Localisation in P.I. Rings Gwyn Bellamy University ID : 0210766 Tutor : Professor C.R. Hajarnavis March 10, 2006 Abstract In order to localise with respect to some multiplicative set S in a noncommu- tative ring, certain conditions must be satisfied by S . When such conditions are satisfied, the resulting localisation is either a left or right localistion (as will be explained) but, in general, not both. The main aim of this paper is to show that in the case where R is a prime Noetherian P.I. ring, all localisations are in fact two sided and can be thought of as a localisation with respect to a certain collection of prime ideals of R. The project is split into seven chapters which we outline below. Chapter 1 - Here we introduce P.I. rings and outline some of their basic properties. Chapter 2 - In this chapter we state and prove two important theorems regarding P.I. rings. These are Kapansky’s Theorem and Posner’s Theorem. Chapter 3 - Using Posner’s Theorem, we introduce the Trace Ring of a prime P.I. ring R. We show that, in this extension ring, localisation is particularly well- behaved. Chapter 4 - Here we introduce, more formally, the idea of localisation in a non- commutative ring. We also introduce the idea of links between prime ideals. Chapter 5 - In this chapter a new class of rings, called F.B.N. rings, are introduced. It will be shown that Noetherian P.I. rings belong to this class. Chapter 6 - Here we develop the idea of Krull/Classical Krull dimensions and show that for F.B.N rings these dimensions coincide. Chapter 7 - Finally, in this chapter we tackle the proof of our main theorem. A description of the link structure of a prime Noetherian P.I. ring is also given.
Transcript
Localisation in P.I. Rings
Gwyn BellamyUniversity ID : 0210766
Tutor : Professor C.R. Hajarnavis
March 10, 2006
Abstract
In order to localise with respect to some multiplicative set S in a noncommu-tative ring, certain conditions must be satisfied by S. When such conditions aresatisfied, the resulting localisation is either a left or right localistion (as will beexplained) but, in general, not both. The main aim of this paper is to show thatin the case where R is a prime Noetherian P.I. ring, all localisations are in fact twosided and can be thought of as a localisation with respect to a certain collection ofprime ideals of R.
The project is split into seven chapters which we outline below.
Chapter 1 - Here we introduce P.I. rings and outline some of their basic properties.Chapter 2 - In this chapter we state and prove two important theorems regardingP.I. rings. These are Kapansky’s Theorem and Posner’s Theorem.Chapter 3 - Using Posner’s Theorem, we introduce the Trace Ring of a primeP.I. ring R. We show that, in this extension ring, localisation is particularly well-behaved.Chapter 4 - Here we introduce, more formally, the idea of localisation in a non-commutative ring. We also introduce the idea of links between prime ideals.Chapter 5 - In this chapter a new class of rings, called F.B.N. rings, are introduced.It will be shown that Noetherian P.I. rings belong to this class.Chapter 6 - Here we develop the idea of Krull/Classical Krull dimensions and showthat for F.B.N rings these dimensions coincide.Chapter 7 - Finally, in this chapter we tackle the proof of our main theorem. Adescription of the link structure of a prime Noetherian P.I. ring is also given.
In this paper we aim to explore some of the properties that rings satisfying polynomialidentities possess. In particular, we aim to develop a proof of the theorem proved byAmiram Braun and Robert B. Warfield Jr. which shows that localisation in primeNoetherian polynomial identity rings is symmetrical. Specifically,
Main Theorem. Let R be a prime Noetherian P.I. ring, and S a right (left) Ore setin R, then there exists a two sided Ore set S ′ such that S ⊆ S ′ and the right (left)localisation of R by S equals the two sided localisation of R by S ′.
Polynomial identity rings can be thought of as being almost commutative and manyof the results for commutative rings, which do not hold for general noncommutativerings, can be seen to be true if the ring satisfies a polynomial identity.
The proofs of many of the theorems quoted below are taken from various books onring theory. When this is the case a reference to the book and the proof involved ismade at the beginning of the proof, e.g.
Proof [McR] 13.1.9
means the proof is adapted from the book [McR], section 13.1.9. Whenever thisis the case every effort has been made to both simplify and clarify the proof, ratherthan copy it directly from the book. If no such reference is made at the start of theresult/proof, then the proof is the author’s own work.
1.2 Basic Properties of P.I. Rings
Definition 1.1. Let Z〈x1 . . . xn〉 be the free ring generated over Z by the noncommutingindeterminants x1 . . . xn. Then we say a ring R is a polynomial identity (P.I.) ring ifthere exists a monic polynomial f ∈ Z〈x1 . . . xn〉\{0} such that f(r1, . . . rn) = 0 for allr1, . . . rn in R. Such an f is called a polynomial identity for the ring R.
The obvious example is A, any commutative ring. Then if we take f(x1, x2) =x1x2−x2x1, we have f(r1, r2) = 0 for all r1, r2 ∈ A. Similarly, if we consider R = M2(A),for some commutative ring A, then, for r1, r2 ∈ R, (r1r2− r2r1)2 will be a scalar matrix,hence f(x1, x2, x3) = (x1x2− x2x1)2x3− x3(x1x2− x2x1)2 will be a polynomial identityfor R which is a P.I. ring.
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Firstly we begin by showing that if R is a P.I. ring then we can choose a ”nice”polynomial identity for it.
Definition 1.2. Let f ∈ Z〈x1, . . . xn〉. Then we say that f is a multilinear polynomialof degree n if
f(x1, . . . xn) =∑
σ∈Sn
aσxσ(1) . . . xσ(n)
where aσ(k) ∈ Z for 1 ≤ k ≤ n. This is equivalent to saying that f(c1x1, . . . cnxn) =c1c2 . . . cnf(x1, . . . xn), for all c1, . . . cn ∈ Z.
Proposition 1.3 ([McR] 13.1.9). If R satisfies a polynomial identity f of degree dthen R also satisfies a multilinear identity g of degree at most d. Furthermore, eachcoefficient of g is also a coefficient of f , and, if f is monic, then so too is g.
[McR] 13.1.9. We can write f(x1, x2, . . . xn) = f1(x1, x2 . . . xn)+f2(x2, . . . xn) where x1
occurs in every monomial of f1. Then by setting x1 = 0 we see that f2, and hence f1 arealso identities of R. Since f is monic, at least one of f1, f2 must be monic so we acceptthe monic one and drop the other. Proceeding by induction, we get a new monic fwho’s coefficients are all coefficients of the original f and each monomial contains everyindeterminant xi, though our new f may have smaller degree.Now if f is not already multilinear then there exists some indeterminant, x1 say, whichhas degree greater than one in one of the monomials of f . Therefore we consider
in which the maximum degree of x1 is one less than what it was in f . Also, thoughthe monomials of g are different to those of f , the coefficient of each monomial in g isthe coefficient of some monomial of f and vice versa, hence g is still monic. Repeating,we can reduce the degree of x1 to one in every monomial. Note that, even though thenumber of indeterminants involved has increased, the degree of every monomial is stillthe same. Proceeding inductively, we may assume that every indeterminant of g hasdegree one in every monomial i.e. g is multilinear.
Now we consider the specific ring Mn(A), where A is some commutative ring.
Proposition 1.4 ([McR] 13.3.2). If A is a commutative ring then Mn(A) satifies nomonic identity of degree less than 2n.
[McR] 13.3.2. Let f be a monic identity for Mn(A), then by the above proposition wemay assume that f is a multilinear polynomial. Hence
f = x1 . . . xt︸ ︷︷ ︸monic leading term
+∑
σ∈St\{1}aσxσ(1) . . . xσ(t)
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Assume that t ≤ 2n − 1. Now choose r1 . . . rt to be the first t elements in the liste11, e12, e22, e23, e33, . . . enn, where eij is the matrix with a one in the i, jth entry andall other entries zero. Then for σ 6= 1, rσ(1) . . . rσ(t) contains a product rirj wherej 6= i, i + 1, i− 1. But then titj = 0 so f(r1, . . . rt) = r1 . . . rt + 0 = e1[ t
2] 6= 0. This is a
contradiction, hence deg(f) ≥ 2n as required.
An important polynomial that often occurs as an identity for some P.I. rings is theone defined below. From the definition we are able to state the following theorem dueto Amistur and Levitzki, which we shall require later.
Definition 1.5. We define the nth standard identity (or standard polynomial) as
sn =∑
σ∈Sn
(−1)sign(σ)xσ(1) . . . xσ(n)
Then sn is a monic polynomial of degree n.
Theorem 1.6 (Amistur, Levitzki). If A is a commutative ring then Mn(A) satisfiesthe polynomial s2n.
Proof. The proof is combinatorial and will be omitted. See [Row2], page 21, for aproof.
Therefore any matrix ring over a commutative ring is a P.I. ring. This together withthe following lemma will give us a major source of P.I. rings.
Lemma 1.7 ([McR] 13.1.7). 1. Any subring or homomorphic image of a P.I. ringis a P.I. ring.
2. A finite direct sum of P.I. rings is a P.I. ring.
[McR] 13.1.7. 1. If f is a monic identity for a ring R then it is also a monic identityfor any subring or homomorphic image of R.
2. If f1, . . . fn are monic identities for R1, . . . Rn then f =∏n
i=1 fi is a monic identityfor
⊕ni=1 Ri.
Therefore any algebra which is finite dimensional over its centre is a P.I. ring.Finally, we note the following definition which will be useful later.
Definition 1.8. Let R, S be rings with R ⊆ S. Then we say that S is a central extensionof R if S is generated as a ring by R and the centre of S.
If R is a P.I. ring, f a multilinear identity for R, and S a central extension of R thenS also satisfies f and hence is also a P.I. ring.
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2 The Theorems of Kaplansky and Posner
The main aim of this section is to prove the important theorems concerning P.I. ringsdue to Kaplansky and Posner. We use Kaplansky’s Theorem to prove Posner’s Theoremwhich shows that a prime P.I. ring is an order in a simple Artinian ring Q.
2.1 Kaplansky’s Theorem
Kaplansky’s Theorem tells us that all primitive P.I. rings have a particularly nice struc-ture - they are all central simple algebras (defined below). To prove Kaplansky’s Theo-rem we will need some results concerning general rings. We shall state and prove thesenow. Firstly a definition:
Definition 2.1. Let R be a ring and M a faithful, simple right R-module. ThenR is said to act densely on M if, for each n ∈ N and v1, . . . vn ∈ M , with v1, . . . vn
linearly independent over D = EndMR (which is a division ring by Schur’s Lemma),and w1, . . . wn ∈ M there exists a r ∈ R such that wi = vir for 1 ≤ i ≤ n.
Theorem 2.2 (The Density Theorem, [Her] 2.1.2). Let R be a primitive ring andM a faithful simple right R-module. Then R acts densely on M .
[Her] 2.1.2. We view M throughout as a vector space over D = EndMR.To prove the theorem it is enough to show that for any finite dimensional subspaceV of M , V 6= M and any m ∈ M\V , there exists an r ∈ R such that V r = 0 andmr 6= 0. To see this, let v1, . . . vn be linearly independent, w1, . . . wn ∈ M , then letVi = SpD(v1, . . . vi−1, vi+1, . . . vn). Then vi ∈ M\Vi, so there exists ri ∈ R such thatviri 6= 0 and Viri = 0. Now viri 6= 0 and M is simple so viriR = M , and there existssi ∈ R with virisi = wi. We set ti = risi and t = t1 + · · ·+ tn which gives us vitj = δijwj
and vit = wi as required.We prove the result by induction on dim(V ). If dim(V ) = 0 then, for m ∈ M\{0}, theremust exist (by the simplicity of M) a r ∈ R with mr 6= 0. Now let V = V0 + wD, wherew /∈ V0 and dim(V0) = dim(V ) − 1. By the inductive hypothesis, if m ∈ M\V0 thenthere exists a ∈ annR(V0) such that ma 6= 0 i.e. mannR(V0) = 0 iff m ∈ annR(V0). Soif m ∈ M\V0, 0 6= mannR(V0) is a submodule of M and hence is M itself. In particular,wannR(V0) = M . We want to show that mannR(V ) 6= 0 for all m ∈ M\V . So assumem ∈ M\V such that mannR(V ) = 0. We define τ : M −→ M by τ(x) = ma wherea ∈ annR(V0) such that x = wa. τ is well defined: if x = 0 then x = wa wherea ∈ annR(V0) ∪ annR({w}) = annR(V ) so τ(x) = ma = 0 since mannR(V ) = 0 byassumption.Now let x = wa ∈ M with a ∈ annR(V0) then xr = (wa)r = w(ar) so τ(xr) =m(ar) = (ma)r = τ(x)r and τ ∈ EndMR = D. Now for any a ∈ annR(V0) we get
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ma = τ(wa) = τ(w)a. So (m−τw)a = 0 i.e. m−τw ∈ V0 and hence m ∈ V0 +Dw = V .This is a contradiction.So for all m ∈ M\V, mannR(V ) = M and hence 0 ( annR(V ).
The following lemma follows directly from the Density Theorem.
Lemma 2.3 ([McR] 13.3.7). Let R be a primitive ring with a faithful simple rightR-module VR and let D = EndVR. Then either dimD V = n and R ∼= Mn(D) for somen < ∞, or else, for each n < ∞, there is a subring Sn of R together with a surjectivering homomorphism Sn → Mn(D).
Proof. If dimD V = n < ∞ then EndDV ∼= Mn(D) (after fixing some basis of V ). Inthis case the Density Theorem says that R → EndDV is an isomorphism. Otherwise,for each n < ∞, choose some subspace Vn of V of dimension n over D. Then letSn = {r ∈ R|Vnr ⊆ Vn}. It can easily be shown that Sn is a subring of R and the actionof Sn on Vn defines a surjective homomorphism into Mn(D) (after fixing some basis ofVn).
Since Kaplansky’s theorem invloves simple algebras we need the following result:
Lemma 2.4 ([McR] 9.6.9 i)). Let U be a simple ring with centre k and V a k-algebra.Then the map A 7→ A⊗U provides a 1-1 correspondence between ideals of V and idealsof V ⊗ U .
[McR] 9.6.9 i). Let A C V , then clearly A ⊗ U C V ⊗ U . Now let B C V ⊗ U and letA = {v ∈ V |v ⊗ 1 ∈ B}. Then A C V . I claim that: if A 6= 0 then A⊗ U = B. Assumenot, then we choose b =
∑ni=1 vi ⊗ ui in B\A ⊗ U which is minimal with respect to n.
Let 0 6= v ⊗ 1 ∈ A ⊗ U and consider b − (1 ⊗ u1)(v ⊗ 1) =∑n
i=2 vi ⊗ ui ∈ B. Theneither
∑ni=2 vi⊗ui is in B\A⊗U or v1⊗u1 belongs to B\A⊗U . In both cases we have
contradicted the minimality of n. So A⊗ U = B.Therefore we need to show that B 6= 0 implies A 6= 0. Again we choose some non-zerob =
∑ni=1 vi ⊗ ui in B with v1, . . . vn linearly independent so as to minimise n. Then
u1 6= 0 and, since U is simple, there exist a1 . . . ak, b1 . . . bk such that∑k
j=1 aju1bj = 1.Hence
∑kj=1(1⊗ aj)b(1⊗ bj) has first term v1 ⊗ 1 but still has n minimal. Then, for all
u ∈ U, (1⊗u)b− b(1⊗u) has fewer than n terms and hence must be zero (since it is stillin B and n is minimal). This says that uui − uiu = 0 for 1 ≤ i ≤ n. Therefore each ui
lies in the centre of U which is k. So 0 6= b = (∑n
i=1 viui)⊗1 and 0 6= ∑ni=1 viui ∈ A.
The above lemma enables us to prove the following result which will be very impor-tant in proving Kaplansky’s theorem. In what follows, [A : B] will be used to denotethe dimension of A as a vector space over B.
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Lemma 2.5 ([McR] 13.3.4). Let R = Mt(D) for some division ring D. Let H be amaximal subfield of D, F the centre of D and VR a simple module so D ∼= EndVR. Then
1. S = R⊗F H is simple and V is a simple S-module with EndVS∼= H.
2. If V has dimension m < ∞ over H then S ∼= Mm(H), [D : H] = [H : F ] = mt and
[R : F ] = m2.
[McR] 13.3.4. 1. The previous Lemma (2.4) shows that S is simple. We give V a S-module structure by defining v(r⊗h) = hvr. This is well defined: (v+w)(r⊗h) =h(v + w)r = hvr + hwr = v(r ⊗ h) + w(r ⊗ h), [v(r ⊗ h)](s⊗ g) = (hvr)(s⊗ g) =g(hvr)s = (gh)v(rs) = v(rs ⊗ gh) = v[(r ⊗ h)(s ⊗ g)]. V is clearly a simpleS-module since any submodule is also a R-submodule.
EndVS = {f : V → V | f is a left H-homo and a right R-homo }
hence EndVS ⊆ EndVR = D, and for α ∈ D we have α ∈ EndVS if and only ifα(v(1⊗h)) = α(v)(1⊗h) for all h ∈ H, v ∈ V . That is, α(hv) = h(αv). So EndVS
is the centraliser of H in D. Since H is a maximal subfield of D, the centraliserof H must be H itself and EndVS
∼= H.
2. EndVS∼= H means that we can view V as a vector space over H. Now assume
dimH V = m < ∞. Then, since V is a faithful R-module, it must be a faithfulS-module. Hence S is a primitive ring and we can apply Lemma 2.3 to concludethat S ∼= Mm(H). Using the fact that dimU ⊗ W = dimU × dimW , for allvector spaces U , V , we have [S : F ] = [R ⊗F H : F ] = [R : F ][H : F ]. But[S : F ] = [S : H][H : F ], hence [S : H] = [R : F ]. Now [R : D] = t2 so m2 =[R : F ] = [R : D][D : F ] = t2[D : F ] and [D : F ] = m2
t2. Similarly, since
V ∼= Dt ∼= Hm, we get m = [V : H] = [V : D][D : H]. So [D : H] = mt . Finally,
[H : F ][D : H] = [D : F ] implies [H : F ] = mt .
In order to state Kaplansky’s Theorem we need the following definition:
Definition 2.6. Let R be a k-algebra. We say that R is a central simple algebra ifthe centre of R is k and R is a simple, finite dimensional, algebra. Note that since Ris finite dimensional, it is an Artinian ring. Hence, by the Artin-Weddeburn Theorem,R ∼= Mn(D) for some division ring D.
Corollary 2.7 ([McR] 13.3.8). Let R = Mt(D) be a central simple algebra, where Dis a division ring with centre F . Let H be a maximal subfield of D. Then:
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1. D⊗F H ∼= Mm(H) where m2 = [D : F ] and R⊗F H ∼= Mn(H) where n2 = (mt)2 =[R : F ].
2. R satisfies s2n but not s2n−1, where sk is the polynomial defined in Definition 1.5.
Proof. 1. D ⊗F H ∼= Mm(H) with [D : F ] = m2 follows from Lemma 2.5, part i).Then, by part ii), we have n2 = [R ⊗F H : H] = [R : F ] = [R : D][D : F ], son2 = t2m2 = (mt)2.
2. Follows from Proposition 1.4 and Theorem 1.6.
Finally, we are ready to prove Kaplansky’s theorem.
Theorem 2.8 (Kaplansky’s Theorem, [McR] 13.3.8). If R is a primitive P.I. ringof minimal degree d then R is a central simple algebra of dimension
(d2
)2over its centre.
[McR] 13.3.8. By Lemma 2.3, either R ∼= Mn(D) or, for all n ∈ N, there exist a subringSn of R and surjective homomorphism Sn −→ Mn(D), where D is some division ring.But, by Proposition 1.4, the minimal degree of Mn(F ) is at least 2n (where F is thecentre of D), hence so too are the minimal degrees of Mn(D) and Sn. This implies thatR is not a P.I. ring. Therefore R ∼= Mn(D), and R is simple.Let VR be a simple module for R and denote by F the centre of D, H a maximalsubfield of D. Then, by Lemma 2.5 i), R ⊗F H = S is simple, V is a simple S-moduleand EndVS
∼= H. If f is a multilinear polynomial identity for R then it is also one for S,hence S is a P.I. ring. Applying the same argument to S as we did to R in the precedingparagraph, we get S ∼= Mt(H) and dimVH = t < ∞. Then Lemma 2.5 ii) says that[R : F ] = t2 < ∞ so R is a central simple algebra.Corollary 2.7 ii) says that R satisfies s2t but not s2t−1. Therefore d = 2t and [R : F ] =(
d2
)2.
2.2 Posner’s Theorem
Using Kaplansky’s Theorem we can prove Posner’s Theorem which says that any primeP.I. ring is an order in a central simple algebra. We begin by introducing central poly-nomials.
Definition 2.9. Let R be a P.I. ring with centre C and g ∈ Z〈x1, . . . xn〉. Then we saythat g is a central polynomial for R if 0 6= g(R) ⊆ C.
An obvious example is R = M2(A) for some commutative ring A and g(x1, x2) =(x1x2 − x2x1)2. Then, for all r, s ∈ R, g(r, s) is a scalar matrix hence lives in the centre
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of R. g(e12, e21) = I2 so g(R) 6= 0 and g is a central polynomial for R.
The obvious question is: Which P.I. rings have a central polynomial? A partialanswer to this is that all rings of the form Mn(A), for some commutative ring A have acentral polynomial. This fact will enable us to prove Posner’s Theorem. We will statethis here as a theorem, but will not give a proof. However, in proving this theorem,a series of polynomials is constructed which have some useful properties. We describethese properties here, they will be required later.
Definition 2.10. Let g(x1, . . . xn) ∈ Z〈x1, . . . xn〉, then we say that g is a t-alternatingpolynomial, for some 1 ≤ t ≤ n, if interchanging any two of the first t indeterimantstakes g to −g. This can be rewritten as
g(xσ(1), . . . xσ(t), xt+1, . . . xn) = sign(σ)g(x1, . . . xt, xt+1, . . . xn) for all σ ∈ St.
Theorem 2.11 ([Row1] 6.1.20). Let A be a commutative ring, R = Mn(A) and Cthe centre of R, then there exists a multilinear, t-alternating polynomial gn, of degree4n2, such that 0 6= gn(R) ⊆ C, with t = n2.
[Row1] 6.1.20. See [Row1], pages 443/444 for a proof ([McR], page 462 contains a similarproof). The proof given is constructive, a family of polynomials gn are defined in termsof Capelli polynomials and it is shown that 0 6= gn(Mn(A)) ⊆ C for all n.
It is the following theorem, which makes use of the idea of central polynomials, thatwe will require in order to prove Posner’s Theorem. But first we need a result to dowith R[x], for semiprime P.I. rings R.
Theorem 2.12. If R is a ring with no nil ideals then R[x] is a semiprimitve ring.
In order to prove the theorem we need the following two results. The first is astandard result in commutative ring theory and won’t be proved here.
Lemma 2.13. Let R be a commutative ring and N(R) the nilpotent radical of R. ThenN(R) equals the intersection of all prime ideals of R.
Proof. See [Reid], page 28.
Lemma 2.14 ([Her] 6.1.2). Let R be a commutative ring with identity, if a0 + a1x +. . . anxn ∈ R[x] is invertible in R[x] then a0 is a unit in R and a1 . . . an are nilpotent.
[Her] 6.1.2. Let p = a0 + a1x + . . . anxn and q = b0 + b1x + . . . bmxm ∈ R[x] such thatpq = 1. Then a0b0 = 1 and a0 is invertible.Now let P be a prime ideal of R, then P [x] is an ideal of R[x] and R[x]/P [x] ∼= (R/P )[x].R/P is an integral domain so if f ∈ (R/P )[x] is a unit then f ∈ R/P . Hence p+P [x] =
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a0 + P [x] and a1x + . . . anxn ∈ P [x]. This means that a1 . . . an are in P . The argumentholds for all primes P of R, hence a1 . . . an are in the intersection of all primes of R. ByLemma 2.13, a1 . . . an are nilpotent.
Now we can prove theorem 2.12.
Proof. Let J(R[x]) be the Jacobson radical of R[x] and M the set of all non-zero elementsof J(R[x]) of minimal degree. Then let LC = { leading coefficient of p : p ∈ M} ∪ {0},it can easily be seen that LC is an ideal of R.I claim LC is a nil ideal of R. Let p ∈ M, p = p0 + · · · + pnxn, we need to show thatpn is nilpotent. Note that ppn − pnp has lower degree than p but is still in J(R[x]),therefore it must be 0. Hence pn commutes with all pi. Repeating with pn−1 showsthat it too commutes with all pi. Inductively, we get that pipj = pjpi for 0 ≤ i, j ≤ n.Now p ∈ J(R[x]) so xp ∈ J(R[x]), and therefore is quasi-regular, so there exists q =q0 + . . . qmxm ∈ R[x] such that (1− xp)q = 1. This gives q0 = 1, −p0q0 + q1 = 0 that isq1 = p2
and qk+1 = fk+1(p0, . . . pk). Therefore, if C is the centre of R, we have p, q ∈C[p0, . . . pn][x], which is a commutative polynomial ring. Hence, by Lemma 2.14 appliedto 1− xp, the coefficients of p are all nilpotent. In particular, pn is nilpotent.
To be able to use the above result we prove the following:
Theorem 2.15 ([McR] 13.2.5). A nonzero P.I. ring with nonzero nil right idealcontains a nonzero nilpotent ideal.
[McR] 13.2.5. We show that if S is a nonzero nil P.I. ring then it contains a nonzeronilpotent ideal I. Then if S is our nil ideal in R, I will be a nonzero nilpotent rightideal of R and hence is contained I + RI which is a nilpotent ideal of R.Therefore, without loss of generality, we assume R is a nonzero nil P.I. ring. We proceedby induction on d, the minimal P.I. degree of R (that is R satisfies some monic polynomialof degree d and no polynomial of lower degree). For d = 2, R will satisfies some monicmultilinear polynomial of the form x1x2 − nx2x1, for some n ∈ Z. Since R is nil we canchoose 0 6= a ∈ R such that a2 = 0. Then Ra = naR and (aR)2 = a(Ra)R = na2R = 0so I = Za + aR is contained in a nonzero nilpotent ideal as required.Now let m > 2 and assume the result holds for m − 1. Let R satisfy the multilin-ear polynomial f(x1, . . . xm). Then we can write f(x1, . . . xm) = f1(x1, . . . xm−1)xm +f2(x1, . . . xm), where no monomial of f2 ends in xm. f1 is a multilinear polynomial ofdegree m − 1. Again we choose 0 6= a in R such that a2 = 0, if r1 . . . rm−1 in R then
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f2(ar1 . . . arm−1, a) = 0 since each monomial of f2 contains a.ark = 0 for some k. Sof1(ar1, . . . arm−1)a = 0.Consider the ideal M = {r ∈ aR|raR = 0} of aR. Then f1 is a monic identity foraR/M because f1(aR)a = 0 ⇒ f1(aR)aR = 0 and f1(aR) ⊆ M . Either aR/M = 0and aR is nilpotent or applying the inductive hypothesis to aR/M , we conclude thatthere exists a nonzero ideal of aR with In ⊆ M for some n ∈ N. Then IaR Cr Rand (IaR)n+1 ⊆ InIaR ⊆ WIaR ⊆ War = 0. So IaR is contained in some nonzeronilpotent ideal of R.
From Theorems 2.12 and 2.15 we are able to deduce the following corollary whichwe will need in order to prove Posner’s theorem.
Corollary 2.16. Let R be a semiprime P.I. ring, then R[x] is a semiprimitive P.I. ring(where x is an indeterminant commuting with all elements of R).
Proof. If R is semiprime then it has no nonzero nilpotent ideals. Therefore, by The-orem 2.15 it has no nozero nil ideals. It then follows from Theorem 2.12 that R[x] issemiprimitive.
The above corollary allows us to prove the following Proposition which is the lastresult we need before tackling Posner’s Theorem (the Proposition is similar to Theorem13.6.4 in [McR], though the proof differs).
Proposition 2.17. Let R be a semiprime P.I. ring with centre C and let I be a nonzeroideal of R. Then I ∩ C 6= 0, that is, any nonzero ideal of R contains a nonzero centralelement.
Proof. If ICR then I[x]CR[x], and if I[x]∩C(R[x]) = I[x]∩C[x] 6= 0 then I∩C 6= 0. Sowe may assume, by Corollary 2.16, that R is a semiprimitive ring. J(R) = 0 means thatthe intersection of all primitive ideals of R is zero. Therefore there exists a primitiveideal P of R such that (I + P )/P 6= 0. Since R is a P.I. ring, Kaplasky’s Theorem (2.8)says that R/P is a simple ring for any primitive ideal P of R. That is, every primitiveideal of R is maximal. It follows that either (I +P )/P = R/P or is zero. Let {Mi|i ∈ Λ}be the set of all maximal ideals of R. Choose some Mi, i ∈ Λ, then Kaplansky’s Theoremsays that R/Mi
∼= Mmi(Di), for some mi ∈ N and division ring Di. If Hi is a maximalsubfield of Di, then Lemma 2.5 tells us that Mmi(Di)⊗Hi
∼= Mni(Hi), for some ni ∈ N.Therefore R/Mi ↪→ Mni(Hi). By the proof of Theorem 2.11, we know that there existsa multilinear polynomial gni of degree 4n2
i such that 0 6= gni(Mni(Hi)) ⊆ C(Mni(Hi)).Since Mni(Hi) is a central extension of R/Mi, gni zero on R/Mi would imply that itwas also zero on Mni(Hi), therefore this does not happen. If d is the minimal de-gree of R, then di, the minimal degree of R/Mi, will be less than or equal to d for alli ∈ Λ, and hence {ni|i ∈ Λ} is bounded (ni = di
2 ). Let n be this maximum. Thus,
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gn(R/M) ⊆ C(R/M) for all maximal ideals M of R. Since the intersection of all theseideals is zero, gn(I) ⊆ C(R). But we know that for at least one M , gn(R/M = I/M) 6= 0,therefore 0 6= gn(I).
Finally, we have the tools necessary to tackle Posner’s theorem.
Theorem 2.18 (Posner’s Theorem, [McR] 13.6.5). Let R be a prime P.I. ring withcentre C. Let S = C\{0}. Then the two sided localisation Q = RS exists. Moreover, Qis a central simple algebra with centre F = CS , R is an order in Q and Q = RF is acentral extension of R.
Proof. Since R is a prime ring, C is an integral domain and hence S is both a left andright denominator set (Definition 4.5). Therefore, by Theorem 4.6 below, Q must exist.Now if I is any ideal of Q then J = I ∩R is clearly an ideal of R and J = 0 if and onlyif I = 0. So let 0 6= I C Q and J = I ∩R. Then, by Proposition 2.17, C ∩ J 6= 0. Hencethere exists 0 6= c ∈ C ∩ I. Then cc−1 = 1 ∈ J and J = Q. Therefore Q is simple.Q is a central extension of R and hence is a P.I. ring. Since it is simple we can applyKaplasky’s Theorem 2.8 and conclude that Q is a central simple algebra.Let ac−1, a ∈ R, c ∈ S be in F the centre of Q and bd−1, b ∈ R, d ∈ S be any elementof Q. Then ac−1bd−1 = bd−1ac−1 implies ab = ba (since we can multiply through bycd ∈ C). Therefore a ∈ C and ac−1 is an element of the field CS . But clearly CS ⊆ F ,so CS = F as required. It follows that Q = RF , since any element of Q has the formac−1 for a ∈ R, c ∈ S.
3 The Trace Ring
Let R be a prime P.I. ring. Then, in order to define the trace ring of R, we create anembedding of R into some matrix ring over a field. This is done using Posner’s Theorem(Theorem 2.18). Let C be the centre of R, S = C\0, Q = RS and F = CS . Then F isa field. Let n = dimF Q, left multiplication by r ∈ R defines a linear map rl : Q −→ Q,and, after fixing a basis for Q, we get a map L : R −→ Mn(F ), r 7→ rl. This is clearly aring homomorphism and kerL = {r ∈ R|rx = 0 for all x ∈ Q}. But 1 ∈ Q so kerL = 0and L is an embedding of R in Mn(F ).
Since rl ∈ Mn(F ), it has a characteristic polynomial, fr ∈ F [x], of degree n withfr(rl) = 0 (it follows that fr(r) = 0 too). Now we let T be the commutative subring ofF generated by C and all coefficients of the polynomials {fr|r ∈ R}. Then we defineT (R) ⊆ Q to be the central extension of R generated by R and T . T (R) has the property
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that every element of R is integral over the centre of T (R). We call T (R) the trace ringof R. T (R) has many useful properties which we will exploit in the next chapter.
3.1 Shirshov’s Theorem and Integrality
Firstly we show that any element of T (R) is integral over T . To be able to do this weneed the following theorem which we will not prove here.
Theorem 3.1 (Shirshov’s Theorem). Let C be a commutative ring, R = C〈x1, . . . xn〉a P.I. ring, and x1, . . . xn all integral over C. Then R is a finitely generated C-moduleand is spanned by all words in x1, . . . xn of length less that m, for some fixed m ∈ N.
Proof. The proof will not be given here but can be found in [Row2], page 206, or [Row1],page 475. The idea of the proof is to define an ordering on words in the xi and thenshow that any word of length greater than m can be written as a sum, over C, of wordsof lower length, or of lower order. This is done by noting first that, since xi is integralover C, there exists an ni such that xk
i can be written as a sum of xli, with l ≤ ni, for
all k > ni. Then, once this has been done, we apply the polynomial identity of R to theword so that it is written as a sum of words lower in the ordering. Induction completesthe proof.
Proposition 3.2. Let R be a prime P.I. ring, T (R) the trace ring of R. Then T (R) isintegral over T .
Proof. To prove this we use the “determinant trick” as described in [Reid], page 43. Letx ∈ T (R), then x =
∑ki=1 tixi, where ti ∈ T and xi ∈ R. Therefore x ∈ T 〈x1, . . . xk〉
with each xi integral over T . So, by Theorem 3.1, T 〈x1, . . . xk〉 is a f.g. T -module andcan be written as Ty1 + . . . Tym, for some yi ∈ T (R) and m ∈ N.We can now apply the “determinant trick”. x belongs to Ty1 + · · ·+Tym so xyi ∈ Ty1 +· · ·+Tym, for each 1 ≤ i ≤ m. This means that xyi =
∑mj=1 aijyj with aij ∈ T . Therefore
xyi−∑m
j=1 aijyj = 0 and∑m
j=1(xδij−aij)yj = 0. We set M = (xδij−aij)ij ∈ Mm(T [x]).If v = (y1, . . . ym)T , then Mv = 0 and MadjMv = 0. But MadjM = (det(M))Im, hencewe can conclude that det(M)yi = 0, for each 1 ≤ i ≤ m. Now 1 ∈ Ty1 + . . . T ym,so 1 =
∑mi=1 tiyi. Then det(M) = det(M).1 =
∑mi=1 ti det(M)yi = 0. But det(M) =
det(xδij − aij) is a monic polynomial in T [x], so x is integral over T .
Corollary 3.3. Let R be a prime P.I. ring, then T (T (R)) = T (R).
3.2 Localisation of T (R)
Next we show that localisation in T (R) can always be thought of as a localisation withregard to some central multiplicative set. This is the result that we hope to extend to
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a general prime P.I. ring in the final chapter. Localisation is not defined until the nextchapter, therefore, if the reader is unfamiliar with noncommutative localisation, it isrecommended that they read the first section of Chapter 4 first. We begin by lookingat the relationship between the primes of T (R) and the primes of its centre Z. TheProposition given below is adapted from Theorem 13.8.14 of [McR].
Proposition 3.4. Let R be a prime Noetherian P.I. ring, T (R) its trace ring and Zthe centre of T (R). Then T (R) satisfies GU, LO and INC over Z.
Proof. We note first that T (R) is integral over Z (Proposition 3.2).
• (LO) Let p be a prime ideal of Z and r ∈ pT (R) ∩ Z. Then r =∑k
i=1 piri
for some ri ∈ R, pi ∈ p, so r ∈ Z〈r1, . . . rk〉. Since each ri is integral over Z,Shirshov’s Theorem 3.1 says that Z〈r1, . . . rk〉 = Zr1 + · · · + Zrl, some l ≥ k.It can easily be checked that p(Zr1 + · · · + Zrl) = pr1 + . . . prl is a subring ofZr1 + · · ·+ Zrl as well as being a Z-submodule. Moreover, r ∈ pr1 + · · ·+ prl andwe can apply the “determinant trick”, as in the proof of Proposition 3.2, to showthat rn + qn−1r
n−1 + · · ·+ q0 = 0, for some qi ∈ p. This shows that rn ∈ p, hencer ∈ p and pT (R)∩Z = p. By Zorn’s Lemma there exists an ideal P in T (R) whichis maximal with respect to P ∩ Z = p. This ideal is prime: let A,B be ideals ofT (R) such that AB ⊆ P . Then (A∩Z)(B ∩Z) ⊆ AB ∩Z ⊆ P ∩Z = p, thereforeeither A ∩ Z ⊆ p or B ∩ Z ⊆ p. So, by maximality of P , either A ⊆ P or B ⊆ Pand P is prime.
• (GU) Let p ⊂ q be prime ideals of Z and P a prime ideal of T (R) such thatP ∩ Z = p (P is guaranteed by LO). Then we quotient out T (R) by P , it isintegral over its centre which is Z/P ∼= Z/p. Then, again by LO, there is a primeideal Q′ of T (R)/P such that Q′ ∩ Z/P = q/P . Let Q be the preimage of Q′ inT (R). Q is a prime ideal, P ⊆ Q, and Q ∩ Z = q.
• (INC) Let P, Q be prime ideals of T (R) such that P ( Q. Then P ∩ Z ( Q ∩ Zif and only if Q/P ∩ Z/P 6= 0 in T (R)/P . T (R)/P is a prime P.I. ring, therefore,by Proposition 2.17, Q/P ∩ Z/P 6= 0 as required.
In order to prove the results below we need some of the properties of t-alternatingpolynomials that were constructed in the proof of Theorem 2.11.
Proposition 3.5 ([McR], 13.5.8). Let f(x1, . . . xt, y1, . . . yr) be a multilinear, t-alternatingpolynomial in Z〈x1, . . . xt, y1, . . . yr〉, with t = n2. Let R = Mn(A), for some commu-tative ring A, T an A-linear transformation of R (viewed as a free A-module), andu1, . . . ut, v1 . . . vr ∈ R. Then
i=0(−1)iaiλt−i, with ai ∈ A, a0 = 1, then for 1 ≤ ak ≤ t we
haveakf(u1 . . . vr) =
∑f(T i(1)u1, . . . T
i(t)ut, v1 . . . vr)
where the summation is over (i(1), . . . i(t)) ∈ {0, 1}t such that∑
j i(j) = k.
[McR], 13.5.8. 1. Since f is a multilinear polynomial, it suffices to prove the resultin the case where u1, . . . ut is a basis of R. In this case, Tui =
2. This result follows from part 1., if we think of λIn − T as an A[λ]-linear transfor-mation of Mn(A[λ]).
3. Using the multiliniarity of f , we can rewrite f((λIn−T )u1, . . . (λIn−T )ut, v1, . . . vr)as a polynomial in λ, whose coefficients are sums of f applied to various elements.Comparing the powers of λ in this with the right hand side gives the requiredrelation.
Lemma 3.6. Let R be a prime Noetherian P.I. ring. Then T (R) is a finite centralextension of R.
Proof. Let 0 6= c ∈ gn(R). Then since gn is an n2-alternating polynomial, for any a thatoccurs as a coefficient of a characteristic polynomial of some r ∈ R, we have ca ∈ gn(R)(this follows from Proposition 3.5 part 3., above). T (R) is generated by R and all sucha, therefore cT (R) ⊆ R. Moreover, c central in R implies that cT (R) is an ideal of Rand hence is finitely generated as an R-module (left or right). c is also a regular elementof T (R) (since it lies in the centre of T (R)), therefore cT (R) is isomorphic to T (R).
Lemma 3.7 ([Row1], 2.12.48). Let R be a ring and R ⊆ S a finite central extension.Then S satisfies LO over R.
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[Row1], 2.12.48. Let P be a prime ideal of R, then, by Zorn’s Lemma, there exists anideal Q of S maximal with respect to the property that Q ∩ S ⊆ P . By the sameargument as in the proof of Proposition 3.4, Q is a prime ideal of S. To show thatQ∩R = P it suffices to show that if R,S are prime rings and 0 6= P a prime ideal of Rthen there exists an ideal A of S such that 0 6= A ∩R ⊆ P . To see this, quotient out Sby Q. Then R/Q ∼= R/(R ∩Q), which is prime since Q is a prime ring (if AB ⊆ R ∩Q,then since S is a central extension, AS, BS C S and ASBS ⊆ Q implies AS ⊆ Q orBS ⊆ Q and A ⊆ R ∩ Q or B ⊆ R ∩ Q). If P/Q 6= 0 then the existence of an ideal Asuch that 0 6= A ∩ (R/Q) ⊆ (P/Q) contradicts the maximality of Q.Hence let R,S be prime rings and P a nonzero prime ideal of R. We can write S =∑m
i=1 aiR with each ai commuting with all of R, moreover we make a1 = 1. Nowwe extend a1 to a maximal R-linearly independent subset of {a1, . . . am}, which afterrelabelling is {a1, . . . an}, 1 ≤ n ≤ m. Since S is prime, annR(ai)aiR = 0 impliesthat annR(ai) = 0 (both aiR and annR(ai) are ideals since ai is central) and henceT =
∑ni=1 a1R is a free left/right R-module. Now we let Mi = {r ∈ R|rai ∈ T}, for
1 ≤ i ≤ m. Then each Mi is an ideal of R and R prime implies M =⋂n
i=1 Mi 6= 0 isan ideal of R. We have PM 6= 0 and, since S is a central extension, PMS C S. Bythe definition of M , PMS ⊆ PT . Since a1 = 1 and T a free module, T = R ⊕ T ′
and PT = P (R ⊕ T ′) = PR ⊕ PT ′ = P ⊕ PT ′. Hence, by Dedekind’s modular law,PMS ∩R ⊆ P ⊕ (T ′ ∩R) = P . Therefore PMS is our required ideal.
The following two Lemmas are adapted from [BS]. They show that any localisationof T (R) is actually a central localisation and this fact will be important in proving ourmain result.
Lemma 3.8 ([BS] Lemma 1). Let R be a prime P.I. ring and S ⊆ R a multiplicativeset of regular elements of R. Then
T (R〈c−1|c ∈ S〉) = T (R)[
1det(c)
∣∣∣∣c ∈ S]
where c−1 is the inverse of c in Q, the quotient ring of R.
[BS] 1. We have cc−1 = 1 so det(c) det(c−1) = 1 and det(c−1) = 1det(c) . If T is the
commutative ring generated by the centre of R〈c−1|c ∈ S〉 and the coefficients ofthe characteristic polynomials of the elements of R〈c−1|c ∈ S〉, then det(c−1) ∈ T ⊆Z(T (R〈c−1|c ∈ S〉)), because it is the final coefficient of the characteristic polynomialof c−1. Hence
T (R)[
1det(c)
∣∣∣∣c ∈ S]⊆ T (R〈c−1|c ∈ S〉)
Conversely, let f(x) = xn+an−1xn−1+· · ·+a1x+det(c) be the characteristic polynomial
of c ∈ S. Then f(x) = f(x)x + det(c) and, by the Cayley-Hamilton Theorem, we have
15
−1det(c) f(c)c = 1. So c−1 = −1
det(c) f(c) ∈ T (R)[
1det(c)
∣∣∣∣c ∈ S]. Therefore
R〈c−1|c ∈ S〉 ⊆ T (R)[
1det(c)
∣∣∣∣c ∈ S]
and
T (R〈c−1|c ∈ S〉) ⊆ T
(T (R)
[1
det(c)
∣∣∣∣c ∈ S])
But, since 1det(c) lies in the centre of T (R) for all c ∈ S, and, by Corollary 3.3, T (R) =
T (T (R)), we have T
(T (R)
[1
det(c)
∣∣∣∣c ∈ S])
= T (R)[
1det(c)
∣∣∣∣c ∈ S].
In order that S being a right Ore set is a sufficient condition for localisation, we nowrestrict ourselves to prime Noetherian P.I. rings (see the first section of Chapter 4 forwhy this is so).
Lemma 3.9 ([BS] Lemma 2). Let R be a prime Noetherian P.I. ring and 0 /∈ S aright (left) Ore set. Then T (R)S = T (RS).
[BS] 2. Firstly, we note that since T (R) is a finite central extension of R, S is also aright Ore set in T (R). Hence the localisation T (R)S exists. Clearly RS = R〈c−1|c ∈ S〉,so by the above Lemma (3.8) we get
T (RS) = T (R)[
1det(c)
∣∣∣∣c ∈ S]⊇ T (R)S
Now we must show that the opposite inclusion holds. If det(c) is invertible in T (R)S ,
for all c ∈ S, then clearly T (R)[
1det(c)
∣∣∣∣c ∈ S]⊆ T (R)S and there is nothing to prove.
So we assume c ∈ S with det(c) not invertible in T (R)S . Therefore det(c)T (R)S =T (R)S det(c) is a proper ideal of T (R)S , hence, by Zorn’s Lemma (since T (R)S has1), there exists a maximal ideal M of T (R)S with det(c)T (R)S ⊆ M . The inclusionsRS ⊆ T (R)S ⊆ T (RS) and Corollary 3.3 imply that T (T (R)S) = T (RS). Therefore, byLemmas 3.6 and 3.7, there exists a prime ideal P of T (RS) such that P ∩ T (R)S = M .But then det(c) ∈ P , which is a contradiction since det(c) is invertible in T (RS). Soevery element of the form det(c), c ∈ S, is invertible in T (R)S and we get T (RS) = T (R)S
as required.
Corollary 3.10. Let R be a prime Noetherian P.I. ring, T (R) its trace ring and 0 /∈ Sa right Ore set of R. Let Z be the centre of T (R). Then there exists a two sided Ore setS ′ ⊆ Z such that T (RS) = T (R)S = T (R)S′.
Proof. Take S ′ = {det(c)|c ∈ S}, then T (R)[
1det(c)
∣∣∣∣c ∈ S]
= T (R)S′ and the rest follows
from Lemmas 3.8 and 3.9 above.
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4 Localisation and Links
4.1 Localisation
In this subsection we quickly introduce the standard ideas of localisation in noncommu-tative rings. This is quite different to the case of commutative rings since additionalconditions need to be satisfied by the multiplicative set S before we can form the ringRS . Moreover, when localisation is possible it will either be a left localisation or a rightlocalisation.
Firstly, we introduce the Ore set and the idea of torsion.
Definition 4.1. Let R be a ring and S ⊆ R a multiplicative set. Then we say that Sis a right Ore set if, for all r ∈ R, x ∈ S, there exists s ∈ R, y ∈ S such that ry = xs.This can be rewritten as rS ∩ xR 6= ∅, for all r ∈ R, x ∈ S. A left Ore set is definedanalogously.
Definition 4.2. Let R be a ring, S ⊆ R a multiplicative set and r ∈ R. We saythat r is S-torsion if there exists a x ∈ S such that rx = 0. If no such x exists wesay that r is S-torsionfree. More generally, if A is a right R-module then we definetS(A) = {a ∈ A|ax = 0 for some x ∈ S}. We show below that if S is a right Ore set,then tS(A) is a submodule of A (this is not true in general).We say that A is S-torsion if tS(A) = A, and A is S-torsionfree if tS(A) = 0.
Lemma 4.3 ([GW], 4.21). Let S be a right Ore set in the ring R. Then, given anyx1, . . . xn ∈ S, there exists r1, . . . rn ∈ R such that x1r1 = · · · = xnrn ∈ S. From this itfollows that if A is a right R-module, then tS(A) is a submodule of A.
[GW], 4.21. We show for the case n = 2, the general result follows by induction. Theright Ore condition applied to x1, x2 gives x1y = x2r for some y ∈ S and r ∈ R. SinceS is a multiplicative set, x1y ∈ S as required.Now let a1, a2 ∈ tS(A), then there exists x1, x2 ∈ S such that a1x1 = a2x2 = 0. By thefirst part, there exists a y ∈ x1R∩x2R∩S, and (a1−a2)y = 0. Given r ∈ R, there existsz ∈ S and s ∈ R such that rz = x1s. Then (a1r)z = (a1x1)s = 0, so a1r ∈ tS(A).
If A is an ideal of R then tS(A) is also an ideal of R.
Definition 4.4. Let R be a ring and S ⊆ R a multiplicative set. A right ring of fractions(or right localisation) for R with respect to S is a ring homomorphism φ : R → S suchthat
1. φ(x) is a unit of S, for all x ∈ S.
2. Each element of S has the form φ(a)φ(x)−1, for some a ∈ R and x ∈ S.
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3. ker(φ) = {r ∈ R|rx = 0 some x ∈ S} = tS(R).
When a localisation of R with respect to S exists, we shall call the homomorphismφ : R → S the localisation map.
Naturally, given a multiplicative set S, we wish to know when such a localisationwith respect to S exists. The answer is a standard result in noncommutative ringtheory, which we state below without proof. Firstly, we must introduce the idea of rightreversibility.
Definition 4.5. Let S be a multiplicative set in a ring R. Then S is right reversible iffor all r ∈ R and x ∈ S such that xr = 0, there exists a y ∈ S such that ry = 0. A rightdenominator set S is a multiplicative set which is both right reversible and a right Oreset.
Theorem 4.6 ([GW], 10.3). Let S be a multiplicative set in a ring R. Then there existsa right ring of fractions for R with respect to S if and only if S is a right denominatorset.
Proof. For a proof see [GW], page 169.
Corollary 4.7. Let R be a ring, S ⊆ R a right denominator set and φ(x1)φ(s1)−1,. . . , φ(xn)φ(sn)−1 elements of RS . Then there exists r1, . . . rn ∈ R and c ∈ S such thatφ(xi)φ(si)−1 = φ(ri)φ(c)−1, for each 1 ≤ i ≤ n.
Proof. The localisation RS exists by Theorem 4.6 above. By Lemma 4.3 there existsc ∈ S and a1, . . . an ∈ R such that c = s1a1 = · · · = snan. φ(c) = φ(si)φ(ai) implies thatφ(ai) is invertible in RS . So set ri = xiai, then φ(ri)φ(c)−1 = φ(xi)φ(ai)(φ(si)φ(ai))−1 =φ(x1)φ(si)−1.
Fortunately for us, in the case where R is Noetherian (or more generally, has ACCon right annihilators of single elements), S a right Ore set implies that S satisfies theright reversibility condition.
Proposition 4.8 ([GW], 10.7). Let S be a right Ore set in a Noetherian ring R, thenS is right reverisble.
[GW], 10.7. Let r ∈ R and x ∈ S such that xr = 0 Then, since we have ACC onright ideals of R, there exists an n ∈ N such that r.ann(xn) = r.ann(xn+1). By the Orecondition on S, there exists s ∈ R and y ∈ S such that ry = xns. Then xn+1s = xry = 0implies s ∈ r.ann(xn+1) = r.ann(xn) and hence ry = xns = 0.
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4.2 Goldie’s Theorem
The set of regular elements of a ring, denoted C(0), is a multiplicatively closed set andit is often desirable to localise with respect to this set i.e. to form as large as possible aring of fractions for R. The question: Under what conditions does this localisation ex-ist? was answered fully by Alfred Goldie, in around 1959. We say the ring is classicallylocalisable when this is the case. Goldie’s Theorem, and the standard results leadingto its proof, are essential tools when it comes to developing theory in connection withlocalisation. Therefore, here we state some important results that we shall require; theproofs will be omitted but can be found in most texts on noncommutative ring theory(see [McR], page 58, or [GW], Chapter 6). Though we don’t define a Goldie ring here,it is sufficient for us to note that a right Noetherian ring is a right Goldie ring.
Lemma 4.9 ([GW], 6.11). Let R be a semiprime right Goldie ring and x ∈ R. Thenthe following conditions are equivalent:
1. x is regular.
2. r.annR(x) = 0.
3. xR ≤e RR.
Proof. For a proof see [GW], page 116.
Proposition 4.10 (Goldie’s Regular Element Theorem, [GW], 6.13). Let R bea semiprime right Goldie ring and I a right ideal of R. Then I is essential in RR if andonly if I contains a regular element.
Proof. For a proof see [GW], page 117.
Corollary 4.11. Let R be a prime right Noetherian ring and I a nonzero ideal of R,then I contains a regular element of R.
Proof. We show that I is essential as a right R-module. Let J be a right ideal of Rwith J ∩ I = 0, then (RJ)I ⊆ R(JI) ⊆ R(J ∩ I) = 0. But RJ is an ideal of R, hence0 = RJ = J and I is essential as a right ideal of R. Hence, by Goldie’s Regular ElementLemma 4.10 above, I contains a regular element of R (since a prime right Noetherianring is semiprime right Goldie).
Theorem 4.12 (Goldie’s Theorem, [GW], 6.15). A ring R has a semisimple clas-sical right quotient ring if and only if R is a semiprime right Goldie ring.
Proof. For a proof see [GW], page 118.
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If A is a right R-module, where R is a semiprime Noetherian ring and C(0) the setof regular elements of R, then we define the torsion submodule of A, written t(A), to betC(0)(A) (since Goldie’s Theorem says that C(0) is a right Ore set). If R is not semiprimeNoetherian then we can’t generally define t(A), since it won’t be a submodule of A butit still make sense to say that A is torsion or torsionfree.
4.3 Links
Firstly, we introduce the notion of links between prime ideals of a ring. There are severaldifferent definitions of what a link is exactly, the one we use is the one given in [GW],page 200. The definition is rather abstract but its usefulness will be apparent later.
Definition 4.13. Let P and Q be prime ideals of Noetherian ring R. We say thatthere is a link from P to Q, written P Ã Q, if there exists an ideal A of R such thatP ∩ Q ) A ⊃ PQ and (P ∩ Q)/A is a (R/P )-(R/Q)-bimodule, which is torsionfree onboth sides. If P Ã Q and A is as above, then (P ∩ Q)/A is called a linking bimodulebetween P and Q.
Let R be a Noetherian ring and Spec(R) the set of all prime ideals of R. Then wemay thing of the elements of Spec(R) as the vertices in a graph, with a directed edgebetween vertices P and Q if and only if P Ã Q. The connected components of thisgraph are called the cliques of Spec(R), and we denote the connected component towhich a vertex P belongs as Cl(P ). Moreover, we define a right link-closed subset Xof Spec(R) to be a set of primes such that Q ∈ X and P Ã Q implies that P ∈ X.This may seem to be a counter intuitive way of defining things but its logic will becomeapparent later.
There is another type of link that we will need in order to prove our main result.This type of link is specific to prime P.I. rings because it involves the trace ring.
Definition 4.14. Let R be a prime P.I. ring, T (R) its trace ring and P, Q prime idealsof R. Let Z be the centre of T (R). Then we say that P and Q are trace linked (“tr-linked”) if there exists prime ideals P ′, Q′ of T (R) such that P ′ ∩ R = P , Q′ ∩ R = Qand P ′ ∩ Z = Q′ ∩ Z. If P and Q are tr-linked then we write this as P Ãtr Q, orequivalently as Q Ãtr P .
We shall show later that, for prime P.I. rings, the two types of link share manyimportant properties. The next two Propositions give a glimpse of this similarity.
Proposition 4.15. Let R be a Noetherian ring, S a right Ore set, P, Q prime ideals ofR such that S ⊆ C(Q) and P Ã Q. Then S ⊆ C(P )
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In order to prove the above lemma, we need to make the following observation (whichis Exercise 10E of [GW]).
Lemma 4.16. Let R be a Noetherian ring, S a right Ore set and J ⊆ I ideals of R. If(I/J)R is S-torsionfree then R(I/J) is S-torsionfree.
Proof. Note first that if S ∩ J 6= ∅ then clearly both R(I/J) and (I/J)R are S-torsionmodules so the result holds. Therefore we consider the case S ∩ J = ∅. Then S/Jis a right Ore set in the Noetherian ring R/J . So, without loss of generality, we mayassume that J = 0. We assume that the left S-torsion subset of I is nonzero (it isnot a submodule since S is right Ore only), and show that this implies that the rightS-torsion submodule is nonzero. So let c ∈ S and 0 6= x ∈ I such that cx = 0. Thenr.ann(c) 6= 0 and we get the ascending chain r.ann(c) ⊆ r.ann(c2) ⊆ . . . . Since R isNoetherian, there exists an n ∈ N such that r.ann(cn) = r.ann(cn+1). Since S is a rightOre set, there exists y ∈ R and d ∈ C such that xd = cny. Then cn+1y = cxd = 0 andhence y ∈ r.ann(cn+1). But r.ann(cn+1) = r.ann(cn), so 0 = cny = xd and x 6= 0 is inthe S-torsion submodule of IR. This completes the proof.
Now we prove Proposition 4.15.
Proof. P Ã Q implies that there exists an ideal A of R such that P ∩Q/A is torsionfreeboth as a left R/P and right R/Q-module. Then S ⊆ C(Q) implies that P ∩ Q/A isS-torsionfree on the right as an R-module. By Lemma 4.16, P ∩Q/A is S-torsionfree onthe left too. I claim that for each c ∈ S we have r.annR/P (c+P ) = 0. Assume not, thenthere exists a c ∈ S, and r ∈ R\P such that cr ∈ P . P ∩Q/A is a faithful (R/P )-moduleso rP ∩Q 6⊆ A. But cr is in P and P ∩Q ⊆ Q means that crP ∩Q ⊆ PQ ⊆ A. This is acontradiction since P ∩Q/A is S-torsionfree on the left. Therefore r.annR/P (c+P ) = 0.Since R/P is a prime Noetherian ring, Lemma 4.9 says that c + P is regular. HenceS ⊆ C(P ).
Lemma 4.17. Let R be a Noetherian ring, P a prime ideal of R and S a right Oreset of R. Then either S ⊆ C(P ) and R/P is S-torsionfree or S 6⊆ C(P ) and R/P isS-torsion.
Proof. If S ⊆ C(P ), then by definition of C(P ), R/P must be S-torsionfree. So assumeS 6⊆ C(P ). Then I = tS(R/P ) 6= 0. I is finitely generated as a left R/P -module, soI = (R/P )x1 + . . . (R/P )xn. There exists a c1 ∈ S such that x1c1 = 0. x2c1 ∈ I sothere exists a c2 ∈ S such that x2c1c2 = 0. By induction, we get c = c1 . . . cn in S suchthat xic = 0 for all 1 ≤ i ≤ n. That is, Ic = 0. Therefore c + P ∈ r.annR/P (T ) = J ,which is an ideal of R/P . But IJ = 0, therefore, since R/P is a prime ring and I 6= 0,we have J = 0 and c ∈ P . So 0 6= c ∈ S ∩ P and R/P is S-torsion.
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Lemma 4.18. Let R be a Noetherian ring, S1,S2 right Ore sets such that RS1 = RS2
and P a prime ideal of R such that S1 ⊆ C(P ). Then S2 ⊆ C(P ).
Proof. We show that if S2 6⊆ C(P ) then S1 6⊆ C(P ). Therefore there exists c2 ∈ S, r ∈R\P such that either c2r = p ∈ P or rc2 = p ∈ P . If c2r = p then we use theright reversibility of S2/P in R/P to get rc′2 = p′ ∈ P . Hence, without loss of gen-erality, rc2 = p ∈ P . RS1 = RS2 implies that there exists x ∈ R, c1 ∈ S1 such thatφ(c2)(φ(x)φ(c1)−1) = 1 (where φ : R → RS1 is the localisation map with respect toS1 and not S2). Then φ(r)φ(c2)(φ(x)φ(c1)−1) = φ(r), φ(px)φ(c1)−1 = φ(r), that is,rc1 − px ∈ tS1(R). Consequently, there exists an s ∈ S1 such that (rc1 − px)s = 0 i.e.rc1s = pxs ∈ P . Therefore S1 6⊆ C(P ).
Compare Proposition 4.15 above with the following Proposition. This Propositionis Lemma 4 of [BW], though the proof we give here is different to the one given in thepaper.
Proposition 4.19 ([BW] 4). Let R be a prime Noetherian P.I. ring and S a right Oreset in R. If P is a prime ideal of R such that S ⊆ C(P ), and P Ãtr Q, for some primeideal Q of R, then S ⊆ C(Q).
Proof. P Ãtr Q implies that there exists primes P ′ and Q′ of T (R) such that P ′ ∩R =P,Q′ ∩ R = Q and P ′ ∩ Z = Q′ ∩ Z = p, where Z is the centre of T (R). Since T (R)is centrally generated over R, S is also a right Ore set for T (R). S ⊆ C(P ) impliesthat R/P is S-torsionfree so there exists x + P 6= P in R/P which is S-torsionfree.P ′ ∩ R = P and x ∈ T (R) implies x + P ′ 6= P ′ is S-torsionfree in T (R)/P ′. ThereforeT (R)/P ′ can’t be S-torsion, thus, by Lemma 4.17, T (R)/P ′ is S-torsionfree and S ⊆C(P ′). We note that, since every nonzero element of Z is invertible in Q, the simplequotient ring of R (and T (R)), every nonzero element of Z is regular in T (R). ThereforeC(P ′) ∩ Z = Z\(Z ∩ P ′) = Z\(Z ∩ Q′) = C(Q′) ∩ Z. By Corollary 3.10, there is a twosided Ore set S ′ ⊆ Z such that T (R)S = T (R)S′ . Then, by Lemma 4.18 above, we haveS ′ ⊆ C(P ′) ∩ Z and hence S ′ ⊆ C(Q′). Applying Lemma 4.18 again, we get S ⊆ C(Q′).Finally, S ⊆ R and Q′ ∩R = Q implies S ⊆ C(Q).
5 F.B.N. Rings
In this section we introduce a new type of ring called Fully Bounded Noetherian (F.B.N.).There are two main reasons for introducing these rings, the first being that there existsa definitive localisation theorem with regards to certain sets of prime ideals of the ring.The second is that we will introduce two notions of dimension of a ring in the nextsection, these give a way of comparing rings, and it turns out that these dimensions areequivalent for F.B.N. rings.
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Definition 5.1. A ring R is right bounded if every essential right ideal of R contains atwo-sided ideal which is essential as a right ideal.
Clearly, a commutative ring is an example of a right bounded ring.
Definition 5.2. A ring R is right fully bounded if, for every prime ideal P of R, R/Pis a right bounded ring. A right F.B.N. ring is a right fully bounded Noetherian ring,a left F.B.N. ring is defined analogously. We say that R is a F.B.N. ring if it is both aleft and right F.B.N. ring.
Fortunately for us, Noetherian P.I rings are F.B.N. rings.
Proposition 5.3. Let R be a Noetherian P.I. ring. Then R is a F.B.N. ring.
Proof. Let P be a prime ideal of R, I an essential right ideal of R/P . Then, since R/Pis a prime P.I. ring, we can apply Posner’s Theorem (2.18) to conclude that R/P ⊆ Q,a central simple algebra. IQ is clearly a right idea of Q, moreover it is essential in Q (if0 6= J Cr Q, then 0 6= J ∩ R and J , IQ have nonzero intersection in R). Since Q is acentral simple algebra and hence semisimple Artinian, IQ must be a direct summand ofQ, so IQ = Q. Therefore there exists a nonzero, regular, central element c in I. cR/Pis then a nonzero ideal of R/P contained in I. Since c is regular, by Lemma 4.9, cR/Pis essential as a right ideal of R/P .
Now we state a result which will be important later in the paper.
Proposition 5.4. Let R be a F.B.N. ring and X the set of all maximal ideals of R. SetN =
⋂M∈X M , then N is the Jacobson radical of R.
Proof. Let J be the Jacobson radical of R, then J is contained in every maximal idealof R (true for any ring). To see this, let M be a maximal ideal of R such that J 6⊆ M ,then J + M = R and j + m = 1 for some j ∈ J,m ∈ M . But j is right quasi-regular,therefore there exists a y with 1 = (1− j)y = my ∈ M . This contradiction shows thatJ ⊆ M for all maximal M .To show the opposite inclusion, we show that every maximal right ideal M of R contains amaximal twosided ideal of R and hence also N . Since J is the intersection of all maximalright ideals, the result will then follow. Therefore let M be a maximal right ideal ofR and set P = r.annR(R/M). Then P is a prime ideal because R/M is a simple rightR-module. If M/P were an essential right ideal of R/P , then there would exist an idealP ) I ⊇ M (since R/P is a right bounded ring). But we’d have (R/M)I = 0, whichcontradicts P = r.annR(R/M), hence M/P isn’t an essential right ideal. Therefore thereexists a nonzero right ideal J such that M ∩ J = 0. Since M is maximal, M ⊕ J = Rand J must be a minimal right ideal of R. This shows that S = soc((R/P )) 6= 0. Butin a semiprime right Noetherian ring, S is an ideal of R and R = S ⊕ T , for some ideal
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T 1. In our case R/P is prime so S = R/P , and R/P is semisimple (and hence simpleArtinian). Therefore P is a maximal ideal of R contained in M , as required.
5.1 A Localisation Theorem
Proposition 4.15 and Proposition 4.19 show that there is some connection between linkedprimes and Ore sets. In particular, if P is a prime in a Noetherian ring R and we wishto localise with respect to some Ore set S ⊆ C(P ), so that P is as “big” as possible inthe resulting ring, then we must have S ⊆ C(X), where X is the set of primes of R inthe smallest right link closure of P , and C(X) =
⋂Q∈X C(Q). Fortunately, it has been
shown (by B. J. Muller and A. V. Jategaonkar), that for sufficiently nice rings 2 andif X is finite, C(X) is itself a right Ore set and hence right localisation with respect toC(N) exists.This result has been extended to the case where X is infinite and we state the theorembelow. In order that the localisation formed is not “trivial” (see [BW], page 333, foran explanation of these trivial cases), we have to impose a second condition on C(X)(in addition to the fact that C(X) is a right Ore set). We say that X satisfies theintersection condition if: given a right ideal I of R which contains an element of C(P )for every P ∈ X it follows that I contains an element of C(X).
Theorem 5.5. Let R be an F.B.N. ring and X a set of incomparable prime ideals inR. Then X is right localisable if and only if X is right link closed and satisfies theintersection condition.
Proof. For a proof of this theorem see [Jat], Theorem 7.1.5 or [McR], Theorem 4.3.17.
Lemma 5.6. We can replace the intersection condition in Theorem 5.5 with the weakercondition: for every ideal I of R, if I ∩ C(P ) 6= ∅ for all P ∈ X, then I ∩ C(X) 6= ∅.
We note here that the proof of this result given in [BW] makes use of the fact thatR is an F.B.N. ring. However, as we show below, this is not necessary.
Proof. We are assuming that, for every two-sided ideal J of R with J ∩C(P ) 6= ∅ for allP ∈ X, we have J ∩ C(X) 6= ∅. Equivalently, if J ∩ C(X) = ∅, then there exists P ∈ Xsuch that J ∩ C(P ) = ∅. Now let I be a right ideal of R such that I ∩ C(X) = ∅ andset J = r.annR(R/I). Then J ⊆ I so J ∩ C(X) = 0 as well. Hence there exists a primeideal P ∈ X such that J ∩ C(P ) = ∅. Therefore (R/I)c 6= 0 for all c ∈ C(P ), and henceI ∩ C(P ) = ∅.
1If the reader is unfamiliar with this result, see [GW], page 1312The theorem holds for Noetherian rings satisfying the right second layer condition, which is a much
weaker condition than being a F.B.N. ring.
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The result mentioned above when X is a finite set now follows from Theorem 5.5and the following Lemma.
Lemma 5.7 ([GW], 7.5). Let N be a proper semiprime ideal in a Noetherian ringR, and let P1, . . . Pn be the prime ideals of R minimal over N . Then P1, . . . Pn areprecisely the primes of R that contain N but are disjoint from C(N), and C(N) =C(P1) ∩ · · · ∩ C(Pn).
Proof. By considering R/N , we may assume N = 0. Since P1 ∩ . . . Pn = 0, we havePi(P1 . . . Pi−1Pi+1 . . . Pn) = 0. Therefore Pi = r.annR(Ii), for some nonzero ideal Ii ofR. Now let x be a regular element of R with r ∈ R such that xr ∈ Pi. Then xrIi = 0implies that rIi = 0, since x is regular, hence r ∈ Pi. Similarly, Pi = l.annR(Ji), forsome nonzero ideal Ji of R. Following the same argument, x regular and r ∈ R suchthat rx ∈ Pi implies that r ∈ Pi. Therefore x ∈ C(Pi). Thus we have shown thatnone of the Pi contains a regular element of R and C(N) ⊆ C(P1) ∩ · · · ∩ C(Pn). Ifx ∈ C(P1) ∩ · · · ∩ C(Pn), then l.annR(x), r.annR(x) ⊆ P1 ∩ · · · ∩ Pn = 0. Therefore x isregular.
Theorem 5.8 ([GW], 14.21). Let R be a F.B.N ring and N a semiprime ideal of R.Then N is right localisable if and only if the set of prime ideals minimal over N is rightlink closed in Spec(R).
The main part of the proof of this theorem is actually the proof of Lemma 13.4 in[CH].
Proof. Let X be the set of prime ideals of R minimal over N , then X = {P1, . . . Pn} is afinite set. By Lemma 5.7, we have C(X) = C(N). Therefore, in order to apply Theorem5.5, it suffices to check that X satisfies the intersection condition described above. LetI be a right ideal of R such that I ∩C(Pi) 6= ∅, for 1 ≤ i ≤ n. We prove by induction onn, so assume there exists a c ∈ I ∩ C(P1) ∩ · · · ∩ C(Pn−1). By Goldie’s Regular ElementLemma 4.10 (GREL), I ∩ C(Pn) 6= ∅ implies that (I + Pn)/Pn is an essential right idealof R/Pn. Since the Pi are incomparable, (Y + Pn)/Pn 6= 0, where Y = P1 . . . Pn−1. Iclaim that (IY + Pn)/Pn is also essential as a right ideal. If Pn ⊆ J is a right ideal ofR with J ∩ IY ⊆ Pn, then J(IY + Pn) ⊆ Pn. But I, Y 6⊆ Pn implies IY + Pn 6⊆ Pn
(by Dedekind’s modular law), therefore J ⊆ Pn i.e. J = Pn. So, by GREL again, thereexists d ∈ IY ∩ C(Pn), d = c + y, where y ∈ IY . Then d ∈ C(P1) ∩ · · · ∩ C(Pn). To seethis - let r ∈ R with (c + y)r = n ∈ N . Since y, n ∈ Y , we have r ∈ Y . But dr ∈ Nmeans that dr ∈ Pn and hence r ∈ Pn. This gives r ∈ Y ∩ Pn = N , as required.
We note here that deriving Theorem 5.8 from Theorem 5.5 isn’t the standard wayof doing things. Theorem 5.8 can be (and normally is) proved directly without the use
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of Theorem 5.5. See [GW], page 246 or [BW], page 114 for the proof.
Let R be a F.B.N. ring, X a right localisable set of prime ideals of R (i.e. X is rightlink closed and satisfies the intersection condition). Then, by Theorem 5.5 above, C(X)is a right Ore set. Let RC(X) denote the localisation of R with respect to C(X). There isa natural correspondence between ideals of R and RC(X), as we shall see in the Lemmabelow (part of the proof of Lemma 5.9 is adapted from the proof of Theorem 10.18 in[GW]).
Lemma 5.9. Let R be a F.B.N. ring, X a right localisable set of prime ideals of R,and RC(X) the resulting localisation. Then there is a natural mapping from ideals ofR to ideals of RC(X), given by I 7→ IRC(X). Conversely, for J an ideal of RC(X),φ−1(J) = {r ∈ R|φ(r) ∈ J} is an ideal of R and J = φ−1(J)RC(X), where φ is thelocalisation map R → RC(X). Moreover, if J is prime then so too is φ−1(J).
Proof. Let x ∈ IRC(X), then x =∑k
i=1 φ(xi)φ(ri)φ(si)−1, where xi ∈ I, ri ∈ R, si ∈C(X). By Corollary 4.7, there exists a common denominator for this sum, so x =(∑k
i=1 φ(xiai))φ(s)−1, for some s ∈ C(X) and ai ∈ R. But each xi belongs to I,therefore x = φ(x′)φ(s)−1, where
∑ki=1 xiai = x′ ∈ I. Now let φ(r)φ(t)−1 be an element
of RC(X), then φ(x′)φ(s)−1φ(r)φ(t)−1 = φ(x′)φ(r′)(φ(s′)φ(t))−1, where r′ ∈ R, s′ ∈ C(X)(this follows from the Ore condition applied to s and r). x′r′ ∈ I and s′t ∈ C(X) soφ(x′)φ(s)−1φ(r)φ(t)−1 ∈ IRC(X).Now consider φ(r)φ(t)−1φ(x′)φ(s)−1: applying the right Ore condition to t, x′, we getx′t′ = ty ∈ I, for some t′ ∈ C(X), y ∈ R. As in the proof of Lemma 4.16, this meansthat there exists a z ∈ C(X) such that yz ∈ I. This gives φ(t)−1φ(x′) = φ(y)φ(t′)−1 =φ(y)(φ(z)φ(z)−1)φ(t′)−1 = φ(yz)φ(t′z)−1. Therefore φ(r)φ(t)−1φ(x′)φ(s)−1 ∈ IRC(X).Finally, if φ(x1)φ(s1)−1, φ(x2)φ(s2)−1 ∈ IRC(X), then φ(x1)φ(s1)−1 + φ(x2)φ(s2)−1 =φ(x1s
′1 + x2s
′2)φ(s)−1 ∈ IRC(X), by Corollary 4.7. Hence IRC(X) is an ideal of RC(X).
Since φ is a ring homomorphism, it is standard that φ−1(J) is an ideal of R. Clearlyφ(φ−1(J))RC(X) ⊆ J , and if x ∈ J , then x = φ(r)φ(s)−1 = (φ(r)1−1)(1φ(s)−1) ∈φ(φ−1(J))RC(X) (since φ(r)1−1 = xφ(s) ∈ J). Hence the inclusion is an equality asrequired. Now assume that J is a prime ideal of RC(X) and let A,B be ideals of R suchthat AB ⊆ φ−1(J). Then RC(X)BRC(X) ⊆ BRC(X) (since BRC(X) is an ideal of RC(X))means that (ARC(X))(BRC(X)) ⊆ ABRC(X) ⊆ φ−1(J)RC(X) = J . Therefore ARC(X) ⊆ Jor BRC(X) ⊆ J and hence A or B ⊆ φ−1(J).
Comparing Lemmas 5.9 and 5.7 shows that the maximal ideals of RC(X) are of theform PRC(X), where P ∈ X.
The next result is similar to those above and will also be required later.
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Lemma 5.10. Let R be a right Noetherian ring, S a right Ore set and P a prime idealof R such that S ⊆ C(P ). Then C(PRS) = C(P )S−1, where C(P )S−1 = {φ(x)φ(s)−1 :x ∈ C(P ), s ∈ S}.Proof. Let φ(x)φ(s)−1 ∈ C(P )S−1 and φ(y)φ(c)−1 ∈ RS (y ∈ R and c ∈ S), suchthat (φ(x)φ(s)−1 + PRS)(φ(y)φ(c)−1 + PRS) = 0. So φ(x)φ(s)−1φ(y)φ(c)−1 ∈ PRS ,φ(x)φ(s)−1φ(y)φ(c)−1 = φ(p)φ(d)−1 where p ∈ P and d ∈ S (it was shown in the proofof Lemma 5.9 that an element of PRS has this form). Using the right Ore condition,φ(s)−1φ(y)φ(c)−1 = φ(y′)φ(s′)−1φ(c)−1 with y′ ∈ R, s′ ∈ S, and we choose a commondenominator so φ(x)φ(y′)φ(z1)φ(e)−1 = φ(p)φ(z2)φ(e)−1, where φ(z1), φ(z2) are unitsin RS and e ∈ S. Then xy′z1 − pz2 ∈ tS(R). Therefore there exists a d ∈ S such that(xy′z1 − pz2)d = 0. That is, x(y′z1d) ∈ P . But x ∈ C(P ) implies that (y′z1d) ∈ P .Therefore φ(y′) = φ(y′z1d)φ(z1d)−1 ∈ PRS , i.e. φ(s)−1φ(y) ∈ PRS and since PRS is anideal we get φ(y)φ(c)−1 ∈ PRS . Hence φ(x)φ(s)−1 ∈ C(PRS) and C(P )S−1 ⊆ C(PRS).Now let φ(x)φ(s)−1 ∈ C(PRS) and y ∈ R such that xy ∈ P . Then (φ(x)φ(s)−1 +PRS)(φ(sy)1−1 + PRS) = 0 and hence φ(sy)1−1 ∈ PRS So we can write φ(sy)1−1 =φ(p)φ(c)−1 with p ∈ P, c ∈ S. The proof of Lemma 4.16 shows that, by the Orecondition, there exists q ∈ P, d ∈ S such that φ(s)−1φ(p) = φ(q)φ(d)−1. Thereforeφ(y)1−1 = φ(q)φ(d)−1 ∈ PRS and yd− q ∈ tS(R). As before, there exists an e ∈ S suchthat (yd − q)e = 0, that is yde ∈ P . But de ∈ S ⊆ C(P ) implies that y ∈ P and soφ(x)φ(s)−1 ∈ C(P )S−1 as required.
6 Krull and Classical Krull dimensions
In this section we introduce the notion of Krull dimension and Classical Krull dimension,these are invariants that we can assign rings. In almost all the cases we will consider,the Krull dimension and Classiacal Krull dimensions will coincide, though showing thatthis is the case will take some work.
6.1 Classical Krull dimension
Classical Krull dimension is defined by transfinite induction and hence will involve or-dinal values. The definition of Classical Krull dimension we give is taken from [GW],page 234.
Definition 6.1. Let R be a ring. We define, by transfinite induction, sets Xα of primeideals of R for each ordinal α. Firstly we let X−1 be the empty set. Next, consider anordinal α ≥ 0, if Xβ has been defined for all ordinals β < α, let Xα be the set of thoseprimes in R such that all prime ideals properly containing P belong to
⋃β<α Xβ. If
some Xγ contains all prime ideals of R then we say that the Classical Kurll dimension
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of R exists, and that this dimension (written Cl.K.dim(R)) equals γ, the smallest suchγ.
Note The following results are not stated in full generality, in many cases we assumethat R is Noetherian or F.B.N. in order to simplify the proof and because we do notrequire the full result. In order to find the fully generality in which these results can beproved, see the reference given at the head of each result.
Lemma 6.2 ([GW] 14.1). Let R be a Noetherian ring, then the Classical Krull di-mension of R exists.
[GW] 14.1. Let the Xα’s be the sets of prime ideals of R as defined above. The cardi-nality of these sets is bounded by the cardinality of P(R), the power set of R. Thereforethe transfinite chain X−1 ⊆ X0 ⊆ X1 ⊆ . . . cannot properly increase forever. So thereexists some ordinal γ such that Xγ = Xγ+1. If Cl.K.dim(R) does not exist then Xγ
does not contain all the prime ideals of R, hence, by the Noetherian condition, thereexists a prime P maximal with respect to not being in Xγ . But then all primes properlycontaining P are in Xγ which implies that P ∈ Xγ+1. Xγ+1 = Xγ gives us the requiredcontradiction.
Proposition 6.3 ([GW] 14.2). Let R be a ring with Cl.K.dim(R) = γ. If α is anynonnegative ordinal strictly less than γ, then there exists a prime P of R such thatCl.K.dim(R/P ) = α. Moreover, if R is right or left Noetherian, then there is a minimalprime P of R such that Cl.K.dim(R/P ) = γ.
[GW] 14.2. From the way that Classical Krull dimension is defined we see that, for aprime P of R, Cl.K.dim(R/P ) = α if and only if P ∈ Xα and P /∈ Xβ for all β < α.So, if there is no prime P such that Cl.K.dim(R/P ) = α, then we have Xα = Xα+1.But this implies that Xβ = Xα for all β > α, and hence Cl.K.dim(R) ≤ α, which is acontradiction. This shows that Cl.K.dim(R) is the supremum of Cl.K.dim(R/P ) as Pvaries over all prime ideals of R. This will be greatest when P is minimal, so in fact it isthe supremum as P varies over all minimal primes of R. If R is right or left Noetherianthen R has finitely many minimal primes, one of which will have Cl.K.dim(R/P ) = γas required.
Theorem 6.4. Let R and S be Noetherian rings, such that R is F.B.N., and SMR abimodule which is finitely generated on both sides. If MR is faithful, then Cl.K.dim(R) ≤Cl.K.dim(S).
Proof. The proof of this theorem makes use of the existence of two-sided affiliated seriesfor M and won’t be given here. For a proof see [GW], page 237 or [Jat] page 231, notingthat a F.B.N. satisfies the second layer condition.
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Corollary 6.5. Let R and S be F.B.N. rings, and SMR a bimodule which is finitelygenerated on both sides. If M is faithful on both sides, then Cl.K.dim(R) = Cl.K.dim(S).
Proof. Follows directly from Theorem 6.4 above.
6.2 Krull dimension
Classical Krull dimension follows from the standard Krull dimension given to commu-tative rings. For noncommutative rings a more general definition which can be appliedto modules as well as rings is needed. Therefore we introduce the Krull dimension of amodule. The definition of Krull dimension that is given is taken from [GW], page 255.As for Classical Krull dimension, the definition is by transfinite induction.
Definition 6.6. Let R be a ring, we define classes Yα of R-modules by transfiniteinduction. Let Y−1 be the class of modules consisting of the zero module only. Next,consider an ordinal α ≥ 0, if Yβ has been defined for all β < α, let Yα be the classof R-modules, such that for every countable decending chain M0 ⊇ M1 ⊇ M2 . . . ofsubmodules of M , we have Mi/Mi+1 ∈
⋃β<α Yβ for all but finitely many i’s. If an
R-module M belongs to some Yα, then the smallest such α is called the Krull dimensionof M and is denoted K.dim(M). If M does not belong to any Yα then we say that Mhas no Krull dimension.
The main aim of this section is to show that, for F.B.N. rings, and hence primeP.I. rings, the Classical Krull dimension of R and the Krull dimension of RR are thesame. We denote the Krull dimension of RR by r.K.dim(R). If I is an ideal of R thenr.K.dim(R/I) will denote the Krull dimension of (R/I)(R/I), whereas K.dim(R/I) willgenerally refer to the Krull dimension of R/I as a right R-module.
Lemma 6.7 ([GW], 15.1). Let M be a module and N a submodule. Then K.dim(M)exists if and only if both K.dim(M/N) and K.dim(N) exist. In this case, K.dim(M) =max{K.dim(M/N), K.dim(N)}.[GW], 15.1. If M has Krull dimension then both N and M/N will have Krull dimension,with K.dim(M) ≥ max{K.dim(M/N), K.dim(N)}, because a chain of submodules of Nis a chain of submodules of M and any chain of submodules of M/N can be written asthe image of a chain of submodules of M under the natural quotient map. We provethe converse by induction on α = max{K.dim(M/N), K.dim(N)}. α = −1 implies M =N = 0. So we assume that α ≥ 0 and that the result holds for all β < α. Let M1 ⊇ M2 ⊇. . . be a descending chain of submodules of M . This gives us descending chains in M/Nand N , by taking Ai = (Mi+N)/(Mi+1+N) and Bi = (Mi∩N) respectively. Therefore,for all but finitely many i, we have K.dim(Ai/Ai+1), K.dim(Bi/Bi+1) < α, and hencemax{K.dim(Ai/Ai+1), K.dim(Bi/Bi+1)} < α. The induction hypothesis implies that
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K.dim(Mi/Mi+1) ≤ max{K.dim(Ai/Ai+1), K.dim(Bi/Bi+1)} < α, for all but finitelymany i. But this is the same as saying that K.dim(M) ≤ α, and the induction iscomplete.
Lemma 6.8 ([GW], 15.5). Let R be a right Noetherian ring, then there exists aminimal prime ideal P of R such that r.K.dim(R) = r.K.dim(R/P ).
[GW], 15.5. Since every right R/P -module is also a right R-module, we have r.K.dim(R/P )≤ K.dim((R/P )R) ≤ K.dim(RR) = r.K.dim(R), for all prime ideals of R. Since R isright Noetherian, there exist minimal prime ideals P1, . . . Pn of R, who’s product is zero.Then for each i = 1, . . . , n, we have
which is ≤ r.K.dim(R/Pi). Therefore, by repeatedly applying Lemma 6.7 above, weget r.K.dim(R) = K.dim(R/(P1P2 . . . Pn)) ≤ max{r.K.dim(R/P1), . . . , r.K.dim(R/Pn)}.Hence r.K.dim(R) ≤ r.K.dim(R/P ), for some minimal prime ideal P of R.
Lemma 6.9 ([GW] 15.6). Let M be a nonzero module with Krull dimension and φ amonomorphism from M to itself. Then K.dim(M) ≥ K.dim(M/φ(M)) + 1.
Proof. Let α = K.dim(M/φ(M)). Since M is nonzero, we have K.dim(M) ≥ 0, so if α =−1 we are done. Therefore, assume α ≥ 0. Consider the descending chain of submodulesM ⊇ φ(M) ⊇ φ2(M) ⊇ . . . . Since φ is injective, we have φn(M)/φn+1(M) ∼= M/φ(M)for all n, and hence each quotient has Krull dimension α. Therefore K.dim(M) > α.
Proposition 6.10. Let R be a right Artinian ring. Then every prime ideal of R ismaximal.
Proof. The Artin-Wedderburn Theorem tells us that if R is a prime right Artinian ringthen it is simple. So assume that R has a nonmaximal prime ideal P . Then R/P isa prime right Artinian ring which is not simple. This contradiction shows that P ismaximal.
The next result is Exercise 15F of [GW].
Lemma 6.11. Let R be a right Noetherian ring then Cl.K.dim(R) ≤ r.K.dim(R).
Proof. Firstly we prove that if P is a prime ideal of R and I some ideal of R withI ) P , then K.dim(R/I) < K.dim(R/P ) as right R-modules. Since R/P is a primeright Noetherian ring, Corollary 4.11 says that I/P contains a regular element c. So we
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define a right R-homomorphism φ : R/P → R/P by x 7→ cx. This is a monomorphismbecause c is regular. Hence, by Lemma 6.9,
r.K.dim(R/P ) > K.dim(
(R/P )φ(R/P )
)
Then φ(R/P ) ⊆ I/P (since c ∈ I), implies that there is a natural epimorphism(R/P )/φ(R/P ) → R/I. Therefore
r.K.dim(R/I) ≤ K.dim(
(R/P )φ(R/P )
)< r.K.dim(R/P )
Now we prove the inequality by induction on α = r.K.dim(R). If α = 0, then R isa right Artinian ring and, by Proposition 6.10, every prime ideal of R is maximal.Hence Cl.K.dim(R) = −1 or 0. Let α > 0 and assume the result holds for all β < α.Since R is right Noetherian, by Proposition 6.3, there exists a minimal prime idealP such that Cl.K.dim(R) = Cl.K.dim(R/P ). Clearly, r.K.dim(R/P ) ≤ r.K.dim(R),so we may assume that R is prime. By Proposition 6.3 again, there exists a primeideal Q of R, Q 6= 0, such that Cl.K.dim(R/Q) = Cl.K.dim(R) − 1. By the firstpart of this proof, r.K.dim(R/Q) < r.K.dim(R), hence, inductively, Cl.K.dim(R/Q) ≤r.K.dim(R/Q). Therefore Cl.K.dim(R) ≤ r.K.dim(R) as required.
Fortunately for us, if we impose the additional restriction that R is a F.B.N. ringthen the above inequality becomes an equality (this is not true for a general rightNoetherian ring - see [GW], page 259 for a counterexample). The proof of Theorem 6.13is a combination of the proofs of Theorem 15.13 in [GW] and Theorem 6.4.8 of [BW].In order to make it readable we split it into two parts.
Lemma 6.12. Let R be a right Noetherian ring and M0 ⊇ M1 ⊇ . . . a descendingchain of right ideals of R. Then there exists a right ideal N of R and n ∈ N such thatMn+k ⊕N is essential in R for all k ≥ 0.3
Proof. If Mi = 0 for some i, then Mk = 0 for all k ≥ i and we can take N to be anyessential right ideal of R. Therefore we may assume Mi 6= 0 for all i. We assume theresult isn’t true and show that this gives us a contradiction. So for each i, let Ni be a rightideal of R such that Mi ⊕Ni ≤e R. Moreover, we demand that Ni ⊇ Ni−1 (if not thenset N ′
i = Ni +Ni−1). Then there exists a k(i) > i such that Mk(i)⊕Ni is not essential inR. Therefore we can create a subsequence i0 = 0, i1 = k(i0), . . . in = k(in−1) . . . , whichhas the property that Min⊕Nin ≤e R and Min+1⊕Nin 6≤e R. Then Min+1⊕Nin+1 ≤e R
3This can be rewritten simply as saying that R has finite rank since it is Noetherian. Hence thereexists an n ∈ N such that the rank of Mn equals the rank of all subsequent right ideals in the chain.Therefore Mn ⊕N ≤e R implies Mn+k ⊕N ≤e R for all k > 0.
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and Min+1 ⊕ Nin 6≤e R implies that Nin+1 ) Nin . This gives us an ascending chain ofright ideals Ni1 ( Ni2 ( . . . . This contradicts the fact the R is right Noetherian.
Theorem 6.13 ([GW] 15.13). Let R be a right F.B.N. ring. Then r.K.dim(R) =Cl.K.dim(R).
Proof. We have already shown in Lemma 6.11 that Cl.K.dim(R) ≤ r.K.dim(R). There-fore it suffices to show that Cl.K.dim(R) ≥ r.K.dim(R). We do this by inductionon α = r.K.dim(R), if α = −1 then R = 0 and contains no prime ideals, henceCl.K.dim(R) = −1. If α = 0 then for all descending chains of right ideals I0 ⊇ I1 ⊇ . . .we eventually get In/In+1
∼= 0 i.e. In = In+1 and R is right Artinian. By Proposition6.10, this implies that every prime ideal of R is maximal, therefore Cl.K.dim(R) = 0.So assume α > 0. By Lemma 6.8, there exists a minimal prime ideal P of R such thatr.K.dim(R) = r.K.dim(R/P ), therefore it is sufficient to show that Cl.K.dim(R/P ) ≥r.K.dim(R/P ). This enables us to assume that R is prime. Let M0 ⊇ M1 ⊇ . . . be adescending chain of right ideals of R. Then to show that r.K.dim(R) ≤ Cl.K.dim(R)we need to show that K.dim(Mi/Mi+1) < Cl.K.dim(R) for all but finitely many i. ByLemma 6.12, there exists a right ideal N of R, and n ∈ N, such that Mn+k ⊕ N ≤e Rfor all k ≥ 0. So let i ≥ n. Then, since Mi ∩N = Mi+1 ∩N = 0, we have
Mi + N
Mi+1 + N∼= Mi
Mi+1
Because R is a prime F.B.N. ring and Mi+1 + N an essential right ideal of R, thereexists a nonzero ideal Ii of R such that Ii ⊆ Mi+1 ⊕N . Therefore ((Mi + N)/(Mi+1 +N))Ii = 0 and hence (Mi/Mi+1)Ii = 0 i.e. Mi/Mi+1 is a right R/Ii-module. ThereforeK.dim(Mi/Mi+1) ≤ r.K.dim(R/Ii). It was shown in the proof of Lemma 6.11 thatIi 6= 0 implies that r.K.dim(R/Ii) < r.K.dim(R). Therefore we can apply the inductivehypothesis to r.K.dim(R/Ii) to conclude that r.K.dim(R/Ii) = Cl.K.dim(R/Ii). ButCl.K.dim(R/Ii) < Cl.K.dim(R), so K.dim(Mi/Mi+1) < Cl.K.dim(R) as required.
Corollary 6.14. Let R, S be F.B.N. rings and RBS a bimodule which is finitely gener-ated and torsionfree on both sides. Then r.K.dim(R) = r.K.dim(S).
Proof. Theorem 6.4 says that Cl.K.dim(R) = Cl.K.dim(S). Then Theorem 6.13 givesr.K.dim(R) = Cl.K.dim(R) = Cl.K.dim(S) = r.K.dim(S).
Corollary 6.15. Let R be a F.B.N. ring and P,Q primes of R such that P Ã Q. Thenr.K.dim(R/P ) = r.K.dim(R/Q).
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Proof. Since P Ã Q, there exists an ideal A such that P ∩ Q/A is an (R/P )-(R/Q)bimodule which is finitely generated and torsionfree on both sides. If P ∩ Q/A is un-faithful as a left R/P -module, then I = annR/P (P ∩Q/A) is a nonzero, twosided idealof R/P . Since R/P is a prime right Noetherian ring, Corollary 4.11 says that I containsa regular element c. But then c(P ∩Q/A) = 0 implies that P ∩Q/A is not torsionfree.So P ∩ Q/A is faithful on the left, the same argument shows that it is also faithful onthe right. Therefore Corollary 6.14 above says that r.K.dim(R) = r.K.dim(S).
Finally, we look at the relationship between the Krull dimension of R/P and T (R)/Q,for a prime ideal P in a prime Noetherian P.I. ring R, and Q a prime lying over P .
Lemma 6.16 ([BW], 2). Let R be a prime Noetherian P.I. ring, T (R) its trace ringand Z the centre of T (R). If P is a prime ideal of R, Q a prime ideal of T (R) such thatQ ∩R = P and p = Q ∩ Z, then r.K.dim(R/P ) = r.K.dim(T (R)/Q) = Cl.K.dim(Z/p).
Proof. We begin by viewing T (R)/Q as a (R/P )-(T (R)/Q)-bimodule. Lemma 3.6 saysthat T (R) is a finite central extension of R and hence Noetherian on both sides. Thisallows us to apply Corollary 6.15 and conclude that r.K.dim(R/P ) = r.K.dim(T (R)/Q).We show that Cl.K.dim(T (R)/Q) = Cl.K.dim(Z/p) by induction on Cl.K.dim(Z/p). IfCl.K.dim(Z/p) = 0, then p is a maximal prime of Z. By Proposition 3.4, T (R) satisfiesLO, GU and INC over Z hence Q ∩ Z = p implies that Q is a maximal ideal of T (R).Therefore Cl.K.dim(T (R)/Q) = 0. Now let Cl.K.dim(Z/p) = α > 0 and assume theresult holds for all β < α. By Proposition 6.3, there exists a prime ideal p′ of Z suchthat p ( p′ and Cl.K.dim(Z/p′) = α − 1. LO impies that there is a prime Q′ ofT (R) such that Q′ ∩ Z = p′, GU then says that Q ( Q′. Finally, GU and INC implythat there is no prime ideal of T (R) lying between Q and Q′ so Cl.K.dim(T (R)/Q′) =Cl.K.dim(T (R)/Q) − 1. The inductive hypothesis tells us that Cl.K.dim(T (R)/Q′) =α − 1, therefore Cl.K.dim(T (R)/Q) = α = Cl.K.dim(Z/p). Then Theorem 6.13 saysthat r.K.dim(T (R)/Q) = Cl.K.dim(T (R)/Q).
7 Localisation in Prime Noetherian P.I. rings
The aim of this section is to build up to a proof of the main result (Main Theorem).We begin by looking at the case where R = T (R).
Proposition 7.1 ([BW], 3). Let R be a prime Noetherian P.I. ring such that R =T (R), P a prime ideal of R and Z the centre of R. Then the following sets are equal 1.Cl(P ), 2. the set of primes Q such that P ∩Z = Q∩Z, 3. the smallest right link closedsubset of Spec(R) containing P . Moreover, this set is finite.
We shall first prove that this set is finite.
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Lemma 7.2. Let R be a prime Noetherian P.I. ring such that R = T (R), P a primeideal of R and Z the centre of R. Let X be the set of all prime ideals Q of R such thatP Ãtr Q, then X is finite.
Proof. Let p = P ∩ Z and Y the set of all prime ideals of R minimal over pR. SinceR is a Noetherian ring, this set is finite. I claim that X ⊆ Y , i.e. the primes of X areminimal over pR. Choose Q ∈ X, then p ⊆ Q implies that pR ⊆ Q. Since p ⊆ Z, pRis an ideal of R. Let Q′ be a prime ideal of R with pR ⊆ Q′ ⊆ Q. Intersecting with Zgives p ⊆ pR ∩ Z ⊆ Q′ ∩ Z ⊆ Q ∩ Z = p, hence Q′ ∩ Z = p. Proposition 3.4 says thatINC holds for R over Z, therefore Q = Q′ is minimal over pR. So we get X ⊆ Y and Xmust be a finite set.
Now we prove Proposition 7.1.
[BW], 3. Let Q be a prime ideal of R such that Q Ã P and p = P ∩ Z. Then, sinceZ\p ⊆ C(P ) is a right Ore set, Proposition 4.15 implies that Z\p ⊆ C(Q). ThereforeQ ∩ Z ⊆ p. But we can apply the same argument to Z\(Z ∩ Q), which is a left Oreset, to get p ⊆ Z ∩ Q. Similarly, if Q′ is a prime ideal such that P Ã Q′, repeatingthe argument gives Q′ ∩ Z = p. Hence Cl(P ) is a subset of the tr-link closure ofP in Spec(R). Moreover, Cl(P ) is finite by Lemma 7.2. Now let Y be the smallestright link-closed subset of Spec(R) containing P . Then Y ⊆ Cl(P ) implies that Y isa finite set. If Y = {P1, P2, . . . Pn = P} and N =
⋂ni=1 Pi, then Lemma 5.7 says that
C(N) = C(P1) ∩ · · · ∩ C(Pn). By Theorem 5.8, C(N) is a right Ore set in R and we canform the localisation RN . Lemma 5.9 shows that the maximal ideals of RN are preciselythose of the form PiRN , for 1 ≤ i ≤ n. Corollary 3.10 shows that there is a twosidedOre set S ⊆ Z such that RN = RS . Therefore if Q is any other prime ideal of R suchthat Q ∩ Z = p, then S ∩ Q = ∅ and QRN 6= RN is a proper ideal of RN . So QRN
is contained in a maximal ideal PiRN of RN . This implies that Q/Pi is C(N)-torsion.But C(N) ⊆ C(Pi) means that R/Pi is S-torsionfree hence Q ⊆ Pi. We have shownabove that Pi ∈ Cl(P ) implies that Pi ∩ Z = p, therefore, since R satisfies INC over Z(Proposition 3.4), it follows that Q∩Z = P ∩Z implies Q ∈ Y . Thus, if X is the tr-linkclosed subset of Spec(R) containing P , we have shown X ⊆ Y ⊆ Cl(P ) ⊆ X, which isfinite.
The next Proposition is the key to our main result because it will allow us to connectthe link structure of R with the link structure of T (R). The Proposition we state andprove below is a weaker result than Proposition 5 of [BW], however it is sufficient forour needs and its proof is considerably simpler. We break the proof into two parts inorder to make it easier to understand.
Lemma 7.3. Let R be a Noetherian and Artinian P.I. ring and S a finite centralextension of R. Let P Ã Q be prime ideals of R. Then there exists prime ideals P ′, Q′
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of S such that P ′∩R = P , Q′∩R = Q and there exists a simple, faithful (S/P ′)-(S/Q′)-bimodule M 4.
The main idea of this proof (to apply the Jordan-Holder Theorem to R ⊗ Rop andS⊗Sop series) comes from [BW], page 329. If the reader is unfamiliar with the Jordan-Holder Theorem and the existence of composition series for modules which are bothArtinian and Noetherian then they may wish to read Chapter 4 of [GW].
Proof. Since P Ã Q, there exists an ideal PQ ⊆ A ( P ∩ Q with P ∩ Q/A a tor-sionfree (R/P )-(R/Q)-bimodule. Therefore we have a sequence of left R⊗Rop-modules0 ⊆ PQ ⊆ A ( P ∩Q ⊆ R ⊆ S, and we let 0 = C0 ⊆ C1 ⊆ · · · ⊆ Cn = S be a compo-sition series of S as a left S ⊗ Sop-module. By applying the Jordan-Holder Theorem torefinements of the two sequences of R⊗Rop-modules defined above, we can conclude thatthere exists an ideal A ⊆ B of R, and R⊗Rop-modules Ci ⊆ D1 ⊆ D2 ⊆ Ci+1 such thatB/A ∼= D2/D1 are simple left R⊗Rop-modules. Since B/A is a submodule of P ∩Q/A,it follows that l.annR(B/A) = P and r.annR(B/A) = Q. Let P ′ = l.annS(Ci+1/Ci),Q′ = r.annS(Ci+1/Ci), then they are prime ideals of S because Ci+1/Ci is a simple leftS ⊗ Sop-module. Ci ⊆ D1 ⊆ D2 ⊆ Ci+1 and P = l.annR(D2/D1), Q = r.annR(D2/D1)imply that P ′∩R ⊆ P , Q′∩R ⊆ Q. But P ′∩R, Q′∩R are prime ideals of R and, sinceR is Artinian, every prime ideal of R is maximal (Proposition 6.10), so it follows thatP ′ ∩R = P , Q′ ∩R = Q as required. We take M to be Ci+1/Ci.
Proposition 7.4. Let R be a prime Noetherian P.I. ring and P Ã Q prime ideals ofR. Then there exist prime ideals P ′, Q′ of T (R) such that P ′ ∩R = P, Q′ ∩R = Q andP ′ Ãtr Q′.
Proof. Since T (R) is a finite central extension of R, PQT (R) is an ideal of T (R). Welet R′ = R/PQ and S = T (R)/PQT (R) so that S is a finite central extension of R′.Let P , Q be the images of P , Q in R′. Then P Ã Q in R′ because PQ ⊆ A. SinceP and Q are precisely the minimal primes of R′ and Corollary 6.15 says that A Ã Bimplies r.K.dim(R′/A) = r.K.dim(R′/B), {P , Q} must be a right link closed set. LetC = C(P ∩ Q) = C(0) be the set of regular elements of R′. Then Theorem 5.8 says that Cis a right Ore set and we can form the localisation R′
C . Since S is a central extension ofR′, C is also a right Ore set in S and SC is a finite central extension of RC . The argumentgiven in the proof of Lemma 7.9 below shows that PR′
C Ã QR′C . Also, since PR′
C , QR′C
are precisely the maximal ideals of R′C and their intersection is 0, the argument given in
the proof of Lemma 7.8 below shows that R′C (and hence SC) is Artinian.
Therefore we can apply Lemma 7.3 and conclude that there exists prime ideals P ′, Q′
in SC such that P ′ ∩R′C = PR′
C and Q′ ∩R′C = QR′
C . We also know that there exists a
4This result says that there exists a “bond” between P ′ and Q′, see [Jat], page 123 for a definitionof a bond between prime ideals.
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simple, faithful (SC/P ′)-(SC/Q′)-bimodule M . Let φ : T (R) → SC be the natural map,who’s restriction to R is the natural map R → R′
C . Then we set P ′ = φ−1(P ′) andQ′ = φ−1(Q′). M becomes a (T (R)/P ′)-(T (R)/Q′)-bimodule by defining (x + P ′)m =(φ(x) + P ′)m and similarly for m(x + Q′). From the way it is defined, M is faithful onboth sides. We use this fact to show that P ′ ∩ Z = Q′ ∩ Z, where Z is the centre ofT (R). Let p = P ′ ∩ Z, then for c ∈ Z\p, (c + P ′)M 6= 0. But c belongs to the centreof T (R), which means that (c + P ′)M = M(c + Q′), and hence P ′ ∩ Z ⊆ Q′ ∩ Z. Thesame argument in reverse gives the required equality.
We can also go the other way, as shown by the next Lemma.
Lemma 7.5. Let R be a F.B.N. and S a finite central extension. If P,Q are primeideals of S such that P Ã Q, then either P ∩R = Q ∩R or P ∩R Ã Q ∩R.
Proof. P Ã Q means that there exists an ideal PQ ⊆ A ⊂ P ∩Q such that P ∩Q/A isa (S/P )-(S/Q)-bimodule which is finitely generated and torsionfree on both sides. LetP ′ = P ∩R and Q′ = Q∩R, then P ′(P ∩Q/A) = (P ∩Q/A)Q′ = 0 imply that P ∩Q/Ais also a (R/P ′)-(R/Q)-bimodule. Clearly it is torsionfree on both sides and, since S isfinitely generated as an R-module, it is also finitely generated on both sides. It follows(by Corollary 6.14) that r.K.dim(R/P ′) = r.K.dim(R/Q′). Now A ∩ R ⊆ P ′, Q′, hencer.K.dim(R/P ′) = r.K.dim(R/Q′) implies that A ∩ R = P ′ ⇔ A ∩ R = Q′. Therefore,either P ′ = Q′ or A ∩ R is a proper subset of both P ′ and Q′. If the latter is the case,then (P ′∩Q′)/(A∩R) is a nonzero finitely generated (R/P ′)-(R/Q′)-bimodule. Since itis isomorphic to P ∩Q∩R/A, it is torsionfree on both sides. Hence P ′ Ã Q′ as required.
Corollary 7.6. Let R be a prime Noetherian P.I. ring and P, Q prime ideals of R suchthat P Ã Q. Then P Ãtr Q.
Proof. This is just a rewording of Proposition 7.4.
Theorem 7.7 ([BW], Theorem D). Let R be a prime Noetherian P.I. ring, and Xa set of prime ideals of R. Then the following are equivalent properties of X:
1. X is left link closed.
2. X is right link closed.
3. X is link closed.
4. X is tr-link closed.
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Proof. By Corollary 7.6, if X has any of properties 1. 2. or 3. then the elements of Xare all trace linked. Therefore it suffices to show that if P, Q are any tracelinked primeideals of R, then there exists primes P1, . . . Pn such that P à P1 à . . . à Pn à Q (asimilar argument shows that there exist Q1, . . . Qm such that Q à Q1 . . . à Qm à P ).P Ãtr Q implies that there exists prime ideals P ′, Q′ of T (R), such that P ′∩Z = Q′∩Zand P ′ ∩ R = P , Q′ ∩ R = Q, where Z is the centre of T (R). Proposition 7.1 saysthat P ′ is in the right link closure of Q′. Therefore there exists prime ideals P ′
1, . . . P′n
of T (R) such that P ′ Ã P ′1 Ã . . . Ã P ′
n à Q′. Let Pi = P ′i ∩ R, then by Lemma 7.5,
either Pi = Pi+1 or Pi à Pi+1, for each 0 ≤ i ≤ n. So, after removing repetitions, weget P à P1 à . . . à Pn′ à Q as required.
We note here that, though the cliques of T (R) are all finite, and there is a clearconnection between linked primes in T (R) and R, it does not follow that the cliques inR are finite. In [BM], Bruno J. Muller gives an example (counter-example 2, page 242)of a prime Notherian P.I. ring which has cliques of infinite cardinality.
7.1 The Main Theorem
Now, at last, we can begin building a proof of our main theorem, which we state againbelow. This next lemma will enable us to show that every localisation in a F.B.N. ringcan be thought of as localisation with respect to some collection of prime ideals. In[BW], the result is shown to follow from Theorem 5.5. However we give a direct proofbelow.
Lemma 7.8. Let R be a F.B.N. ring and X the set of all maximal ideals of R. ThenC(X) is precisely the set of units of R.
Proof. Clearly every unit of R is contained in C(X). Set J =⋂
M∈X M , then, byProposition 5.4, J is the Jacobson radical of R. It suffices to show that elements ofC(X) are invertible in R/J . To see this, let x ∈ R such that x + J is invertible, thenthere exists a y in R such that (1 − xy) ∈ J . This means that 1 − (1 − xy) is rightinvertible in R, that is, (xy)z = 1 for some z ∈ R. Therefore yz is a right inverse for xin R. Similarly, since 1− (1− yx) = yx is left invertible in R, x also has a left inverse.Therefore x is an invertible element of R. By Proposition 6.3, there exists a minimalprime P of R/J such that Cl.K.dim(R/J) = Cl.K.dim((R/J)/P ). But every primeideal of R/J is maximal, so 0 = Cl.K.dim((R/J)/P ) = Cl.K.dim(R/J). Since R/J isa F.B.N., ring we have Cl.K.dim(R/J) = r.K.dim(R/J) = 0 (by Theorem 6.13), andhence R/J is a right Artinian ring. Therefore all regular elements of R/J are invertible(consider the ascending chain xR ⊇ x2R ⊇ · · · ⊇ xnR ⊇ . . . , for each regular x ∈ R/J).But the regular elements of R/J are C(X) + J , as required.
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Lemma 7.9. Let R be a F.B.N. ring, S a right Ore set and φ : R → RS the localisationmap of R with respect to S. Let Q′ be a maximal ideal of RS , Q = φ−1(Q′) and P aprime ideal of R such that P Ã Q. Then PRS is a maximal ideal of RS .
Proof. We note that Lemma 5.9 shows that Q is indeed a prime ideal of R. Also, sinceQ′ is a maximal ideal of RS , S ⊆ C(Q) and, by Proposition 4.15, S ⊆ C(P ). P Ã Qmeans that there exists an ideal PQ ⊆ A ( P ∩ Q such that P ∩ Q/A is a (R/P )-(R/Q)-bimodule which is torsionfree and faithful on both sides. Let (P ∩Q/A)S−1 bethe corresponding (RS/PRS)-(RS/QRS)-bimodule. Then, since Lemma 5.10 says thatC(PRS) = C(P )S−1 and C(QRS) = C(Q)S−1, it is clear that (P∩Q/A)S−1 is torsionfreeon both sides. By the same argument as in the proof of Corollary 6.15, (P ∩Q/A)S−1
must also be faithful on both sides. Corollary 6.5 tells us that Cl.K.dim(RS/PRS) =Cl.K.dim(RS/QRS). Since QRS is maximal, it follows that PRS is maximal.
Proposition 7.10 ([BW], 11). Let R be a F.B.N. ring and S a right Ore set in R.Then there is a right localisable set X of prime ideals of R such that S ⊆ C(X) andRC(X) = RS .
The proof of Proposition 7.10 is a combination of the proof given in [BW], page 334and Lemmas 7.8 and 7.9 above.
Proof. Let X ′ be the set of all maximal ideals in RS and X = {φ−1(M)|M ∈ X ′} (whereφ : R → RS is the localisation map). Then Lemma 5.9 says that X consists of primeideals and Lemma 7.9 shows that it is right link closed. Since φ(C(X)) = C(X ′), Lemma7.8 says that these are precisely the units in RS . By definition of localisation, φ(S)consists of units of RS , hence it follows that S ⊆ C(X). Let I be an ideal of R suchthat I ∩ C(X) = ∅, then φ(I)RS is a proper ideal of RS and hence is contained in somemaximal ideal of RS . Therefore I ⊆ P for some P ∈ X and I ∩C(P ) = ∅ i.e. X satisfiesthe weaker, symmetric, intersection condition (and hence, by Lemma 5.6, the standardintersection condition). Theorem 5.5 says that C(X) is right Ore and we can form thelocalisation RC(X). Clearly RC(X) = RS as required.
The main theorem now follows easily.
Main Theorem ([BW], Theorem B). Let R be a prime Noetherian P.I. ring, andS a right (left) Ore set in R, then there exists a two sided Ore set S ′ such that S ⊆ S ′and the right (left) localisation of R by S equals the two-sided localisation of R by S ′.[BW], Theorem B. By Proposition 7.10, there exists a set X of right localisable primeideals of R such that RC(X) = RS . We know that X is right link closed and satisfiesthe weaker, symmetric, intersection condition. Theorem 7.7 says that X is also a leftlink closed set, so we apply Theorem 5.5 again to conclude that C(X) is also a left Oreset.
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The main theorem tells us that, if R is a prime Noetherian P.I. ring and S a left orright Ore set, then in the localisation RS any element can be written both as φ(r)φ(s)−1
or φ(s′)−1φ(r′), for some r, r′ ∈ R and s, s′ ∈ S.
