Bachelorarbeit

Zur Erlangung des akademischen Grades Bachelor of Science

Dimension theory for commutative rings

Eingereicht von: Daniel Heiß Eingereicht bei: Prof. Dr. Niko Naumann

Universität Regensburg Fakultät für Mathematik

Ausgabetermin: 22.08.2013 (Semester 5) Abgabetermin: 25.09.2013 Contents

Contents

I Introduction 1

II Primary Decomposition 2 2.1 Primary ideals ...... 2 2.2 The Lasker-Noether decomposition theorem ...... 4

III Krull’s principal ideal theorem 6 3.1 The n-th symbolic powers ...... 6 3.2 Krull’s principal ideal theorem ...... 7

IV A ﬁrst boundary for dim(R[X]) 11

V Dimension theory of K[X1,...,Xn] 14

VI Dimension theory of Noetherian rings 16 6.1 Dimension under polynomial extensions ...... 16 6.2 Non-transcendental extensions ...... 17 Summary of non-transcendental extensions ...... 21

VII Prüfer domains and a note on Seidenberg’s F -rings 21 7.1 Dimension theory of Prüfer domains ...... 21 7.2 A note on F -rings ...... 22 Summary of polynomial extensions ...... 27

VIII A journey to the power series ring 27 Summary and comparison with polynomial extensions ...... 29

References iii

ii 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 n vectors that are needed to generate the vector space. For the euclidean space R this coin- cides 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 log(3) dimension 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 Z0 ( Z1 ( ... ( Zn 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 km the chain of irreducible closed subsets of Y correspond by Hilbert’s Nullstellensatz to a chain of prime ideals p0 ( p1 ( ... ( pn of k[X1,...,Xm] 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[X1,...,Xm]/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[X1,...,Xm]/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 princi- pally [BMRH73] and we will give an elementary proof that dim k[X1,...,Xn] = n for any ﬁeld k or more generally any Artinian ring. Next we pass from Artinian rings to Noethe- rian 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 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,X1,...,Xn 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. √ n o n Deﬁnition 2.1. Let a ⊆ R be an ideal. Then we call a := x ∈ R ∃n ∈ N: x ∈ a the radical of 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 pn = p holds for all n ≥ 1.

(iii) Let I be a ﬁnite set and for all i ∈ I let qi ⊆ R be an ideal. Then we have pT T √ i∈I qi = 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

2 II Primary Decomposition

p T Example 2.3. (i) For the nilradical of R we have N(R) := (0) = p∈Spec(R) p. (ii) For a prime number p ∈ we see pT (pn) = p(0) = (0) 6= (p) = T p(pn). Z n∈N n∈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.

n (ii) The primary ideals of Z are (0) and (p ) 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. √ 2 So q ⊆ R is primary. But we have p := q = (X,Y ) and p ( 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 −Z2) and p := (X,¯ Z¯). Then p is prime since R/p =∼ K[Y ], but p 2.2 X¯ · Y¯ = Z¯2 ∈ p2 and X/¯ ∈ p2, Y/¯ ∈ p2 = p.

So p2 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. In particular the powers of maximal ideals are primary. Proof. The second assertion follows from the ﬁrst since for any n ≥ 1 and any maximal √ ideal m ∈ MaxSpec(R) we have mn 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. 2

3 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 = xnyn ∈ q. Since x∈ / p we have xn ∈/ q so – since q is primary – there is m ≥ 1 such that (yn)m = √ ynm ∈ q. Thus y ∈ q = p. 2

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

n o

(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 2.10. 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 m2 = (X2,XY,Y 2) is primary, but m2 = m2 + (X) ∩ m2 + (Y ). 2

T Deﬁnition 2.11. A representation a = i qi of an ideal a ⊆ R as a ﬁnite intersection of primary ideals qi ⊆ R is called a primary decomposition. It is said to be irredundant or reduced if the following conditions hold: T (i) qi 6⊆ qj ∀j. i6=j √ √ (ii) qj 6= qi ∀i 6= 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 2.12. Let R be Noetherian and a ⊆ R an ideal. Then there exist ﬁnitely many T irreducible ideals qi ⊆ R such that a = i qi. 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. 2

4 II Primary Decomposition

Lemma 2.13. 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:(bs) becomes s∈N stationary. So there is some n ≥ 1 such that

(∗) q:(bn) = q:(bn+1).

Claim: q = q + (bn) ∩ q + (a) So the lemma follows since bn, a∈ / q, hence q is not irreducible. Proof of the claim: Since “⊆” is clear, it suﬃces to prove the reverse: Let x ∈ q + (bn) ∩ q + (a). Then there are u, v ∈ q, y, z ∈ R such that

(†) x = u + ybn = v + za.

(†) (†) So we have: bx = bv + zab ∈ q =⇒ q 3 bx = bu + ybn+1. (∗) =⇒ ybn+1 ∈ q =⇒ y ∈ q:(bn+1) = q:(bn) =⇒ ybn ∈ q. =⇒ x = u + ybn ∈ q and the claim follows. 2

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. T Proof. By Lemma 2.12 and Lemma 2.13 we have a = i qi for ﬁnitely many primary ideals qi. So we just have to make sure that this representation is irredundant or that we can make it irredundant. T First if there is some ideal qj such that i6=j qi ⊆ qj we omit qj in the representation. It T is clear that we still have a = i6=j qi. If on the other hand there are i 6= j such that √ √ qi = qj then we replace qi and qj by ˜qij := qi ∩ qj. So we just have to make sure that

˜qij is primary. Tn More generally we show: If q1,..., qn are p-primary then q := i=1 qn is p-primary. pTn 2.2 Tn √ T For this we ﬁrst observe that i=1 qi = 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∈ / qj but qj is p-primary and √ xy ∈ qj, so it follows y ∈ qj = p. 2

T Deﬁnition 2.15. Let a ⊆ R be an ideal admitting a primary decomposition a = qi. √ i The radicals pi := qi are called the associated primes or the prime ideals belonging to a

or simply the primes of a. If pi is minimal in the set of associated primes, then pi 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 III Krull’s principal ideal theorem

The following theorem will be used repeatedly:

Theorem 2.16. Let {pi}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 pi. 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 ⊆ pi ⊆ p. T (⇐) We have p ⊇ a = qi. Then by [AM69, 1.11] there exists i such that qi ⊆ p and it √ √ i follows that pi = qi ⊆ p = p by Lemma 2.2. 2

Corollary 2.17. Let a ⊆ R be an ideal and p ∈ Spec(R).

(i) Let furthermore b ⊆ a be an ideal and denote by a¯, p¯ the image of a, p respectively in R/b. Then p¯ is an isolated prime of a¯ 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−1p is an isolated prime of S−1a 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]). 2

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−1R.

−1 −1 (i) For an ideal a ⊆ R let S a := ϕ∗(a) denote the image of a in S R.

(ii) For an ideal b ⊆ S−1R we write b ∩ R := ϕ∗(b) for the pre-image of b in R.

−1 (iii) For a prime ideal p ∈ Spec(R) we write aRp := (R \ p) 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. 2

6 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

(n) n p := p Rp ∩ R

to be the n-th symbolic power of p.

Lemma 3.3. The ideal p(n) ⊆ R is p-primary. n n Proof. First pRp is maximal in Rp, so by Lemma 2.7 we have p Rp = (pRp) is pRp- 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). p Finally one immediately veriﬁes p(n) = p. 2

(n+1) (n) Corollary 3.4. If R is a Noetherian domain and p 6= (0) then p ( p ∀n ≥ 1. (j) j (n+1) (n) Proof. Ideal theory provides p Rp = p Rp for all j ≥ 1, so if p = p for some n ≥ 1, n+1 n then p Rp = p Rp in the local Noetherian ring (Rp, pRp) yielding that Rp is an Artinian ([AM69, 8.6]) domain, hence it is a ﬁeld. Contradiction to p 6= (0). 2

Example 3.5. In the situation of 3.4 all conditions are necessary:

2 n (i) Let R := Z/4Z and (0) 6= (2) =: p. Then clearly p = (0) = p for all n ≥ 2.

(ii) Consider the integral ring extension Z ,−→ R := Z¯C and some (0) 6= p ∈ Spec(R) (then p is necessarily maximal). Observe the local ring (Rp, pRp): For any α ∈ pRp √ √ we evidently have α ∈ Rp. Now α is not a unit in Rp otherwise α would be. This √ √ √ 2 2 states α ∈ pRp and it follows pRp 3 α = α α ∈ (pRp) hence pRp = (pRp) .

As Rp is a domain, this additionally shows that Rp is not Noetherian (nor is R).

3.2 Krull’s principal ideal theorem

Deﬁnition 3.6. Let p ∈ Spec(R). By ht(p) := dim(Rp) 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 p0 ( p1 ( ... ( pn = 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 6.3.

7 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 −→ Rp.

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) 6= (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. (n) Set qn := v and obtain by Lemma 3.3 and Corollary 3.4 an inﬁnitely decreasing chain of primary ideals (∗) q1 ) ... ) qn ) qn+1 ) ... So we get a decreasing chain qn + (a) n of ideals containing (a). 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 qn + (a) and therefore also qn + (a) become stationary. So there is n ≥ 1

such that qn + (a) = qn+1 + (a) yielding qn ⊆ qn+1 + (a).

Thus for an arbitrary chosen x ∈ qn there exist y ∈ qn+1 ⊆ qn, z ∈ R such that x =

y + za =⇒ za = x − y ∈ qn (+). Evidently a∈ / v, otherwise we get (a) ⊆ v ( p in contradiction to (♦). √ So since qn is primary and a∈ / qn (+) implies z ∈ qn.

We get: qn ⊆ qn+1 + aqn and since the other inclusion is obvious, it follows

(†) qn = qn+1 + aqn.

0 Now pass to R := R/qn+1 in which we denote by ¯q the image of qn. The equation (†) implies: ¯q =a ¯ · ¯q =⇒ ¯q = (¯a) · ¯q. Now ¯q is ﬁnitely generated as R0 is Noetherian. Furthermore we have Jac(R0) = p¯ since p is the only prime ideal of R. This gives (¯a) ⊆ p¯ = Jac(R0), so Nakayama’s lemma [AM69, 2.6]

supplies ¯q = (0)¯ . So qn = qn+1. to (∗). 2

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 2.16. 2

Corollary 3.10. 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 ⊇ p1 ⊇ p2 be a chain of prime ideals of R. Then pass to the Noetherian

8 III Krull’s principal ideal theorem

domain R/p2 and get p¯ ⊇ p¯1 ⊇ (0)¯ . Corollary 2.17 gives that p¯ is an isolated prime of (¯a).

If (¯a) = (0)¯ Proposition 3.9 provides ht(p¯) = 0 and thus p¯ = (0)¯ (so p = p2).

If (¯a) 6= (0)¯ it follows by Theorem 3.8 that ht(p¯) = 1 yielding p¯1 = (0)¯ or p¯1 = p¯. 2

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 3.11. Let R be Noetherian, p0 ) ... ) pm a chain of prime ideals and let qi (for i ∈ I) be ﬁnitely many prime ideals not containing p0. Then there are prime ideals 0 0 0 p0 ) p1 ) ... ) pm−1 ) pm such that none of the pj are contained in any of the qi. Proof. It suﬃces to prove the claim for m = 2. S Since p0 6⊆ qi ∀i ∈ I we have: p0 6⊆ i qi ∪ p2 (see [AM69, 1.11]). S So there exists an element x ∈ p0 : x∈ / i qi ∪ p2.

Set a := p2 + (x) and denote by F := {p˜i} the set of associated primes of a (existence by 0 Theorem 2.14). Since a ⊆ p0 there is by Theorem 2.16 an isolated prime p1 ∈ F such that 0 0 0 0 p1 ⊆ p0. Further as x ∈ a ⊆ p1 we have by construction: p1 6⊆ qi ∀i ∈ I and p1 ) p2. 0 So the only thing left to show is that p0 ) p1:

Observe the Noetherian domain R/p2 in which a¯ is obviously a principal ideal, generated 0 0 by x¯. By Corollary 2.17 p1 is an isolated prime of a¯. Thus if we had p0 = p1 it would follow that p¯0 is an isolated prime of a¯ hence by Corollary 3.10 we have ht(p¯0) ≤ 1, but we have ¯ by assumption p0 ) p1 ) p2 yielding p¯0 ) p¯1 ) p¯2 = (0). 2

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 = (x1, . . . , xr) and b := (x1, . . . , xr−1). If p is among the isolated primes {vi} of b we are through by induction hypothesis.

So assume p is not contained in any vi. (In fact: b ⊆ a ⊆ p and vi are the minimal primes containing b.) Now let m := ht(p) and assume m 6= 0. So by deﬁnition there is a chain of prime ideals

(∗) p =: p0 ) p1 ) ... ) pm−1 ) pm.

Since p0 is not contained in any of the vi we can by Lemma 3.11 assume that pm−1 6⊆ vi ∀i. Now pass to the ring R/b in which a¯ is a principal ideal, generated by the residue class of

xr. By Corollary 2.17 p¯0 is an isolated prime of a¯ hence Corollary 3.10 supplies ht(p¯0) ≤ 1.

Recall that pm−1 6⊆ vi ∀i.

Thus pm−1 + b 6⊆ v¯i while the latter are the isolated primes of (0)¯ ( ♦).

9 III Krull’s principal ideal theorem

Claim: p¯0 is an isolated prime of pm−1 + b.

Proof : Obviously we have pm−1 + b ⊆ p¯0. Then let ˜q be an isolated prime of pm−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 p0 is an isolated prime of pm−1 + b (Corollary 2.17) and therefore p0/pm−1 is

an isolated prime of (pm−1 + b)/pm−1 which is generated by r − 1 elements in R/pm−1.

The induction hypothesis implies that ht(p0/pm−1) ≤ r − 1. (†) Moreover the order-preserving one-to-one correspondence of ideals (see [AM69, 1.1]) supplies

ht(p0/pm−1) = m − 1 (see (∗)). Therefore by (†) m − 1 ≤ r − 1 hence m ≤ r. 2

Corollary 3.13. (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 3.14. 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 a1, . . . , ai ∈ p such that every isolated prime pj of (a1, . . . , ai) is of height i.

Since ht(pj) = i < h = ht(p) we have p 6⊆ pj ∀j. It follows that there is an element S ai+1 ∈ p\ j pj. Now an arbitrary chosen isolated prime v of (a1, . . . , ai+1) contains some pj by Theorem 2.16 and this inclusion is proper since ai+1 ∈ v \ pj. Remember that ht(pj) = i,

so we have ht(v) ≥ i + 1, but v is an isolated prime of (a1, . . . , ai+1) hence by Theorem 3.12 we get ht(v) = i + 1.

So successively this procedure supplies elements a1, . . . , ah ∈ p such that the isolated primes

vj of (a1, . . . , ah) are of height h. Again Theorem 2.16 yields that p contains some vj and since both ideals are of height h we have equality and the claim follows. 2

2 3 ∼ 2 3 2 3 Example 3.15. Consider R := C[X,Y ]/(Y − X ) = C[t , t ] and m := (t , t ). By 2 3 Theorem 3.8 ht (Y − X ) = 1 and as we will see in section V dim C[X,Y ] = 2, so dim(R) = 1 yielding ht(m) = 1. Thus ht(mRp) = 1, but mRm cannot be generated by less

than two elements. Especially Rm is not regular. Cf. Corollary 3.13. 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.

10 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

nX i o p[X] := aiX ∈ R[X] ai ∈ p ∀i ⊆ R[X]. i

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−1R)[X] =∼ S−1(R[X]). Proof. (i) In the ﬁrst assertion p[X] ⊆ pR[X] is clear and so we just have to show “⊇”: P P j Let α ∈ pR[X]. Then α = i αiβi with αi ∈ p, βi ∈ R[X], thus βi = j γijX , so P j α = i,j αiγij X ∈ p[X]. | {z∈p } For the second assertion observe

η : R[X] − (R/p)[X] X i X i aiX 7−→ a¯iX i i

Apply the fundamental theorem on homomorphisms observing ker(η) = p[X] = pR[X]. (ii) Observe

R u with

−1 v ( S R R[X] −1 −1 _ η1 : S R −→ S (R[X]), α 7−→ α η1 η2 X X ai η : R[X] −→ (S−1R)[X], a Xi 7−→ Xi 2 i 1 i i

u g ) −1 o −1 (S R)[X] / S (R[X]) f

X We set f(X) := 1 and by universal properties the existences are clear and it is easily veriﬁed that g ◦ f = id, f ◦ g = id. 2

An immediate consequence of Lemma 4.1 (ii) is:

11 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)−1R[X] =∼ (R \ (0))−1R[X] = Quot(R)[X]

Therefore there is an order-preserving one-to-one correspondence n o ∼ Spec Quot(R)[X] −→ p ∈ Spec R[X] p ∩ R = (0) .

−1 ∗ In section III we used that for the ring homomorphism ϕ: R −→ S 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]. In particular it follows that in general there is no equality in the latter assertion in (i).

Proof. (i) 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). n o

Claim: P := p[X] + (X) = f ∈ R[X] ev0(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 3 ev0(ab) = ev0(a) ev0(b) while ev0(a) ∈/ p, but p is prime yielding ev0(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 ev0(x) = ev0(f + gX) = ev0(f) + 0 ev0(g) = ev0(f) ∈ p. P i Pn i−1 Conversely let x = i aiX ∈ R[X] such that ev0(x) ∈ p. Then x = a0+X i=1 aiX ∈ p[X] + (X). Finally for the last assertion observe that p[X] + (X) ∩ R = p. 2

For an integral ring extension R,−→ S and prime ideals q1 ⊆ q2 in S the equation q1 ∩ R = q2 ∩R implies q1 = q2 (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 IV A first boundary for dim(R[X])

Theorem 4.4. Let R be a ring and P1 ( P2 ( P3 be a chain of three distinct prime ideals in R[X]. Then P1 ∩ R 6= P3 ∩ R.

Proof. Let p := P1 ∩ R. Then we have p[X] ⊆ P1. Assume that P3 ∩ R = p. Factoring ¯ ¯ ¯ out p of R and p[X] of R[X] we obtain a chain of prime ideals P1 ( P2 ( P3 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 Q1 ( Q2 ( Q3 in Quot(R/p)[X] yielding dim Quot(R/p)[X] ≥ 2. (Note that Quot(R/p)[X] is a PID) 2

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 4.4.2

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 p0 ( p1 ( ... ( pn in R. Now consider the chain

p0[X] ⊆ ... ⊆ pn[X] ⊆ pn[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 P0 ( P1 ( ... ( Pm be a chain of prime ideals in R[X]. Then an immediate consequence of Theorem 4.4 is that P0 ∩ R P2 ∩ R P4 ∩ R ... P m ∩ R is a ( ( ( ( 2b 2 c m chain of length b 2 c of distinct prime ideals in R. m This gives b 2 c ≤ 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. 2

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:

m Corollary 4.8. dim(R) + m ≤ dim R[X1,...,Xm] ≤ 2 dim(R) + 1 − 1 ∀m ∈ N.

13 V Dimension theory of K[X1,...,Xn]

V Dimension theory of K[X1,...,Xn]

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[X1,...,Xn]) = 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 ) P1 ) P2. Now clearly p2 := P2 ∩ R ⊆ P ∩ R = p, but ht(p) = 0 yielding p2 = p and so all prime ideals

in (∗) have the same contraction in R contradicting Theorem 4.4. 2initial step (m := ht(p) 6= 0) 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. 4.3 If in the ﬁrst case q = p then p[X] = q[X] ⊆ 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]).2

Remark. One might get the impression that Lemma 5.1 can be proved easier by just assum- ing 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:

P2 ( P3 ( P4

( (

... ( P1 ( P2 ( P3 ( P4

where both chains cannot be reﬁned ( [Kun80, p. 55]). In this situation there is no prime

ideal between P3 and P4, but nevertheless we could have ht(P4) ≥ ht(P3) + 2 (if the

pictured chain is the longest chain of prime ideals ending with P3).

Corollary 5.2. Let P ⊆ R[X1] be any prime ideal such that P ) (p := P ∩ R)[X1]. Then

14 V Dimension theory of K[X1,...,Xn]

in the ring R[X1,...,Xn] we have ht(P[X2,...,Xn]) = ht(p[X1,...,Xn]) + 1. Proof. First we have

P[X2,...,Xn] ∩ R[X2,...,Xn] = (P ∩ R)[X2,...,Xn] = p[X2,...,Xn].

Now observe that P ) p[X1] implies P[X2,...,Xn] ) p[X1,...,Xn] and we are through by Lemma 5.1 applied to the ring R[X2,...,Xn][X1]. 2

Theorem 5.3. Let P ⊆ R[X1,...,Xn] be some prime ideal and p := P ∩ R ∈ Spec(R).

Then ht(P) ≤ ht(p[X1,...,Xn]) + n. Proof. Proof by induction on n. (n = 1) This is Lemma 5.1. (n − 1 n) Let q := P ∩ R[X1]. If we have q = p[X1] then we are through by induction hypothesis applied to the ring R[X1][X2,...,Xn]. If in the other case we have p[X1] ( q then by Corollary 5.2 we get ht(q[X2,...,Xn]) = ht(p[X1,...,Xn]) + 1. Again the induction

hypothesis on R[X1][X2,...,Xn] gives ht(P) ≤ (n − 1) + ht(q[X2,...,Xn]). Together we

have thus shown ht(P) ≤ ht(q[X2,...,Xn]) + (n − 1) = ht(p[X1,...,Xn]) + n. 2

This leads us to the main theorem of this section:

Theorem 5.4. Let K be a ﬁeld. Then dim(K[X1,...,Xn]) = n.

Proof. Consider the chain of prime ideals of K[X1,...,Xn]:

(0) ( (X1) ( (X1,X2) ( ... ( (X1,...,Xn).

First it is clear that all these ideals are prime since for each 1 ≤ k ≤ n we have ∼ K[X1,...,Xn]/(X1,...,Xk) = K[Xk+1,...,Xn]. This yields dim(K[X1,...,Xn]) ≥ n.

Conversely let P ∈ Spec(K[X1,...,Xn]) be some prime ideal. Then P ∩ K ∈ Spec(K) = {(0)}. Theorem 5.3 implies ht(P) ≤ ht (0) + n = n. Hence dim(R[X1,...,Xn]) ≤ n. 2

Remark. • In the proof of Theorem 5.4 we see that the result of Theorem 5.3 cannot

be improved: Choose P := (X1,...,Xn) and note ht(P) = n while p = (0).

• To show dim(K[X1,...,Xn]) ≥ 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 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[X1,...,Xn]) = n.

Proof. By Corollary 4.8 we have dim(R[X1,...,Xn]) ≥ n. To see the reverse inequality

let P ∈ Spec(R[X1,...,Xn]) 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[X1,...,Xn]) = 0

as well, so like in the proof of Theorem 5.4 we conclude dim(R[X1,...,Xn]) ≤ 0 + n = n.2

Remark 5.6. Note that this is (in the case n > 1) a real improvement of the general n boundary given in Corollary 4.8 where we stated that dim R[X1,...,Xn] ≤ 2 − 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[X1,...,Xn]) = 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[X1,...,Xn] = 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 ⊆ P ∩ R ⊆ p[X] ∩ R . | {z } | ={za } =:q | ={zp }

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. 2

Now we can proof the main result of this section:

16 VI Dimension theory of Noetherian rings

Proof (Theorem 6.1). By Hilbert’s basis theorem (see [AM69, 7.5]) R[X1,...,Xn−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 a1, . . . , ar ∈ R such that p is an isolated

prime of a := (a1, . . . , ar). 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. 2

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 ∼ 1 ∼ R[X]/m = R 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 evx : R[X] − R[x],X 7−→ x. Then we have R[X]/ ker(evx) = R[x], so ker(evx) is a prime ideal and by deﬁnition dim(R[x]) = dp(ker(evx)). Now as mentioned

in Remark 3.7 ht(ker(evx)) + dp(ker(evx)) ≤ dim(R[X]) and thus we obtain dim(R[x]) ≤

dim(R[X]) − ht(ker(evx)). By assumption we have ker(evx) 6= (0) hence ht(ker(evx)) ≥ 1. Together with dim(R[X]) = dim(R) + 1 the claim follows. 2

This raises the question how badly the equation “dim(R[x]) = dim(R)” can fail if x is not integral.

17 VI Dimension theory of Noetherian rings

n o

Deﬁnition 6.5. For j ∈ N with j ≤ dim(R) let Fj(R) := p ∈ Spec(R) ht(p) = j . T We simply write Fj if it is clear which domain R we are dealing with and by Fj we

mean the intersection of all prime ideals in the set Fj.

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:

T Lemma 6.6. There exists a domain R[x] with dim R[x] < j iﬀ Fj(R) 6= (0). T 1 1 Proof. Suppose there is 0 6= z ∈ Fj. Then R is a domain with dim R < j. n o z z To see this let S := zn n ∈ . Then R 1 = S−1R. Now choose an arbitrary prime N z ideal q ∈ Spec(S−1R) and set p := q ∩ R. Remember that q = S−1p and ht(p) = ht(q).

Assume ht(p) ≥ j, then there is a prime ideal p0 ∈ Spec(R) with ht(p0) = j and p0 ⊆ p. Thus by deﬁnition of z we have z ∈ p yielding p ∩ S 6= ∅. Hence ht(q) = ht(p) < j.

T If on the other hand we have Fj = (0) then we show that dim R[x] ≥ j for any domain R[x] (with x 6= 0). By Theorem 6.1 this assertion is certainly true if x is transcendental, so m suppose there are a0, . . . , am ∈ R with am 6= 0 and amx + ... + a0 = 0. n o n −1 Set S := am n ∈ N . Now x is integral over S R since we have a a xm + m−1 xm−1 + ... + 0 = 0. am am

Therefore it follows dim(S−1R) = dim (S−1R)[x] = dim S−1(R[x]) (∗).

By assumption there is a prime ideal p ∈ Spec(R) with ht(p) = j such that am ∈/ p hence p ∩ S = ∅. This gives j = ht(p) = ht(S−1p) yielding dim(S−1R) ≥ j and the claim follows (∗) since now we have j ≤ dim(S−1R) = dim S−1(R[x]) ≤ dim R[x]. 2

Corollary 6.7. There is a domain R,−→ R[x] with dim R[x] dim(R) if and only if T Fdim(R) 6= (0).

Note that not all maximal ideals need be of height dim(R) as we have seen in Example 6.3.

T Example 6.8. (i) Let R be a domain with dim(R) = 1 and F1 6= (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 h 1 i course Z(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 K closure. Further let B := A be the integral closure of A in K. Now the set

18 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−1B 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) 6= P ∈ Spec(B) we have P ∩ S = ∅ iff P ∩ A = (X). Proof : If P ∩ S = ∅ we have P ∩ A ⊆ (X). Moreover P ∩ A 6= (0) as P= 6 (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. 2claim T By the claim it follows that (if there are height-one prime ideals in R) we have F1(R) ⊇ (X) 6= (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: √ n For this let n ≥ 2 be an integer, ζ1, . . . , ζn ∈ C be the n-th roots of unity and Y := X + 1 ∈ n B be some root of T − (X + 1) ∈ A[T ]. Then we obtain n ideals p1,..., pn ⊆ A[Y ] by

setting pi := (X,Y − ζi). Now consider:

∼ ∼ n A[Y ]/pi = A[Y ]/(X,Y − ζi) = A[T ]/(X,T − (X + 1),T − ζi) ∼ n ∼ n ∼ n ∼ = C[X,T ]/(X,T − (X + 1),T − ζi) = C[T ]/(T − ζi,T − 1) = C/(ζn − 1) = C

So we have pi ∈ Spec(A[Y ]) for all i. Clearly A[Y ] ,−→ B is integral, so Spec(B) − Spec(A[Y ]) is surjective providing P1,..., Pn ∈ Spec(B) such that Pi ∩ A[Y ] = pi and thus

Pi ∩ A = pi ∩ A =: ℘i ∈ Spec(A).

Since pi 6= (0) and A,−→ A[Y ] is integral we have ℘i 6= (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 P1,..., Pn ∈ Spec(B)

with Pi ∩ A = (X) for all i. Repeating this construction for all n ≥ 2 we obtain inﬁnitely many such prime ideals and we are through. 2 T Proof (6.8 (ii)). Since all maximal ideals have height n, the intersection Fn equals the Jacobson radical Jac(R) = T MaxSpec(R), which in turn equals the nilradical N(R) since T R is Jacobson. Finally R is a domain, hence (0) = N(R) = Jac(R) = Fn. 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]). 2

19 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 q1,..., qn ∈ Spec(R) are the only height-one prime ideals contained in p. S S Evidently we have p 6⊆ qi for each i, hence p 6⊆ i qi. Choose x ∈ p \ i qi 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∈ / qi we obviously have p˜ 6= qi for all i. 2

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 6.10. We always have dim R[x] ≥ dim(R) − 1. T Proof. We prove Fn−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) = p0 ( ... ( pn = 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 p2 and for each j ∈ J we have dim(R/p˜j) = n − 1. T We choose z ∈ Fn−1 and an arbitrary j ∈ J. 0 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(p0) ≥ n − 1. So – by deﬁnition of z – we have z ∈ p0 yielding z¯ ∈ p¯. Since p¯ was arbitrary chosen in the set of prime ideals of height n − 2 we have z¯ ∈ T 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 6= 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 The-

orem 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 2.14. T Thus we have z = 0 and since z was arbitrary chosen it follows Fn−1 = (0). Hence by Lemma 6.6 we are through. 2

Corollary 6.11. Let R be a domain and let x1, . . . , xn be elements such that R[x1, . . . , xn] 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 {xi}i∈I such that Z[xi]i∈I is a ﬁeld. Nor can we hope for any sort of induction argument (consider Z(p)[X]).

20 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

Rp is a valuation domain.

Remark 7.2. • There are various ways to deﬁne valuation and Prüfer domains. For example there are 22 equivalent conditions for a Prüfer domain given in [BG06, The- orem 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[X1,...,Xn]) = 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[X1,...,Xn]) ≤ dim(R) + n and therefore it suﬃces to verify

ht(M) ≤ dim(R) + n for any maximal ideal M ∈ MaxSpec(R[X1,...,Xn]). But we have seen ht(M) ≤ ht (M ∩ R)[X1,...,Xn] + n for any maximal ideal, so the

assertion above is clear if we show ht(p[X1,...,Xn]) ≤ ht(p) for any prime ideal p ∈ Spec(R). We have a one-to-one correspondence

n o −1 ∼ Spec (R \ p) R[X1,...,Xn] −→ P ∈ Spec(R[X1,...,Xn]) P ∩ R ⊆ p ,

so clearly every chain of primes in R[X1,...,Xn] descending from p[X1,...,Xn] corresponds

21 VII Prüfer domains and a note on Seidenberg’s F -rings

to some in the localization, so we may assume that R is local with unique maximal ideal p.

Further by deﬁnition of R we know that Rp is a valuation domain, so its principal ideals are totally ordered by inclusion. That means that the elements of R are totally ordered by divisibility. (∗)

Now we prove ht(p[X1,...,Xn]) ≤ ht(p):

For this let (0) 6= Q ∈ Spec(R[X1,...,Xn]) be an arbitrary prime ideal contained in

p[X1,...,Xn].

Claim: Q = (Q ∩ R)[X1,...,Xn].

Thus every prime ideal contained in p[X1,...,Xn] is an extended prime ideal of R, hence

ht(p[X1,...,Xn]) ≤ ht(p) as desired.

Proof of the claim: It is clear that Q[X1,...,Xn] ⊆ Q. For the reverse let 0 6= f ∈ Q ⊆

R[X1,...,Xn] be arbitrary chosen. Now (∗) supplies some a ∈ R and g ∈ R[X1,...,Xn] such that f = ag and one of the coeﬃcients of g is a unit.

This means in particular that g∈ / p[X1,...,Xn] ⊇ Q 3 f = ag. But Q is a prime ideal, so

a ∈ Q and we further had a ∈ R, so a ∈ Q ∩ R yielding ag ∈ (Q ∩ R)[X1,...,Xn]. 2

C Example 7.4. The ring of algebraic integers R := Z is not Noetherian (see Example 3.5 (ii)), but it is a Prüfer domain, so dim(R[X1,...,Xn]) = n + 1. Proof. In [FS01, p. 93] we ﬁnd the more general theorem: If R is a Prüfer domain, L K := Quot(R) and K,−→ L any algebraic ﬁeld extension of K then the integral closure R C is a Prüfer domain. Apply this theorem to R := Z and L := Q . 2

Remark to the proof : Alternatively it is easily shown (some straight forward calculations with the class number of certain intermediate rings) that R is a Bezout domain and therefore a Prüfer domain.

Remark. As mentioned in the introduction to this section the assertion of Theorem 7.3 is also true for semi-hereditary rings. A ring is called semi-hereditary if its ﬁnitely generated ideals are projective, but these rings can also be characterized as follows: A ring is semi-

hereditary if the total quotient ﬁeld is regular and the quotient ring Rm is a valuation domain for any maximal ideal m. See [End61, Theorem 2] for a proof. We see that the deﬁnitions of semi-hereditary rings and Prüfer domains are quite similar. In fact one can see that a Prüfer domain is a semi-hereditary ring that is furthermore a domain (for details: see [BG06]). So we may consider semi-hereditary rings as the non-integral generalization of Prüfer domains. However the proof of Theorem 7.3 does only depend on the fact that

Rp is a valuation domain which is also true for semi-hereditary rings.

7.2 A note on F -rings

In Remark 4.7 we mentioned Seidenberg’s result on the existence of rings R that fail to fulﬁll the condition dim(R[X]) = dim(R) + 1. In this subsection we brieﬂy examine this fact. A more detailed discussion would go beyond the scope of this thesis. For the sake of simplicity

22 VII Prüfer domains and a note on Seidenberg’s F -rings

R denotes a domain throughout this subsection. In fact: If there is a ring failing the condition above then by factoring out an appropriate height-zero prime ideal we obtain a domain that fails the condition.

Deﬁnition 7.5. A domain R is called an F -ring if dim(R) = 1 and dim(R[X]) 6= 2.

Theorem 7.6. If R is a domain with dim(R) = n and dim(R[X]) 6= n + 1 then there is a

height-one prime ideal p ∈ Spec(R) such that either Rp is an F -ring or dim(R/p) = m and dim (R/p)[X] 6= m + 1 for some m < n. Proof. In the ﬁrst case assume there is some height-one prime ideal p ∈ Spec(R) such that ht(p[X]) ≥ 2. Then there is some prime ideal P ∈ Spec(R[X]) such that (0) ( P ( p[X]. Then by virtue of the one-to-one correspondence we obtain a chain of prime ideals

−1 −1 −1 (0) ( (R \ p) P ( (R \ p) p[X] ( (R \ p) p[X] + (X).

So Rp is an F -ring. In the other case let (∗) ht(p[X]) = 1 for all p ∈ Spec(R) with ht(p) = 1 and let

(†) (0) = P0 ( P1 ( ... ( Pn+2

be a chain of prime ideals in R[X]. If P1 ∩ R =: p 6= (0) then dim(R/p) ≤ n − 1 and ∼ dim (R/p)[X] ≥ n + 1 since (R/p)[X] = R[X]/pR[X] and pR[X] ⊆ P1. 0 So assume p = (0), but then p := P2 ∩ R is not (0) by Theorem 4.4. Now choose some p˜ ∈ Spec(R) such that p˜ ⊆ p0 and ht(p˜) = 1. 0 Thus by (∗) we have ht(p˜[X]) = 1 hence p˜[X] ( P2, so in (†) replace P1 by P1 := p˜[X] and 0 we are in the previous case that P1 ∩ R 6= (0). 2

Corollary 7.7. If there is a domain R with dim(R) = n and dim(R[X]) 6= n + 1 then there exists an F -ring. Proof. Apply Theorem 7.6 repeatedly. Note that (in the notation of Theorem 7.6) m = 0 is impossible (as it is shown in Theorem 5.5 or Theorem 4.6). 2

This restricts the question about the existence of rings with dim(R[X]) 6= dim(R) + 1 to the existence of F -rings. And further we only need to examine integrally closed domains:

Lemma 7.8. R is an F -ring if and only if its integral closure R¯ is one. Proof. We prove that if the ring extension A,−→ B is integral then so is A[X] ,−→ B[X], equivalent to B[X] ⊆ C where C is the integral closure of A[X] in Quot(B[X]). To see this P i let f ∈ B[X] be arbitrary chosen. Then f = i biX with bi ∈ B. Since bi are integral over A they are integral over A[X] hence bi ∈ C and as X ∈ A[X] clearly X ∈ C. By [AM69, 5.3] P i C is a ring, so f = i biX ∈ C. So the lemma follows since we have shown dim(R) = dim(R¯) and dim(R[X]) = dim(R¯[X]).2

23 VII Prüfer domains and a note on Seidenberg’s F -rings

Proposition 7.9. If R is an F -ring, then there is an F -ring having just one proper prime ideal.

Proof. Let R be an F -ring. Then there are prime ideals (0) ( P1 ( P2 ( P3 in R[X]. Set p := P3 ∩ R and the one-to-one correspondence supplies that Rp[X] is three-dimensional

and of course dim(Rp) = 1, so Rp is the desired F -ring. 2

So searching for domains with dim(R[X]) 6= dim(R) + 1 we may concentrate on integrally closed local F -rings. Note that Rp is integrally closed if R is (see [Sta13, Lemma 0307]).

Remark 7.10. Another characterization of valuation domains is that for any element x ∈ −1 a Quot(R) we have either x ∈ R or x ∈ R: Let x = b ∈ Quot(R) (with b 6= 0). We have a | b or b | a as R is a valuation domain, so either x ∈ R or x−1 ∈ R. Conversely let a b (a), (b) be non-zero principal ideals. Then either b ∈ R yields (a) ⊆ (b) or a ∈ R yields the converse.

Theorem 7.11. Let (R, m) be an integrally closed local domain and x ∈ Quot(R) such that x, x−1 ∈/ R. Then mR[x] is a prime ideal, but not maximal. n Proof. Observe (∗) mR[x]+(x) 6= (1): Otherwise we obtain 1 = a0 +a1x+...+anx with −1 n n−1 ai ∈ R, a0 ∈ m. Therefore x would be a root of (a0 − 1)X + a1X + ... + an ∈ R[X] −1 ∗ yielding that x is integral over R (note that (a0 − 1) ∈ R as (R, m) is local and a0 ∈ m), but R is integrally closed.

Claim 1 : For any monic polynomial f ∈ R[X] we have f(x) := evx(f) ∈/ mR[x]. Proof : The claim is clear if deg(f) = 0 (then f = 1; remember (∗)), so assume deg(f) ≥ 1. First f(x) ∈/ R since otherwise f − f(x) = 0 would be an equation of integral dependence for x, but x∈ / R = R¯. Now assume f(x)−1 ∈ R. Then clearly f(x)−1 ∈/ R∗ (as f(x) ∈/ R), so f(x)−1 ∈ m yielding 1 ∈ f(x) · m ⊆ mR[x] to (∗). So f(x)−1 is not not contained in

R. As above it follows mR[f(x)] + (f(x)) 6= (1). Now x is a root of f − f(x) ∈ R[f(x)][X], so R[x] is integral over R[f(x)]. This means in particular that for any prime ideal P of R[f(x)] that contains mR[x] + (f(x)) there is a prime ideal P of R[x] such that P ∩ R[f(x)] = P. Thus mR[x] + f(x)R[x] 6= (1). Since f + 1 is also a monic polynomial of positive degree we further obtain that

mR[x] + (f(x) + 1)R[x] 6= (1). This shows f(x) ∈/ mR[X] as desired. 2claim −1 Claim 2 : evx (mR[x]) = mR[X] (where evx : R[X] − R[x]). Proof :“⊇” is clear. To see the reverse let f ∈ R[X] such that f∈ / mR[X]. Then we write

f = f1 + f2 where f2 ∈ mR[X] and no coeﬃcient of f1 is contained in m. Let c denote −1 −1 the leading coeﬃcient of f1. Then c f1 is monic, so by claim 1 we have c f1(x) ∈/ mR[x]

yielding f1(x) ∈/ mR[x], so f(x) ∈/ mR[x]. 2claim 2 We know (section IV) that mR[X] is a prime ideal, but not maximal, so by claim 2 we easily show that mR[x] is prime, but not maximal:

Let a, b ∈ R[x]. As evx is surjective we have pre-images a0, b0 ∈ R[X]. Assume ab ∈ mR[x]. −1 Then mR[x] 3 ab = evx(a0) evx(b0) = evx(a0b0) hence a0b0 ∈ evx (mR[x]) = mR[X], so

24 VII Prüfer domains and a note on Seidenberg’s F -rings

P i suppose without loss of generality a0 ∈ mR[X]. Thus a0 = i γiX with γi ∈ m and there- P i fore clearly evx(a0) = i γix ∈ mR[x]. That mR[x] is not maximal follows similarly. 2

Finally we get a necessary and suﬃcient condition for a domain to be an F -ring.

Theorem 7.12. A domain R with dim(R) = 1 is not an F -ring if and only if every quotient

ring R¯p of its integral closure R¯ is a valuation domain for any prime ideal p ∈ Spec(R¯). Proof. By Lemma 7.8 we may assume R is integrally closed and by Proposition 7.9 there

is some p ∈ Spec(R) with Rp is an F -ring, so Rp is not a Prüfer domain (Theorem 7.3). In

particular Rp is not a valuation domain.

To see the if-part let Rp be not a valuation domain, so by Remark 7.10 there is some −1 x ∈ Quot(Rp) with x, x ∈/ Rp. Then by Theorem 7.11 we obtain dim(Rp[x]) ≥ 2. This

yields dim(Rp[X]) ≥ 3 (similar arguments as in the proof of Lemma 6.4). So we are through 4.1 as 3 = dim(Rp[X]) = dim(R[X]p) ≤ dim(R[X]). 2

Especially every integrally closed domain with only one proper prime ideal that is not a valuation domain is an F -ring. It is yet not clear whether there exist such rings at all. Krull in [Kru36, p. 670f.] gave the following:

Example 7.13. Let K := C(X,Y ) and f1 R := f = ∈ K f2(0, y) 6= 0, f(0, y) ∈ C, gcd(f1, f2) = 1 . f2

Then R is a one-dimensional integrally closed local domain that is not a valuation domain. Remark: Whenever we write an element f ∈ R as f = f1 , then we assume f , f ∈ [X,Y ] f2 1 2 C and gcd(f1, f2) = 1. As C[X,Y ] is factorial we may talk about divisibility and the gcd. R is a subring of K as one easily veriﬁes. For example let f = f1 , g = g1 ∈ R. Proof. f2 g2 Then f − g = f1g2−f2g1 . As X f and X g it follows X f g (as X ∈ [X,Y ] is prime). f2g2 - 2 - 2 - 2 2 C Further (f − g)(0, y) = f(0, y) − g(0, y) ∈ C, so f − g ∈ R. n o

Since K is a ﬁeld, R is a domain. Next m := f ∈ R f(0, y) = 0 ⊆ R is a prime ideal f1(0,y) −1 f2 of R. Now let f ∈ R \ m. Then 0 6= f(0, y) = , so f1(0, y) 6= 0. Further for f = f2(0,y) f1 −1 −1 −1 ∗ ∗ we have f (0, y) = f(0, y) ∈ C, so f ∈ R yielding f ∈ R , thus R \ m = R , hence (R, m) is local. Now let (0) 6= p ∈ Spec(R) be some prime ideal of R (necessarily contained in m). Then r X g1 r+1 choose some g ∈ p and write g = with gcd(X, g1) = 1. Then any element in m can g2 √ √ √ be divided by g yielding mr+1 ⊆ p ⊆ m. So we obtain m = mr+1 ⊆ p ⊆ m = m, hence √ √ p = m, but p = p as p ∈ Spec(R), so (R, m) is one-dimensional. Further let h = h1 ∈ (X,Y ) such that hn + r hn−1 + ... + r = 0 with r ∈ R (that means h2 C 1 n i ri(0, y) ∈ C), so X - h2 and h(0, y) ∈ C. Together it follows h ∈ R, hence R is integrally closed. Eventually R is not a valuation domain because X,XY ∈ R but neither X | XY nor XY | X (in R!). 2

25 VII Prüfer domains and a note on Seidenberg’s F -rings

So we ﬁnally have seen that indeed there are rings that fail the condition “dim(R[X]) = dim(R) + 1”. In fact there are many of them: Given a domain R with dim(R) = n and dim(R[X]) = m Seidenberg constructs in [Sei54, Thm. 1] a domain S with dim(S) = n + 1 and dim(S[X]) = m + 2. Further he gives a similar construction that supplies a domain T with dim(T ) = n + 1 and dim(T [X]) = m + 1 in [Sei54, Thm. 2]. With these two constructions he is able to prove (using induction) that for any natural number n and any m within n + 1 and 2n + 1 there is a domain O with dim(O) = n and dim(O[X]) = m ( [Sei54, Thm. 3]). The constructions are quite technical and similar to those in Theorem 7.14. To conclude this outlook on Seidenberg’s work we may mention the following theorem concerning one-dimensional domains:

Theorem 7.14. Let R be a one-dimensional domain, then: (i) If R is not an F -ring, then dim R[X1,...,Xn] = n + 1. (ii) If R is an F -ring, then n + 2 ≤ dim R[X1,...,Xn] ≤ 2n + 1. Moreover for any m with n + 2 ≤ m ≤ 2n + 1 there is an F -ring such that dim R[X1,...,Xn] = m. Proof (sketch). (i) Similar as in the proof of Lemma 7.8 we may assume that R is inte-

grally closed. Then by Theorem 7.12 Rp is a valuation domain for any p ∈ Spec(R), hence R is a Prüfer domain yielding the claim by Theorem 7.3.

(ii) Let P0 ( ... ( Ps be a chain of prime ideals in R[X1,...,Xn]. If Ps ∩ R = (0) the one-to-one correspondence (see Corollary 4.2) supplies prime ideals P0 ( ... ( Ps in Quot(R)[X1,...,Xn] yielding s ≤ n and we are through. If otherwise there is some

1 ≤ i ≤ n with Pi ∩ R = (0) and Pi+1 ∩ R =: p 6= (0), then p is maximal as dim(R) = 1. Again the one-to-one correspondence gives (∗) i ≤ n as in the case above. Now consider ∼ R[X1,...,Xn]/Pi+1 = (R/p)[X1,...,Xn] and note that R/p is again a ﬁeld as p is maximal, so we have s − (i + 1) ≤ n. Together with (∗) it follows s ≤ 2n + 1.

To see the lower bound note that we have by assumption dim(R[X1]) = 3 and any further indeterminate raises the dimension by at least one. Finally we will sketch the proof of the “moreover”-part: 0 0 Let K be a ﬁeld, X,Y1,...,Ym be indeterminates, K := K(Y1,...,Ym),L := K (X) and let ν be a discrete rank-one valuation of L obtained by setting

i i+1 n ν(aiX + ai+1X + ... + anX ) := i

0 (where aj ∈ K , ai 6= 0). Further let R be the set of elements whose residues are ﬁnite and

in K. Then one can show that R is an F -ring and for all m ≤ n the ring R[X1,...,Xn] is (m + n + 1)-dimensional. Varying m within 1 and n yields the claim. For details see [Sei54, Theorem 7]. 2

26 VIII A journey to the power series ring

Summary of polynomial extensions

Let n := dim(R) and for some m ≥ 1 let S := R[X1,...,Xm]. We have seen: n = 1 Noetherian Prüfer n = 0 F -ring general not F -ring m m + 1 ≥ m + 2 ≥ m + n dim(S) m + n m + n (= m + n) (= m + n) ≤ 2m+1 ≤ 2m(n+1)−1 Furthermore in the F -ring-case or in the general case (restricted to m = 1) there is for any quantity d within the boundary some R such that S fulﬁlls dim(S) = d. In the general case with m > 1 this assertion is wrong (e.g. section IV), but also note that the upper bound in the F -ring-case is stronger than the one of the general case applied to n = 1 (if m > 1).

VIII A journey to the power series ring

To conclude the thesis we give in this section a very brief overview of another sort of ring extensions: We study the Krull dimension of power series rings over certain classes of rings. One of the crucial diﬀerences between power series rings and polynomial rings is that Lemma 4.1 (i) does not extend to power series rings. This means: For an ideal a ⊆ R we have aR X ⊆ a X and in general there is no equality. This “little diﬀerence” has a J K J K surprising consequence:

Example 8.1 ([Wat12, §5]). Let R := Q × Q × ... be a countable product of Q. Then clearly dim(R) = 0. Further let a ⊆ R be the ideal generated by

(1, 0, 0,...), (0, 1, 0,...), (0, 0, 1, 0,...),...

So a is the set of elements of R that have ﬁnite support. Hence if f ∈ aR X then f has J K ﬁnite support. Now observe

g := (1, 0, 0,...) + (0, 1, 0,...)X + (0, 0, 1, 0,...)X2 + ... ∈ a X ⊆ R X . J K J K Clearly gn (for any n ≥ 1) does not have finite support hence g ∈ a X \ paR X and in J K J K Theorem 8.3 we will see that this fact implies dim R X = ∞ (even though dim(R) = 0!). J K Definition 8.2. (i) An ideal a ⊆ R is called an SFT -ideal (short for strong finite type) if there is some n ∈ N and a finitely generated ideal b ⊆ a contained in a such that an ∈ b for each a ∈ a.

(ii) A ring R is said to satisfy the SFT-property if each ideal a ⊆ R is an SFT-ideal.

The relevance of this property when studying the dimension is given by the following theorem by Arnold:

27 VIII A journey to the power series ring

Theorem 8.3. The following conditions are equivalent and imply that dim R X = ∞: J K (i) R does not satisfy the SFT-propery.

(ii) There is some ideal a ⊆ R such that a X 6⊆ paR X . J K J K Proof. See [Arn73a, Theorem 1].

Remark. More precisely Arnold shows that in the case that R is not SFT, there exist an inﬁnitely ascending chain of prime ideals.

So the SFT-property is necessary for the fact that the dimension does not rise to infinity. Therefore it is reasonable to ask whether it is sufficient; and furthermore: if the dimension is finite one might ask whether it obeys the same boundaries that we found for the polynomial case. Starting with an obvious subclass of SFT-rings it looks like it:

Theorem 8.4. Let R be Noetherian, then dim R X1,...,Xn = dim(R) + n. J K Proof (sketch). As R X is Noetherian if R is (see [AM69, p. 81]) we may – by means of J K induction – reduce the proof and just verify dim R X = dim(R) + 1. J K Let m := dim(R). Then there is a chain of prime ideals p0 ( ... ( pm of R. Lifting this chain into R X we obtain J K

p0 X ( ... ( pm X ( pm X + (X) J K J K J K and dim R X ≥ m + 1. J K Now an arbitrary maximal ideal M ∈ MaxSpec R X is of the form M = m + (X) with J K m ∈ MaxSpec(R). Let l := ht(m) ≤ m. Then Theorem 3.14 supplies elements a1, . . . , al ∈ R

such that m is an isolated prime ideal of a := (a1, . . . , al), but then M is an isolated prime ideal of a+(X) while the latter is generated by l+1 elements yielding ht(M) ≤ l+1 ≤ n+1 by Theorem 3.12. 2

The other kind of rings that have the same behavior are the Prüfer domains. So one might hope that for a Prüfer domain we also have dim R X1,...,Xn = dim(R) + n. This J K is unfortunately not the case. In fact the ring R in the Example 8.1 is semi-hereditary (remember the close relation between Prüfer domains and semi-hereditary rings). However:

Theorem 8.5. Let R be a Prüfer domain. Then the following conditions are equivalent:

(i) R is an SFT-ring.

(ii) dim R X = dim(R) + 1. J K (iii) dim R X < ∞. J K Proof. See [Arn73b, Theorem 3.8].

28 VIII A journey to the power series ring

So the SFT-property is also suﬃcient for Prüfer domains when we adjoin one power series variable. At this point one might hope that this can be extended to several variables with some inductive argument, but unfortunately the property “Prüfer domain” does not survive power series extensions. However in another paper Arnold showed:

Theorem 8.6. Let R be a Prüfer domain that satisﬁes the SFT-property. Then we have dim R X1,...,Xn = n · dim(R) + 1. J K Proof. See [Arn82, Theorem 3.6].

This theorem in particular supplies that if R is a ﬁnite-dimensional Prüfer domain that satisﬁes the SFT-property, then R X1,...,Xn also is an SFT-ring. Finally we turn to the J K last class of rings for which we have proved “good behavior” in the polynomial case:

Theorem 8.7. Let R be a SFT-ring with dim(R) = 0, then dim R X1,...,Xn = n. J K Proof. See [CCD96, Corr. 2.2]. 2

Summary and comparison with polynomial extensions

Let R be a ring with m := dim(R). Further let 1 ≤ n ∈ N, S := R X1,...,Xn , d := dim(S) 0 J K and T := R[X1,...,Xn] as well as d := dim(T ).

Power series rings

Non-SFT SFT d = ∞

other Noetherian m = 0 Prüfer d ≥ m + n d = m + n d = n (∗) 5=

m 6= 1 m = 1 d = mn + 1 d = n + 1

(1) (2)

d0 = m + n d0 = n + 1 m 6= 1 m = 1

m + n ≤ d0 ≤ d0 = m + n d0 = n n Prüfer 2 (m + 1) − 1 Noetherian m = 0 other

Polynomial rings (∗) A one-dimensional Prüfer domain fulfills the SFT-property iff it is Dedekind ([Arn82]). (1) Same lower bound, but there is no upper bound known in the power series case. (2) Here holds equation iff n = 1.

29 References

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[Arn73b] Jimmy T. Arnold. Power series rings over Prüfer domains. Paciﬁc J. Math., 44:1–11, 1973.

[Arn82] Jimmy T. Arnold. Power series rings with ﬁnite Krull dimension. Indiana Univ. Math. J., 31(6):897–911, 1982.

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