69, 7.5]) R[X 1,..., X n 1 ] is Noetherian if R is, so it suffices to prove that dim(r[x]) = dim(r) + 1. Moreover by Theorem 4.6 we just have to verify that dim(r[x]) dim(r) + 1: For this let P Spec(R[X]) be an arbitrary prime ideal and p := P R. Further let r := ht(p). Now Theorem 3.14 supplies elements a 1,..., a r R such that p is an isolated prime of a := (a 1,..., a r ). Then in Lemma 6.2 we have shown that p[x] is an isolated prime of a[x] = ar[x]. But obviously ar[x] is generated by r elements so Theorem 3.12 gives ht(p[x]) r = ht(p). Alltogether with Theorem 5.3 we obtain ht(p) ht(p[x]) + 1 ht(p) + 1 dim(r) + 1. Since P was an arbitrary prime ideal of R[X] the theorem follows. Example 6.3 (cf. Remark 3.7). Observe a discrete valuation ring R with uniformizing element π. Then by Theorem 6.1 dim(r[x]) = 2 as dim(r) = 1. Further for the prime ideal m := (πx 1) of R[X] we have ht(m) = 1 by Theorem 3.8 and dp(m) = 0 since R[X]/m = R [ 1 π ] = Quot(R). So we have ht(m) + dp(m) dim ( R[X] ) (even though R[X] is universally catenary; details omitted). Note that this provides an example of a ring with maximal ideals that have not the same height! 6.2 Non-transcendental extensions After this journey through the dimension theory of polynomial rings we make a little break and observe what happens to the dimension when we adjoin an element x that is not transcendental. The first class of such elements are the ones that are integral over R. It is shown in [AM69, 5] that for these elements we have dim(r) = dim(r[x]), so throughout the rest of this section we can restrict our study to elements that are neither transcendental nor integral over R. In particular we do not study rings that happen to be fields as any element that is not transcendental over a field is automatically integral over it. Furthermore we restrict our attention to the case where R is a domain, so in the following R always denotes a Noetherian domain with dimension at least one. We will see that transcendental elements are the only elements that raise the dimension: Lemma 6.4. Let x be non-transcendental over R. Then dim(r[x]) dim(r). Proof. Observe ev x : R[X] R[x], X x. Then we have R[X]/ ker(ev x ) = R[x], so ker(ev x ) is a prime ideal and by definition dim(r[x]) = dp(ker(ev x )). Now as mentioned in Remark 3.7 ht(ker(ev x )) + dp(ker(ev x )) dim(r[x]) and thus we obtain dim(r[x]) dim(r[x]) ht(ker(ev x )). By assumption we have ker(ev x ) (0) hence ht(ker(ev x )) 1. Together with dim(r[x]) = dim(r) + 1 the claim follows. This raises the question how badly the equation dim(r[x]) = dim(r) can fail if x is not integral. 17

20 VI Dimension theory of Noetherian rings Definition 6.5. For j N with j dim(r) let F j (R) := { p Spec(R) } ht(p) = j. We simply write F j if it is clear which domain R we are dealing with and by F j we mean the intersection of all prime ideals in the set F j. But before we give an answer to our question, we first treat another problem: We ask whether there are any ring extensions R R[x] with dim(r) < dim(r[x]) at all. In the following we give a necessary and sufficient condition and moreover we give an element z with dim(r[z]) < dim(r) if such an element exists: Lemma 6.6. There exists a domain R[x] with dim ( R[x] ) < j iff F j (R) (0). Proof. Suppose there is 0 z F j. Then R [ ] ( [ 1 z is a domain with dim R 1 ]) } z < j. To see this let S := {z n n N. Then R [ ] 1 z = S 1 R. Now choose an arbitrary prime ideal q Spec(S 1 R) and set p := q R. Remember that q = S 1 p and ht(p) = ht(q). Assume ht(p) j, then there is a prime ideal p 0 Spec(R) with ht(p 0 ) = j and p 0 p. Thus by definition of z we have z p yielding p S. Hence ht(q) = ht(p) < j. If on the other hand we have F j = (0) then we show that dim ( R[x] ) j for any domain R[x] (with x 0). By Theorem 6.1 this assertion is certainly true if x is transcendental, so suppose there { are a 0,...}, a m R with a m 0 and a m x m a 0 = 0. Set S := a n m n N. Now x is integral over S 1 R since we have x m + a m 1 a m x m a 0 a m = 0. Therefore it follows dim(s 1 R) = dim ( (S 1 R)[x] ) = dim ( S 1 (R[x]) ) ( ). By assumption there is a prime ideal p Spec(R) with ht(p) = j such that a m / p hence p S =. This gives j = ht(p) = ht(s 1 p) yielding dim(s 1 R) j and the claim follows since now we have j dim(s 1 R) ( ) = dim ( S 1 (R[x]) ) dim ( R[x] ). Corollary 6.7. There is a domain R R[x] with dim ( R[x] ) dim(r) if and only if Fdim(R) (0). Note that not all maximal ideals need be of height dim(r) as we have seen in Example 6.3. Example 6.8. (i) Let R be a domain with dim(r) = 1 and F 1 (0). Then there exists an element x such that R[x] is a field. Examples for such rings are domains with finitely many prime ideals. In particular any local one-dimensional domain will do, so Z (p) evidently is such a ring (and of [ ] course Z 1 (p) p = Q). But there are also rings with an infinite number of prime ideals and nevertheless fulfilling the conditions above: Let A := C[X] where X is an indeterminate and K := Quot(A) be the algebraic closure. Further let B := A K be the integral closure of A in K. Now the set 18

21 VI Dimension theory of Noetherian rings S := C[X] \ (X) A is multiplicatively closed and we obtain (as we will show) a ring R := S 1 B that has the desired properties. To the contrary an example where the dimension cannot decrease: (ii) Let R be Jacobson with dim(r) =: n and further assume that ht(m) = n for each maximal ideal m MaxSpec(R). Then dim ( R[x] ) = dim(r) for any ring extension R R[x] where x is not transcendental. This is true for any principal ideal domain with an infinite number of prime ideals like Z and K[X] where K denotes a field. In particular: Given a principal ideal domain the number of prime ideals suffices to distinguish how the dimension behaves. Proof (6.8 (i)). First A B is integral, hence dim(b) = dim(a) = 1, so dim(r) 1. Claim: For all (0) P Spec(B) we have P S = iff P A = (X). Proof : If P S = we have P A (X). Moreover P A (0) as P (0) and A B is integral. So (0) P A (X) implies P A = (X) since dim(a) = 1. Conversely if P A = (X), then clearly P S = by definition of S. claim By the claim it follows that (if there are height-one prime ideals in R) we have F 1 (R) (X) (0). So we just have to show that there are infinitely many height-one prime ideals in R which is (by ideal theory and the claim above) equivalent to showing that there are infinitely many prime ideals in B all lying over (X) in A: For this let n 2 be an integer, ζ 1,..., ζ n C be the n-th roots of unity and Y := n X + 1 B be some root of T n (X + 1) A[T ]. Then we obtain n ideals p 1,..., p n A[Y ] by setting p i := (X, Y ζ i ). Now consider: A[Y ]/p i = A[Y ]/(X, Y ζi ) = A[T ]/(X, T n (X + 1), T ζ i ) = C[X, T ]/(X, T n (X + 1), T ζ i ) = C[T ]/(T ζ i, T n 1) = C/(ζ n n 1) = C So we have p i Spec(A[Y ]) for all i. Clearly A[Y ] B is integral, so Spec(B) Spec(A[Y ]) is surjective providing P 1,..., P n Spec(B) such that P i A[Y ] = p i and thus P i A = p i A =: i Spec(A). Since p i (0) and A A[Y ] is integral we have i (0) for all i and obviously X i. This yields i = (X) as dim(a) = 1, so all-in-all we have n prime ideal P 1,..., P n Spec(B) with P i A = (X) for all i. Repeating this construction for all n 2 we obtain infinitely many such prime ideals and we are through. Proof (6.8 (ii)). Since all maximal ideals have height n, the intersection F n equals the Jacobson radical Jac(R) = MaxSpec(R), which in turn equals the nilradical N(R) since R is Jacobson. Finally R is a domain, hence (0) = N(R) = Jac(R) = F n. Evidently in a principal ideal domain R every maximal ideal m is of height ht(m) = 1 = dim(r), so the only thing to show for the latter assertion is that a principal ideal domain with infinitely many prime ideals is Jacobson. This is well-known (see [Sta13, Lemma 00G4]). 19

22 VI Dimension theory of Noetherian rings The previous results not only answer the question about the existence of ring extensions under which the dimension decreases, but also translates the question how much the dimension can decrease at all into an ideal-theoretic question about the intersection of prime ideals with lower height. The next result ensures that we have a sufficient supply of them: Lemma 6.9. Let p Spec(R) be a prime ideal with ht(p) 2. Then there are infinitely many height-one prime ideals contained in p. Proof. Assume q 1,..., q n Spec(R) are the only height-one prime ideals contained in p. Evidently we have p q i for each i, hence p i q i. Choose x p \ i q i and let p be some isolated prime of (x). Krull s principal ideal theorem 3.8 provides ht( p) = 1 and Theorem 2.16 supplies p p, but since x / q i we obviously have p q i for all i. Remark. The proof above does extend to rings that are not a domain (by factoring out a height-zero prime ideal contained in p). This shows that whenever a ring R has only finitely many prime ideals (in particular a finite ring) then dim(r) 1. Now we can give an answer to our question and we see that the dimension cannot plummet: Corollary We always have dim ( R[x] ) dim(r) 1. Proof. We prove F n 1 = (0) by induction on n := dim(r) 1. The case n = 1 is clear since (0) is the only prime ideal of height 0. Now let m Spec(R) with ht(m) = n > 1 and (0) = p 0... p n = m be a chain of prime ideals in R. By Lemma 6.9 there are infinitely many height-one prime ideals { p j } j J contained in p 2 and for each j J we have dim(r/ p j ) = n 1. We choose z F n 1 and an arbitrary j J. By ideal theory any prime ideal p of height n 2 in R/ p j is the image of some p Spec(R) with ht(p ) n 1. So by definition of z we have z p yielding z p. Since p was arbitrary chosen in the set of prime ideals of height n 2 we have z Fn 2 (R/ p j ), but this intersection is ( 0) by induction hypothesis. This gives z p j for any j J. Now assume z 0. Then for any j p j is an isolated prime of (z) (by Theorem 2.16 p j contains some isolated prime of (z) while these are of height one by Theorem 3.8, remember ht( p j ) = 1 as well) providing (z) has an infinitely number of isolated prime ideals. This contradicts the Lasker-Noether decomposition theorem Thus we have z = 0 and since z was arbitrary chosen it follows F n 1 = (0). Hence by Lemma 6.6 we are through. Corollary Let R be a domain and let x 1,..., x n be elements such that R[x 1,..., x n ] is a field. Then dim(r) n. Note that we do not have the reverse of the corollary above: Clearly dim(z) = 1 but there is no finite set of elements {x i } i I such that Z[x i ] i I is a field. Nor can we hope for any sort of induction argument (consider Z (p) [X]). 20

23 VII Prüfer domains and a note on Seidenberg s F -rings Summary For a Noetherian domain R with n := dim(r) and some x which is not transcendental over R we have either dim(r[x]) = n or dim(r[x]) = n 1. The latter equation holds if and only if the intersection of all height-n maximal ideals does not equal (0). VII Prüfer domains and a note on Seidenberg s F - rings 7.1 Dimension theory of Prüfer domains Let s leave the class of Noetherian rings for a while. We have established a general boundary for the dimension of R[X] and we asked whether there are classes of rings which behave like dim(r[x]) = dim(r) + 1. This question lead us to Noetherian rings, but actually those are not the only rings with this property. In this section we will see that semi-hereditary rings (in particular Prüfer domains) behave the very same way. Definition 7.1. (i) A domain R such that all principal ideals of R are totally ordered by inclusion is called a valuation domain. (ii) A prüfer domain R is a domain such that for any prime ideal p Spec(R) the ring Remark 7.2. R p is a valuation domain. There are various ways to define valuation and Prüfer domains. For example there are 22 equivalent conditions for a Prüfer domain given in [BG06, Theorem 1.1]. We defined the valuation and Prüfer domains by the property that we need in the following proof. The definition of a Prüfer domain looks similar to those of a Dedekind domain. In fact a domain is Dedekind if and only if it is a Noetherian Prüfer domain, so we might think of Prüfer domains as the non-noetherian generalization of Dedekind domains. Theorem 7.3. Let R be a Prüfer domain. Then dim(r[x 1,..., X n ]) = dim(r) + n. Proof. This proof resembles the one of Theorem 6.1, so for details we refer to section VI. It is enough to prove dim(r[x 1,..., X n ]) dim(r) + n and therefore it suffices to verify ht(m) dim(r) + n for any maximal ideal M MaxSpec(R[X 1,..., X n ]). But we have seen ht(m) ht ( (M R)[X 1,..., X n ] ) + n for any maximal ideal, so the assertion above is clear if we show ht(p[x 1,..., X n ]) ht(p) for any prime ideal p Spec(R). We have a one-to-one correspondence Spec ( (R \ p) 1 R[X 1,..., X n ] ) {P Spec(R[X 1,..., X n ]) } P R p, so clearly every chain of primes in R[X 1,..., X n ] descending from p[x 1,..., X n ] corresponds 21