Generalized Krull's Principal Ideal Theorem

DEPARTMENTOF MATHEMATICS,IITBOMBAY

Generalized Krull’s Principal Ideal Theorem

Sai Krishna P M S

November 2020

1 INTRODUCTION

All rings are assumed to be commutative with unity.

In this report, we will describe a proof of the generalized Krull’s Principal Ideal Theorem. We will prove this without using dimension theorem. The proof involves some basics results of primary ideals and the notion of symbolic powers of a prime ideal. We will start by making some deﬁnitions and stating some results which will be used directly or indirectly in the proof of the main theorems. We will ﬁrst prove the Krull’s Principal Ideal Theorem and then use it to prove the generalized Krull’s Principal Ideal Theorem. This theorem is also known as Krull’s Height Theorem.

2 PRELIMINARIES

2.1 DEFINITIONS

Primary ideal A proper ideal I of a ring R is said to be primary if ab I and a I b pI. ∈ ∉ ⇒ ∈ Clearly if I is a primary ideal then pI is a prime ideal. Let pI P, then the ideal I is called = P-primary.

Artinian rings A ring R is called Artinian if every descending chain of ideals becomes stationary. Equivalently a ring R is called Artinian if every non empty family of ideals of R has a minimal element.

1 Minimal Prime ideal A prime ideal P is said to be a minimal prime ideal over an ideal I if it is minimal among all prime ideals containing I.

A Prime ideal P is said to be a minimal prime ideal of a ring R if it is a minimal prime ideal over the zero ideal.

Height of a prime ideal Let P be a prime idea. We deﬁne the height of P as follows.

ht P Sup ©r a chain P P P P where P is aprimeidealª. = |∃ = 0 ⊃ 1 ⊃ ··· ⊃ r i

Height of an ideal Let I be an ideal of R. Then the height of I is deﬁned as follows.

ht I In f ©ht P P is aprimeideal containing Iª. = |

Krull Dimension The Krull dimension of a ring R is deﬁned as follows.

dim R Sup ©ht P P spec(R)ª. = | ∈

Remarks:

• The height of a prime ideal P is equal to the dimension of the ring RP .

• If the height of a prime ideal is zero then it is a minimal prime ideal of the ring R.

• The above deﬁnition of height of an ideal can be modiﬁed by considering only the minimal prime ideals over I.

• The above deﬁnition of dimension can be modiﬁed by considering only the maximal ideals of a ring R.

3 SUPPORTING RESULTS

3.1 PRIMARY DECOMPOSITION THEOREMFORIDEALS

Let R be a Noetherian ring and I be a proper ideal of R.

2 • Then there exist primary ideals Q ,Q ,...,Q of R such that I Q Q Q . 1 2 n = 1 ∩ 2 ··· ∩ n

p T • Let Pi Qi . The primary ideals above can be chosen such that Qi j i Q j and Pi ’s = ⊃ 6= are distinct. Then I Q Q Q is called an irredundant primary decomposition = 1 ∩ 2 ··· ∩ n of R.

• If Pi ,Qi are chosen as above then P1,P2,...,Pn are unique.

• The uniquely determined prime ideals P1,P2,...,Pn are called associated prime ideals of I and they are precisely the set of prime ideals among the ideals (I : x) where x R. ∈

• If P j is a minimal associated prime ideal of I then the corresponding primary ideal Q j is unique and is given by Q IR R. j = PJ ∩ Remark:

• For a Noetherian ring R, minimal associated primes of an ideal I are precisely the minimal prime ideals over I.

3.2 SYMBOLICPOWERSOFAPRIMEIDEAL

Let I P n where P is a prime ideal of a Noetherian ring R. = Let I Q Q Q be a irredundant primary decomposition of I with P pQ . = 1 ∩ 2 ··· ∩ k i = i

• I Q Q Q = 1 ∩ 2 ··· ∩ k • pI pQ pQ pQ = 1 ∩ 2 ··· ∩ k • P P P P = 1 ∩ 2 ··· ∩ k • P P for 1 i k ⊆ i ≤ ≤ • P ...P P P P P hence P P for some j. 1 k ⊆ 1 ∩ 2 ··· ∩ k = j ⊆ • Thus P P is the minimal associated prime ideal of I. So, by the primary decompo- = j sition theorem the corresponding primary ideal Q P nR R. We denote the ideal J = P ∩ P nR R by P (n) called the nth symbolic power of the prime ideal P. P ∩

• P (n) P nR R. = P ∩ • Notice that P (1) P (2) P (3) .... ⊇ ⊇

3 3.3 SOMEUSEFUL PROPERTIES

• Any minimal prime ideal of a ring R is contained in the set of zero divisors of R.

Proof: Let P be a minimal prime ideal of R. Then RP has a unique prime ideal(PRP ). It follows that (pR ) PR and thus every element of P is a zero divisor of R. P = P • For any ideal I of a Noetherian ring R there exists a positive integer N such that (pI)N I. ⊆

Proof: Let pI a ,...a where ani I for 1 i l. Choosing N n n ... n 1, =< 1 l > i ∈ ≤ ≤ = 1 + 2 + + l + it follows that (pI)N I. ⊆ • If I is P-Primary then P is the minimal prime of I.

Proof: I is P-Primary hence pI P. Let I Q Q Q be a irredundant primary = = 1 ∩ 2 ··· ∩ k decomposition of I with P pQ . Then by taking radicals and by similar argument as i = i before we can conclude that P is the unique minimal prime ideal of I.

• For any ideal I of a ring R if pI m is an maximal ideal then I is m-primary. =

Proof: Suppose ab I and b pI m.Then, notice that pI m pI (b) which ∈ ∈ = = ⊂ + means that pI (b) R which means that 1 I (b) hence it follows that a I. + = ∈ + ∈ • For any Ideal I of a Noetherian ring R if mn I m for some maximal ideal m and some ⊆ ⊆ positive integer n. Then I is m-primary.

Proof : Since I m so, pI m and since mn I thus m pI. So we have that pI m ⊆ ⊆ ⊆ ⊆ = and by the above result I is m-primary.

• If I is an ideal of a Noetherian local ring (R,m) such that m is minimal prime ideal of I then I is m-primary.

Proof: Let I Q Q Q be a irredundant primary decomposition of I with = 1 ∩ 2 ··· ∩ k P pQ . Since m is the unique maximal ideal so P m for all i and since m is i = i i ⊆ the minimal prime ideal of I it follows that P m for all i. Hence I is m-primary. i = • If the maximal ideal of a local ring is nilpotent then it is the unique prime ideal of the local ring.

Proof:Let P be any other prime ideal then mn (0) P. It follows that m P and hence = ⊆ ⊆ P m. =

4 3.4 SUPPORTING THEOREMS

Nakayama Lemma Let M be a ﬁnitely generated R-module and I be an ideal contained in J(R), the Jacobson radical of R,such that IM M. Then M 0. = = Theorem A ring R is Artinian if and only if it is Noetherian and zero dimensional.

Theorem Let R S be a integral ring extension. Then dim R dim S. ⊆ = Prime Avoidance Lemma n Let I be an ideal with I i 1Pi , Pi prime ideal. Then I Pi for some i. ⊆ ∪ = ⊆ The result holds even if the ﬁrst two ideals P1 and P2 are not prime ideals.

4 MAIN THEOREMS

4.1 PRINCIPAL IDEAL THEOREM

Let R be a Noetherian Ring and P be a minimal prime of an ideal (a) of R. Then ht P 1. ≤ Consequently, ht (a) 1. ≤

Proof: ht P 1 dim(R ) 1 ≤ ⇔ P ≤ By using the correspondence of ideals and prime ideals of a ring and its localization we can replace R by a local ring. Thus we have a Noetherian local ring R and maximal ideal P is the minimal prime of the ideal (a).

By using the correspondence of prime and maximal ideal in a ring and its quotient it follows that R/(a) has a unique prime/maximal ideal. Thus dim R/(a) 0 and R/(a) is Noetherian = hence R/(a) is Artinian.

Let P be any prime ideal of R properly contained in P. Then ht P 0 ht P 1. 1 1 = ⇔ ≤ Clearly a P since P is the minimal prime ideal of (a).Consider the following descending ∉ 1 chain of ideals.

(a) P (a) P (2) .... + 1 ⊇ + 1 ⊇ This corresponds to a descending chain of ideals in the ring R/(a). Since R/(a) is Artinian so the corresponding descending of ideals in R/(a) becomes stationary.Hence there exists a (n) (n 1) positive integer n such that (a) P (a) P + . + 1 = + 1

5 (n) (n 1) (n 1) (n) So, any x P can be written as x ab y for some b R and y P + . Since P + P ∈ 1 = + ∈ ∈ 1 1 ⊆ 1 so ab P (n). P (n) is P -Primary and We have that ab P (n) and a P . It follows that b P (n) ∈ 1 1 1 ∈ 1 ∉ 1 ∈ 1 This gives that (n) (n) (n 1) P (a)P P + . 1 = 1 + 1

(n) (n) (n 1) (n) (n 1) (n) (n 1) P (a)P P + P /P + (a)P /P + . 1 = 1 + 1 ⇒ 1 1 = 1 1 (n) (n 1) (n) (n 1) By applying Nakayama’s Lemma to the R-module P1 /P1 + , we get that P1 P1 + .We (n) n (n 1) (n 1) = know that P1 P1 RP1 R and P1 + P1 + RP1 R. Using the bijection between contracted = ∩ n = n 1 ∩ and extended ideals we get that P R P + R . 1 P1 = 1 P1

n 1 n n 1 n P + P P and localisation preserves ﬁnite product of ideals so we get P + R P R P R , 1 = 1 1 1 P1 = 1 P1 1 P1 hence P nR P R P nR . Applying Nakayama’s Lemma to the R -module P nR , we get 1 P1 = 1 P1 1 P1 P1 1 P1 (P R )n P nR 0. So the maximal ideal of the local ring is nilpotent which means that 1 P1 = 1 P1 = P1RP1 is the unique prime ideal of RP1 .

Thus, dim R 0 which means that ht P 0. This implies that ht P 1. P1 = 1 = ≤

4.2 KRULL’S HEIGHT THEOREM

Let R be a Noetherian Ring and P be a minimal prime of an ideal (a1,...,ar ) of R. Then ht P r . Consequently, ht (a ,...,a ) r . ≤ 1 r ≤ Proof : ht P dim R . So, by using the correspondence of ideals and prime ideals of a ring and its = P localization we can replace R by a local ring.

Thus, we have a Noetherian local ring R and its maximal ideal P is the minimal prime of the ideal (a1,a2,...,ar ). It follows that (a1,a2,...,ar ) is P-Primary.

We will use induction. The case r 1 is the Principal Ideal Theorem. Let r 1 and suppose = > the result holds for an ideal generated by (r 1) elements. −

Let P1 be any prime ideal properly contained in P and is maximal among the prime ideals of R properly contained in P.ht P r 1 ht P r . We will show that ht P r 1.To show 1 ≤ − ⇔ ≤ 1 ≤ − that ht P r 1 we will try to produce an ideal generated by (r 1) elements for which P is a 1 ≤ − − 1 minimal prime ideal and then use the induction hypothesis.

Since P is minimal over (a ,a ,...,a ) hence not all a P . Suppose a P . Then, P 1 2 r i ∈ 1 1 ∉ 1 1 ⊂ P (a ) P. By way of choosing P it follows that P is the minimal prime over P (a ). Since P 1+ 1 ⊆ 1 1+ 1 is the unique maximal ideal of R it follows that P (a ) is P-Primary. Thus, pP (a ) P.R is 1 + 1 1 + 1 = Noetherian so there exists a positive integer n such that P n P (a ). Hence, for i 2,3,...,r ⊆ 1 + 1 = we have that an y ax where y P and x R. i = i + i i ∈ 1 i ∈

6 This means that there exist a large enough positive integer l such that (a ,a ,...,a )l 1 2 r ⊆ (a1, y2,..., yr ). (a1,a2,...,ar ) is P-Primary. So, as before there exist an integer m such that P m (a ,a ,...,a ).This means that P ml (a , y ,..., y ).Since P ml (a , y ,..., y ), it follows ⊆ 1 2 r ⊆ 1 2 r ⊆ 1 2 r that (a1, y2,..., yr ) is P-Primary and hence P is the minimal prime of (a1, y2,..., yr ).

By taking quotient with (y2,..., yr ) it follows that P/(y2,..., yr ) is the minimal prime of (a , y ,..., y )/(y ,..., y ). Hence, by the Principal Ideal Theorem we get that ht (P/(y ,..., y )) 1 2 r 2 r 2 r ≤ 1.P /(y ,..., y ) is properly contained in P/(y ,..., y ) so, ht P /(y ,..., y ) 0. 1 2 r 2 r 1 2 r =

This implies that P1 is a minimal prime ideal of (y2,..., yr ).By the induction hypothesis, ht P r 1. Hence, we can conclude that ht P r . 1 ≤ − ≤

5 CONSEQUENCES

• Let R be a ring and P be a minimal prime over the ideal (a) where a is a nonzero divisor.

If ht P 0, then P is a minimal prime ideal of R, but P contains a nonzero divisor so P = cannot be a minimal ideal of R. Hence, ht P 1. ≥

• Using this we can make the following version of Krull’s Principal Ideal Theorem.

Let R be a Noetherian Ring and P be a minimal prime of an ideal (a) of R. Then ht P 1. ≤ Consequently, ht (a) 1. Moreover, if a is a nonzero divisor then ht P 1 and hence ≤ = ht (a) 1. =

Regular sequence A sequence of elements a ,a ,...,a in a ring R is said to be regular if (a ,a ,...,a ) R, a is 1 2 r 1 2 r 6= 1 a nonzero divisor of R and ai is a nonzero divisor of R/(a1,a2,...,ai 1) for i 2,...,r . − =

By taking quotient with (a1,a2,...,ar 1) and using induction we can conclude the following − version of Krull’s Height Theorem.

Let R be a Noetherian Ring and P be a minimal prime of an ideal (a1,...,ar ) of R. Then ht P r . Consequently, ht (a ,...,a ) r . Moreover, if a ,a ,...,a is a regular sequence then ≤ 1 r ≤ 1 2 r ht P r and hence ht (a ,a ,...,a ) r. = 1 2 r = Few consequences of the Height Theorem

• The height of any prime ideal of a Noetherian ring is ﬁnite.

• The dimension of a Noetherian local ring is ﬁnite.

7 • Let K be a ﬁeld. Consider the ideal m (X , X ,..., X ) of K [X , X ,..., X ]. Notice that = 1 2 n 1 2 n X , X ,..., X is a regular sequence hence it follows that ht m n. 1 2 n = a a • Let K be a ﬁeld and K be an algebraic closure of K . Then, K [X1, X2,..., Xn] is integral over K [X , X ,..., X ]. Hence, dim K [X , X ,..., X ] dim K a[X , X ,..., X ]. 1 2 n 1 2 n = 1 2 n By Hilbert’s Nullstellensatz, any maximal ideal of K a[X , X ,..., X ] is of the form (X 1 2 n 1 − λ , X λ ,..., X λ ) where λ K a. 1 2 − 2 n − n i ∈ Notice that X λ , X λ ,..., X λ is a regular sequence. So, by Krull’s Height Theo- 1 − 1 2 − 2 n − n rem ht (X λ , X λ ,..., X λ ) n. Thus, dim K [X , X ,..., X ] dim K a[X , X ,..., X ] 1− 1 2− 2 n− n = 1 2 n = 1 2 n = n.

As a consequence we can conclude that the dimension of any ﬁnitely generated K - algebra is ﬁnite.

Theorem Let R be a Noetherian ring. Then, R is a UFD if and only if every height one prime ideal is principal. Proof : Suppose R is a UFD and let P be a prime ideal such that ht P 1. Then, clearly P is a non-zero = ideal. So, by using unique factorization property and that P is a prime ideal it follows that there exists an irreducible element a P. Hence, (a) P. ∈ ⊆ The ideal (a) is a prime ideal and ht (a) 1 since R is a domain. But ht P 1, hence it ≥ = follows that (a) P. = Now, Suppose every height one prime ideal is principal. Since R is Noetherian, so it is a factorization domain. It sufﬁces to show that every irreducible element of R is a prime element.

Let a an irreducible element of R and P be a minimal prime over (a). By Krull’s Height Theorem, ht P 1. Since R is a domain hence ht P 1. ≤ = So P is a principal ideal. Let P (b). Then we have that (a) (b) and (a) is an irreducible = ⊆ element, hence (a) (b). So, (a) is a prime ideal, hence a is a prime element. = Theorem Let R be a Noetherian ring and I an ideal of R such that ht I n. Then there exists a ,a ,...,a = 1 2 n ∈ I such that ht(a ,a ,...,a ) r 1 r n. 1 2 r = ∀ ≤ ≤ Proof :

We will use induction on r . Let ht I 1. Let P ,P ,...,P be the minimal prime ideals of R. ≥ 1 2 l

8 Since ht I 1 and height of a minimal prime ideal of R is zero, it follows that I P for ≥ ⊂ j 1 j l. By prime avoidance, it follows that I P P P . ≤ ≤ ⊂ 1 ∪ 2 ··· ∪ l

Choose a I (P P P ). It follows that ht (a) 1. But, by Principal Ideal Theorem ∈ − 1 ∪ 2 ··· ∪ l ≥ ht(a) 1. Hence, we get that ht (a) 1. This proves the result for r 1. ≤ = =

Suppose i 1 and we have chosen a1,a2,...,ai 1 I such that ht (a1,a2,...,ai 1) i 1. > − ∈ − = − Let Q1,Q2,...,Qk be the minimal primes over (a1,a2,...,ai 1). By Krull’s Height Theorem, − ht Q i 1 i ht I. Hence, it follows that I Q for 1 j k. By prime avoidance, j ≤ − < = ⊂ j ≤ ≤ I Q Q Q . ⊂ 1 ∪ 2 ··· ∪ k

Choose a I (Q Q Q ). By, Height Theorem, ht (a ,a ,...,a ) i. Let P 0 be any i ∈ − 1 ∪ 2 ··· ∪ k 1 2 i ≤ minimal prime ideal over (a1,a2,...,ai ). Since P 0 contains ai , so it cannot be a minimal prime over (a1,a2,...,ai 1) and since ht (a1,a2,...,ai 1) i 1. Thus ht P 0 i 1. It follows that − − = − > − ht (a ,a ,...,a ) i. 1 2 i = This completes the induction and the result holds true for 1 r n. ≤ ≤

REFERENCES

[1] Sudhir R. Ghorpade, Lectures on Commutative Algebra http://www.math.iitb.ac.in/ srg/Lecnotes/AfsPuneLecNotes.pdf

[2] N S Gopala Krishnan, Commutative Algebra, University Press.

[3] I. Kaplansky, Commutative Rings, The University of Chicago Press,1974.