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 noncommutative 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 wellbehaved.
Chapter 4 - Here we introduce, more formally, the idea of localisation in a noncommutative 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.
Contents
1 Introduction
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Basic Properties of P.I. Rings . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2 The Theorems of Kaplansky and Posner
2.1 Kaplansky’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Posner’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
7
3 The Trace Ring
3.1 Shirshov’s Theorem and Integrality . . . . . . . . . . . . . . . . . . . . .
3.2 Localisation of T (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
12
12
4 Localisation and Links
4.1 Localisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Goldie’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
17
19
20
5 F.B.N. Rings
5.1 A Localisation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
24
6 Krull and Classical Krull dimensions
6.1 Classical Krull dimension . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Krull dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
27
29
7 Localisation in Prime Noetherian P.I. rings
7.1 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
37
1
Introduction
1.1
Introduction
In this paper we aim to explore some of the properties that rings satisfying polynomial
identities possess. In particular, we aim to develop a proof of the theorem proved by
Amiram Braun and Robert B. Warfield Jr. which shows that localisation in prime
Noetherian polynomial identity rings is symmetrical. Specifically,
Main Theorem. Let R be a prime Noetherian P.I. ring, and S a right (left) Ore set
in R, then there exists a two sided Ore set S 0 such that S ⊆ S 0 and the right (left)
localisation of R by S equals the two sided localisation of R by S 0 .
Polynomial identity rings can be thought of as being almost commutative and many
of the results for commutative rings, which do not hold for general noncommutative
rings, 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 on
ring theory. When this is the case a reference to the book and the proof involved is
made 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 this
is the case every effort has been made to both simplify and clarify the proof, rather
than copy it directly from the book. If no such reference is made at the start of the
result/proof, then the proof is the author’s own work.
1.2
Basic Properties of P.I. Rings
Definition 1.1. Let Zhx1 . . . xn i be the free ring generated over Z by the noncommuting
indeterminants x1 . . . xn . Then we say a ring R is a polynomial identity (P.I.) ring if
there exists a monic polynomial f ∈ Zhx1 . . . xn i\{0} such that f (r1 , . . . rn ) = 0 for all
r1 , . . . 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 ) =
x1 x2 −x2 x1 , 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, (r1 r2 − r2 r1 )2 will be a scalar matrix,
hence f (x1 , x2 , x3 ) = (x1 x2 − x2 x1 )2 x3 − x3 (x1 x2 − x2 x1 )2 will be a polynomial identity
for R which is a P.I. ring.
1
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 ∈ Zhx1 , . . . xn i. Then we say that f is a multilinear polynomial
of degree n if
X
f (x1 , . . . xn ) =
aσ xσ(1) . . . xσ(n)
σ∈Sn
where aσ(k) ∈ Z for 1 ≤ k ≤ n. This is equivalent to saying that f (c1 x1 , . . . cn xn ) =
c1 c2 . . . cn f (x1 , . . . xn ), for all c1 , . . . cn ∈ Z.
Proposition 1.3 ([McR] 13.1.9). If R satisfies a polynomial identity f of degree d
then R also satisfies a multilinear identity g of degree at most d. Furthermore, each
coefficient 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 are
also identities of R. Since f is monic, at least one of f1 , f2 must be monic so we accept
the monic one and drop the other. Proceeding by induction, we get a new monic f
who’s coefficients are all coefficients of the original f and each monomial contains every
indeterminant xi , though our new f may have smaller degree.
Now if f is not already multilinear then there exists some indeterminant, x1 say, which
has degree greater than one in one of the monomials of f . Therefore we consider
g(x1 , . . . xn , xn+1 ) = f (x1 + xn+1 , x2 , . . . xn ) − f (x1 , x2 , . . . xn ) − f (xn+1 , x2 , . . . xn )
in which the maximum degree of x1 is one less than what it was in f . Also, though
the monomials of g are different to those of f , the coefficient of each monomial in g is
the 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 the
number of indeterminants involved has increased, the degree of every monomial is still
the same. Proceeding inductively, we may assume that every indeterminant of g has
degree 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 no
monic identity of degree less than 2n.
[McR] 13.3.2. Let f be a monic identity for Mn (A), then by the above proposition we
may assume that f is a multilinear polynomial. Hence
X
f=
x1 . . . xt
+
aσ xσ(1) . . . xσ(t)
| {z }
monic leading term
σ∈St \{1}
2
Assume that t ≤ 2n − 1. Now choose r1 . . . rt to be the first t elements in the list
e11 , e12 , e22 , e23 , e33 , . . . enn , where eij is the matrix with a one in the i, j th entry and
all other entries zero. Then for σ 6= 1, rσ(1) . . . rσ(t) contains a product ri rj where
j 6= i, i + 1, i − 1. But then ti tj = 0 so f (r1 , . . . rt ) = r1 . . . rt + 0 = e1[ t ] 6= 0. This is a
2
contradiction, hence deg(f ) ≥ 2n as required.
An important polynomial that often occurs as an identity for some P.I. rings is the
one defined below. From the definition we are able to state the following theorem due
to Amistur and Levitzki, which we shall require later.
Definition 1.5. We define the nth standard identity (or standard polynomial) as
sn =
X
(−1)sign(σ) xσ(1) . . . xσ(n)
σ∈Sn
Then sn is a monic polynomial of degree n.
Theorem 1.6 (Amistur, Levitzki). If A is a commutative ring then Mn (A) satisfies
the polynomial s2n .
Proof. The proof is combinatorial and will be omitted. See [Row2], page 21, for a
proof.
Therefore any matrix ring over a commutative ring is a P.I. ring. This together with
the following lemma will give us a major source of P.I. rings.
Lemma 1.7 ([McR] 13.1.7).
is a P.I. ring.
1. Any subring or homomorphic image of 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 identity
for any subring or homomorphic image of R.
Q
2. If f1L
, . . . fn are monic identities for R1 , . . . Rn then f = ni=1 fi is a monic identity
for 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 extension
of 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 then
S also satisfies f and hence is also a P.I. ring.
3
2
The Theorems of Kaplansky and Posner
The main aim of this section is to prove the important theorems concerning P.I. rings
due to Kaplansky and Posner. We use Kaplansky’s Theorem to prove Posner’s Theorem
which 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 structure - they are all central simple algebras (defined below). To prove Kaplansky’s Theorem we will need some results concerning general rings. We shall state and prove these
now. Firstly a definition:
Definition 2.1. Let R be a ring and M a faithful, simple right R-module. Then
R 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 = vi r for 1 ≤ i ≤ n.
Theorem 2.2 (The Density Theorem, [Her] 2.1.2). Let R be a primitive ring and
M 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 subspace
V of M , V 6= M and any m ∈ M \V , there exists an r ∈ R such that V r = 0 and
mr 6= 0. To see this, let v1 , . . . vn be linearly independent, w1 , . . . wn ∈ M , then let
Vi = SpD (v1 , . . . vi−1 , vi+1 , . . . vn ). Then vi ∈ M \Vi , so there exists ri ∈ R such that
vi ri 6= 0 and Vi ri = 0. Now vi ri 6= 0 and M is simple so vi ri R = M , and there exists
si ∈ R with vi ri si = wi . We set ti = ri si and t = t1 + · · · + tn which gives us vi tj = δij wj
and vi t = wi as required.
We prove the result by induction on dim(V ). If dim(V ) = 0 then, for m ∈ M \{0}, there
must exist (by the simplicity of M ) a r ∈ R with mr 6= 0. Now let V = V0 + wD, where
w ∈
/ V0 and dim(V0 ) = dim(V ) − 1. By the inductive hypothesis, if m ∈ M \V0 then
there exists a ∈ annR (V0 ) such that ma 6= 0 i.e. mannR (V0 ) = 0 iff m ∈ annR (V0 ). So
if 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 assume
m ∈ M \V such that mannR (V ) = 0. We define τ : M −→ M by τ (x) = ma where
a ∈ annR (V0 ) such that x = wa. τ is well defined: if x = 0 then x = wa where
a ∈ annR (V0 ) ∪ annR ({w}) = annR (V ) so τ (x) = ma = 0 since mannR (V ) = 0 by
assumption.
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
4
ma = τ (wa) = τ (w)a. So (m − τ w)a = 0 i.e. m − τ w ∈ 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 right
R-module VR and let D = EndVR . Then either dimD V = n and R ∼
= Mn (D) for some
n < ∞, or else, for each n < ∞, there is a subring Sn of R together with a surjective
ring homomorphism Sn → Mn (D).
Proof. If dimD V = n < ∞ then EndD V ∼
= Mn (D) (after fixing some basis of V ). In
this case the Density Theorem says that R → EndD V is an isomorphism. Otherwise,
for each n < ∞, choose some subspace Vn of V of dimension n over D. Then let
Sn = {r ∈ R|Vn r ⊆ Vn }. It can easily be shown that Sn is a subring of R and the action
of Sn on Vn defines a surjective homomorphism into Mn (D) (after fixing some basis of
Vn ).
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 ideals
of 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 let
A = {v ∈ V |v ⊗ 1 ∈ B}. P
Then A C V . I claim that: if A 6= 0 then A ⊗ U = B. Assume
with respect to n.
not, then we choose b = ni=1 vi ⊗ ui in B\A ⊗ U which is minimal
Pn
Let 0 6=
v
⊗
1
∈
A
⊗
U
and
consider
b
−
(1
⊗
u
)(v
⊗
1)
=
v
1
i=2 i ⊗ ui ∈ B. Then
Pn
either i=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
Pn we need to show that B 6= 0 implies A 6= 0. Again we choose some non-zero
b = i=1 vi ⊗ ui in B with v1 , . . . vn linearly independent so as to minimise n. Then
P
u1 6= 0 and, since U is simple, there exist a1 . . . ak , b1 . . . bk such that kj=1 aj u1 bj = 1.
P
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 still
in B and n is minimal). This says that uui −P
ui u = 0 for 1 ≤ i ≤ n.P
Therefore each ui
n
lies in the centre of U which is k. So 0 6= b = ( i=1 vi ui )⊗1 and 0 6= ni=1 vi ui ∈ A.
The above lemma enables us to prove the following result which will be very important in proving Kaplansky’s theorem. In what follows, [A : B] will be used to denote
the dimension of A as a vector space over B.
5
Lemma 2.5 ([McR] 13.3.4). Let R = Mt (D) for some division ring D. Let H be a
maximal 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 ] =
2
[R : F ] = m .
m
t
and
[McR] 13.3.4.
1. The previous Lemma (2.4) shows that S is simple. We give V a Smodule 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 simple
S-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 if
α(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 centraliser
of 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 faithful
S-module. Hence S is a primitive ring and we can apply Lemma 2.3 to conclude
that S ∼
= Mm (H). Using the fact that dim U ⊗ W = dim U × dim W , for all
vector 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 =
2
[R : F ] = [R : D][D : F ] = t2 [D : F ] and [D : F ] = mt2 . Similarly, since
V ∼
= Dt ∼
= H m , 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 if
the centre of R is k and R is a simple, finite dimensional, algebra. Note that since R
is 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 D
is a division ring with centre F . Let H be a maximal subfield of D. Then:
6
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 ], so
n2 = t2 m2 = (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. ring
¡ ¢2
of minimal degree d then R is a central simple algebra of dimension d2 over its centre.
[McR] 13.3.8. By Lemma 2.3, either R ∼
= Mn (D) or, for all n ∈ N, there exist a subring
Sn 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 the
centre of D), hence so too are the minimal degrees of Mn (D) and Sn . This implies that
R 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 maximal
subfield of D. Then, by Lemma 2.5 i), R ⊗F H = S is simple, V is a simple S-module
and 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 preceding
paragraph, we get S ∼
= Mt (H) and dim VH = t < ∞. Then Lemma 2.5 ii) says that
2
[R : F ] = t < ∞ 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 ] =
¡ d ¢2
2 .
2.2
Posner’s Theorem
Using Kaplansky’s Theorem we can prove Posner’s Theorem which says that any prime
P.I. ring is an order in a central simple algebra. We begin by introducing central polynomials.
Definition 2.9. Let R be a P.I. ring with centre C and g ∈ Zhx1 , . . . xn i. Then we say
that 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 ) =
(x1 x2 − x2 x1 )2 . Then, for all r, s ∈ R, g(r, s) is a scalar matrix hence lives in the centre
7
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 partial
answer to this is that all rings of the form Mn (A), for some commutative ring A have a
central polynomial. This fact will enable us to prove Posner’s Theorem. We will state
this 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 describe
these properties here, they will be required later.
Definition 2.10. Let g(x1 , . . . xn ) ∈ Zhx1 , . . . xn i, then we say that g is a t-alternating
polynomial, for some 1 ≤ t ≤ n, if interchanging any two of the first t indeterimants
takes 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 C
the centre of R, then there exists a multilinear, t-alternating polynomial gn , of degree
4n2 , 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 similar
proof). The proof given is constructive, a family of polynomials gn are defined in terms
of 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, that
we will require in order to prove Posner’s Theorem. But first we need a result to do
with 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 a
standard 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. Then
N (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 + a1 x +
. . . an xn ∈ 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 + a1 x + . . . an xn and q = b0 + b1 x + . . . bm xm ∈ R[x] such that
pq = 1. Then a0 b0 = 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] =
8
a0 + P [x] and a1 x + . . . an xn ∈ P [x]. This means that a1 . . . an are in P . The argument
holds for all primes P of R, hence a1 . . . an are in the intersection of all primes of R. By
Lemma 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 elements
of 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 + · · · + pn xn , we need to show that
pn is nilpotent. Note that ppn − pn p 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 shows
that it too commutes with all pi . Inductively, we get that pi pj = pj pi 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 + . . . qm xm ∈ R[x] such that (1 − xp)q = 1. This gives q0 = 1, −p0 q0 + q1 = 0 that is
q1 = p20 . Inductively, assume qk = fk (p0 , . . . pk−1 ), then
−
k
X
pl qk−l + qk+1 = 0
so
qk+1 =
k
X
pl fk−l (p0 , . . . pk−l )
l=0
l=0
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 applied
to 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 ideal
contains a nonzero nilpotent ideal.
[McR] 13.2.5. We show that if S is a nonzero nil P.I. ring then it contains a nonzero
nilpotent ideal I. Then if S is our nil ideal in R, I will be a nonzero nilpotent right
ideal 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 proceed
by induction on d, the minimal P.I. degree of R (that is R satisfies some monic polynomial
of degree d and no polynomial of lower degree). For d = 2, R will satisfies some monic
multilinear polynomial of the form x1 x2 − nx2 x1 , for some n ∈ Z. Since R is nil we can
choose 0 6= a ∈ R such that a2 = 0. Then Ra = naR and (aR)2 = a(Ra)R = na2 R = 0
so 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 multilinear 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 of
degree m − 1. Again we choose 0 6= a in R such that a2 = 0, if r1 . . . rm−1 in R then
9
f2 (ar1 . . . arm−1 , a) = 0 since each monomial of f2 contains a.ark = 0 for some k. So
f1 (ar1 , . . . arm−1 )a = 0.
Consider the ideal M = {r ∈ aR|raR = 0} of aR. Then f1 is a monic identity for
aR/M because f1 (aR)a = 0 ⇒ f1 (aR)aR = 0 and f1 (aR) ⊆ M . Either aR/M = 0
and aR is nilpotent or applying the inductive hypothesis to aR/M , we conclude that
there exists a nonzero ideal of aR with I n ⊆ M for some n ∈ N. Then IaR Cr R
and (IaR)n+1 ⊆ I n IaR ⊆ W IaR ⊆ W ar = 0. So IaR is contained in some nonzero
nilpotent ideal of R.
From Theorems 2.12 and 2.15 we are able to deduce the following corollary which
we 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 Theorem 2.15 it has no nozero nil ideals. It then follows from Theorem 2.12 that R[x] is
semiprimitive.
The above corollary allows us to prove the following Proposition which is the last
result we need before tackling Posner’s Theorem (the Proposition is similar to Theorem
13.6.4 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 nonzero
ideal of R. Then I ∩ C 6= 0, that is, any nonzero ideal of R contains a nonzero central
element.
Proof. If I CR then I[x]CR[x], and if I[x]∩C(R[x]) = I[x]∩C[x] 6= 0 then I ∩C 6= 0. So
we may assume, by Corollary 2.16, that R is a semiprimitive ring. J(R) = 0 means that
the intersection of all primitive ideals of R is zero. Therefore there exists a primitive
ideal 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 primitive
ideal 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 Theorem
says that R/Mi ∼
= Mmi (Di ), for some mi ∈ N and division ring Di . If Hi is a maximal
subfield 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 exists
a multilinear polynomial gni of degree 4n2i 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 it
was also zero on Mni (Hi ), therefore this does not happen. If d is the minimal degree of R, then di , the minimal degree of R/Mi , will be less than or equal to d for all
i ∈ Λ, and hence {ni |i ∈ Λ} is bounded (ni = d2i ). Let n be this maximum. Thus,
10
gn (R/M ) ⊆ C(R/M ) for all maximal ideals M of R. Since the intersection of all these
ideals 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 with
centre C. Let S = C\{0}. Then the two sided localisation Q = RS exists. Moreover, Q
is a central simple algebra with centre F = CS , R is an order in Q and Q = RF is a
central extension of R.
Proof. Since R is a prime ring, C is an integral domain and hence S is both a left and
right 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 only
if I = 0. So let 0 6= I C Q and J = I ∩ R. Then, by Proposition 2.17, C ∩ J 6= 0. Hence
there 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 apply
Kaplasky’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 element
of Q. Then ac−1 bd−1 = bd−1 ac−1 implies ab = ba (since we can multiply through by
cd ∈ 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 form
ac−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 an
embedding 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 is
a 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 a
ring homomorphism and ker L = {r ∈ R|rx = 0 for all x ∈ Q}. But 1 ∈ Q so ker L = 0
and L is an embedding of R in Mn (F ).
Since rl ∈ Mn (F ), it has a characteristic polynomial, fr ∈ F [x], of degree n with
fr (rl ) = 0 (it follows that fr (r) = 0 too). Now we let T be the commutative subring of
F generated by C and all coefficients of the polynomials {fr |r ∈ R}. Then we define
T (R) ⊆ Q to be the central extension of R generated by R and T . T (R) has the property
11
that every element of R is integral over the centre of T (R). We call T (R) the trace ring
of 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 we
need the following theorem which we will not prove here.
Theorem 3.1 (Shirshov’s Theorem). Let C be a commutative ring, R = Chx1 , . . . xn i
a P.I. ring, and x1 , . . . xn all integral over C. Then R is a finitely generated C-module
and 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 then
show that any word of length greater than m can be written as a sum, over C, of words
of lower length, or of lower order. This is done by noting first that, since xi is integral
over C, there exists an ni such that xki 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 the
word so that it is written as a sum of words lower in the ordering. Induction completes
the proof.
Proposition 3.2. Let R be a prime P.I. ring, T (R) the trace ring of R. Then T (R) is
integral over T .
Proof. To prove this we
Pkuse the “determinant trick” as described in [Reid], page 43. Let
x ∈ T (R), then x = i=1 ti xi , where ti ∈ T and xi ∈ R. Therefore x ∈ T hx1 , . . . xk i
with each xi integral over T . So, by Theorem 3.1, T hx1 , . . . xk i is a f.g. T -module and
can be written as T y1 + . . . T ym , for some yi ∈ T (R) and m ∈ N.
We can now apply the “determinant trick”. x belongsPto T y1 + · · · + T ym so xyi ∈ T y1 +
· · ·+TP
ym , for each 1 ≤ i ≤P
m. This means that xyi = m
j=1 aij yj with aij ∈ T . Therefore
m
m
xyi − j=1 aij yj = 0 and j=1 (xδij −aij )yj = 0. We set M = (xδij −aij )ij ∈ Mm (T [x]).
If v = (y1 , . . . ym )T , then M v = 0 and M adj M v = 0. But M adj M = (det(M ))Im , hence
we can P
conclude that det(M )yi = 0, for each 1P≤ i ≤ m. Now 1 ∈ T y1 + . . . T ym ,
m
so 1 = m
i=1 ti yi . Then det(M ) = det(M ).1 =
i=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 with
regard to some central multiplicative set. This is the result that we hope to extend to
12
a general prime P.I. ring in the final chapter. Localisation is not defined until the next
chapter, therefore, if the reader is unfamiliar with noncommutative localisation, it is
recommended that they read the first section of Chapter 4 first. We begin by looking
at the relationship between the primes of T (R) and the primes of its centre Z. The
Proposition 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 Z
the 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).
Pk
• (LO) Let p be a prime ideal of Z and r ∈ pT (R) ∩ Z. Then r =
i=1 pi ri
for some ri ∈ R, pi ∈ p, so r ∈ Zhr1 , . . . rk i. Since each ri is integral over Z,
Shirshov’s Theorem 3.1 says that Zhr1 , . . . rk i = Zr1 + · · · + Zrl , some l ≥ k.
It can easily be checked that p(Zr1 + · · · + Zrl ) = pr1 + . . . prl is a subring of
Zr1 + · · · + Zrl as well as being a Z-submodule. Moreover, r ∈ pr1 + · · · + prl and
we can apply the “determinant trick”, as in the proof of Proposition 3.2, to show
that rn + qn−1 rn−1 + · · · + q0 = 0, for some qi ∈ p. This shows that rn ∈ p, hence
r ∈ p and pT (R) ∩ Z = p. By Zorn’s Lemma there exists an ideal P in T (R) which
is maximal with respect to P ∩ Z = p. This ideal is prime: let A, B be ideals of
T (R) such that AB ⊆ P . Then (A ∩ Z)(B ∩ Z) ⊆ AB ∩ Z ⊆ P ∩ Z = p, therefore
either A ∩ Z ⊆ p or B ∩ Z ⊆ p. So, by maximality of P , either A ⊆ P or B ⊆ P
and P is prime.
• (GU) Let p ⊂ q be prime ideals of Z and P a prime ideal of T (R) such that
P ∩ Z = p (P is guaranteed by LO). Then we quotient out T (R) by P , it is
integral over its centre which is Z/P ∼
= Z/p. Then, again by LO, there is a prime
ideal Q0 of T (R)/P such that Q0 ∩ Z/P = q/P . Let Q be the preimage of Q0 in
T (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 ∩ Z
if 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-alternating
polynomials 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-alternating
polynomial in Zhx1 , . . . xt , y1 , . . . yr i, with t = n2 . Let R = Mn (A), for some commutative ring A, T an A-linear transformation of R (viewed as a free A-module), and
u1 , . . . ut , v1 . . . vr ∈ R. Then
13
1. f (T u1 , . . . T ut , v1 , . . . vr ) = det(T )f (x1 , . . . xt , y1 , . . . yr )
2. f ((λIn − T )u1 , . . . (λIn − T )ut , v1 , . . . vr ) = det(λIn − T )f (x1 , . . . xt , y1 , . . . yr )
P
3. If det(λIn − T ) = ti=0 (−1)i ai λt−i , with ai ∈ A, a0 = 1, then for 1 ≤ ak ≤ t we
have
X
ak f (u1 . . . vr ) =
f (T i(1) u1 , . . . T i(t) ut , v1 . . . vr )
P
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 toP
prove the result
in the case where u1 , . . . ut is a basis of R. In this case, T ui = tj=1 Tij uj , with
Tij ∈ A and
X
X
f (T u1 , . . . T ut , v1 , . . . vr ) = f (
T1j uj , · · ·
Ttj uj , v1 , . . . vr )
=
X
T1σ(1) . . . Ttσ(t) f (uσ(1) , . . . uσ(t) , v1 , . . . vr )
σ∈St
=
X
sign(σ)T1σ(1) . . . Ttσ(t) f (u1 , . . . ut , v1 , . . . vr )
σ∈St
= det(T )f (u1 , . . . ut , v1 , . . . vr )
2. This result follows from part 1., if we think of λIn − T as an A[λ]-linear transformation 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 required
relation.
Lemma 3.6. Let R be a prime Noetherian P.I. ring. Then T (R) is a finite central
extension of R.
Proof. Let 0 6= c ∈ gn (R). Then since gn is an n2 -alternating polynomial, for any a that
occurs 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 such
a, therefore cT (R) ⊆ R. Moreover, c central in R implies that cT (R) is an ideal of R
and hence is finitely generated as an R-module (left or right). c is also a regular element
of 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.
14
[Row1], 2.12.48. Let P be a prime ideal of R, then, by Zorn’s Lemma, there exists an
ideal Q of S maximal with respect to the property that Q ∩ S ⊆ P . By the same
argument as in the proof of Proposition 3.4, Q is a prime ideal of S. To show that
Q ∩ R = P it suffices to show that if R, S are prime rings and 0 6= P a prime ideal of R
then there exists an ideal A of S such that 0 6= A ∩ R ⊆ P . To see this, quotient out S
by 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 or
BS ⊆ Q and A ⊆ R ∩ Q or B ⊆ R ∩ Q). If P/Q 6= 0 then the existence of an ideal A
such that 0 6= A ∩ (R/Q) ⊆ (P/Q) contradicts the maximality of Q.
Hence
Pm let R, S be prime rings and P a nonzero prime ideal of R. We can write S =
i=1 ai R with each ai commuting with all of R, moreover we make a1 = 1. Now
we extend a1 to a maximal R-linearly independent subset of {a1 , . . . am }, which after
relabelling is {a1 , . . . an }, 1 ≤ n ≤ m. Since S is prime, annR (ai )ai R = 0 implies
that P
annR (ai ) = 0 (both ai R and annR (ai ) are ideals since ai is central) and hence
T = ni=1 a1 R is a free left/right R-module. Now we let Mi = {r ∈ R|ra
Tn i ∈ T }, for
1 ≤ i ≤ m. Then each Mi is an ideal of R and R prime implies M = i=1 Mi 6= 0 is
an ideal of R. We have P M 6= 0 and, since S is a central extension, P M S C S. By
the definition of M , P M S ⊆ P T . Since a1 = 1 and T a free module, T = R ⊕ T 0
and P T = P (R ⊕ T 0 ) = P R ⊕ P T 0 = P ⊕ P T 0 . Hence, by Dedekind’s modular law,
P M S ∩ R ⊆ P ⊕ (T 0 ∩ R) = P . Therefore P M S is our required ideal.
The following two Lemmas are adapted from [BS]. They show that any localisation
of T (R) is actually a central localisation and this fact will be important in proving our
main result.
Lemma 3.8 ([BS] Lemma 1). Let R be a prime P.I. ring and S ⊆ R a multiplicative
set of regular elements of R. Then
¯
·
¸
1 ¯¯
−1
T (Rhc |c ∈ Si) = T (R)
c∈S
det(c) ¯
where c−1 is the inverse of c in Q, the quotient ring of R.
1
. If T is the
[BS] 1. We have cc−1 = 1 so det(c) det(c−1 ) = 1 and det(c−1 ) = det(c)
−1
commutative ring generated by the centre of Rhc |c ∈ Si and the coefficients of
the characteristic polynomials of the elements of Rhc−1 |c ∈ Si, then det(c−1 ) ∈ T ⊆
Z(T (Rhc−1 |c ∈ Si)), because it is the final coefficient of the characteristic polynomial
of c−1 . Hence
¯
·
¸
1 ¯¯
T (R)
c ∈ S ⊆ T (Rhc−1 |c ∈ Si)
det(c) ¯
Conversely, let f (x) = xn +an−1 xn−1 +· · ·+a1 x+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
·
−1 ¯
det(c) f (c)c
= 1. So
c−1
=
−1 ¯
det(c) f (c)
¯
¯
1 ¯
det(c) ¯c
∈ T (R)
¸
∈ S . Therefore
¯
¸
1 ¯¯
Rhc |c ∈ Si ⊆ T (R)
c ∈ S and
det(c) ¯
¯
µ
·
¸¶
1 ¯¯
−1
T (Rhc |c ∈ Si) ⊆ T T (R)
c∈S
det(c) ¯
·
−1
But, since
1
det(c)
lies in the centre of T (R) for all c ∈ S, and, by Corollary 3.3, T (R) =
¯
¯
µ
·
¸¶
·
¸
¯
¯
1 ¯
1 ¯
T (T (R)), we have T T (R) det(c)
c
∈
S
=
T
(R)
c
∈
S
.
¯
det(c) ¯
In order that S being a right Ore set is a sufficient condition for localisation, we now
restrict ourselves to prime Noetherian P.I. rings (see the first section of Chapter 4 for
why this is so).
Lemma 3.9 ([BS] Lemma 2). Let R be a prime Noetherian P.I. ring and 0 ∈
/ S a
right (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 a
right Ore set in T (R). Hence the localisation T (R)S exists. Clearly RS = Rhc−1 |c ∈ Si,
so by the above Lemma (3.8) we get
¯
¸
·
1 ¯¯
c ∈ S ⊇ T (R)S
T (RS ) = T (R)
det(c) ¯
Now we must show that the opposite
inclusion
¯
·
¸ holds. If det(c) is invertible in T (R)S ,
¯
1 ¯
for all c ∈ S, then clearly T (R) det(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 has
1), there exists a maximal ideal M of T (R)S with det(c)T (R)S ⊆ M . The inclusions
RS ⊆ T (R)S ⊆ T (RS ) and Corollary 3.3 imply that T (T (R)S ) = T (RS ). Therefore, by
Lemmas 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 ). So
every 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 ∈
/S
a right Ore set of R. Let Z be the centre of T (R). Then there exists a two sided Ore set
S 0 ⊆ Z such that T (RS ) = T (R)S = T (R)S 0 .
¯
¸
·
¯
1 ¯
0
Proof. Take S = {det(c)|c ∈ S}, then T (R) det(c) ¯c ∈ S = T (R)S 0 and the rest follows
from Lemmas 3.8 and 3.9 above.
16
4
Localisation and Links
4.1
Localisation
In this subsection we quickly introduce the standard ideas of localisation in noncommutative rings. This is quite different to the case of commutative rings since additional
conditions need to be satisfied by the multiplicative set S before we can form the ring
RS . Moreover, when localisation is possible it will either be a left localisation or a right
localisation.
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 S
is 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 defined
analogously.
Definition 4.2. Let R be a ring, S ⊆ R a multiplicative set and r ∈ R. We say
that r is S-torsion if there exists a x ∈ S such that rx = 0. If no such x exists we
say that r is S-torsionfree. More generally, if A is a right R-module then we define
tS (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 any
x1 , . . . xn ∈ S, there exists r1 , . . . rn ∈ R such that x1 r1 = · · · = xn rn ∈ S. From this it
follows 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. The
right Ore condition applied to x1 , x2 gives x1 y = x2 r for some y ∈ S and r ∈ R. Since
S is a multiplicative set, x1 y ∈ S as required.
Now let a1 , a2 ∈ tS (A), then there exists x1 , x2 ∈ S such that a1 x1 = a2 x2 = 0. By the
first part, there exists a y ∈ x1 R ∩ x2 R ∩ S, and (a1 − a2 )y = 0. Given r ∈ R, there exists
z ∈ S and s ∈ R such that rz = x1 s. Then (a1 r)z = (a1 x1 )s = 0, so a1 r ∈ 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 such
that
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.
17
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 homomorphism
φ : R → S the localisation map.
Naturally, given a multiplicative set S, we wish to know when such a localisation
with respect to S exists. The answer is a standard result in noncommutative ring
theory, which we state below without proof. Firstly, we must introduce the idea of right
reversibility.
Definition 4.5. Let S be a multiplicative set in a ring R. Then S is right reversible if
for all r ∈ R and x ∈ S such that xr = 0, there exists a y ∈ S such that ry = 0. A right
denominator set S is a multiplicative set which is both right reversible and a right Ore
set.
Theorem 4.6 ([GW], 10.3). Let S be a multiplicative set in a ring R. Then there exists
a right ring of fractions for R with respect to S if and only if S is a right denominator
set.
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 that
φ(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 exists
c ∈ S and a1 , . . . an ∈ R such that c = s1 a1 = · · · = sn an . φ(c) = φ(si )φ(ai ) implies that
φ(ai ) is invertible in RS . So set ri = xi ai , 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 ACC
on right annihilators of single elements), S a right Ore set implies that S satisfies the
right reversibility condition.
Proposition 4.8 ([GW], 10.7). Let S be a right Ore set in a Noetherian ring R, then
S is right reverisble.
[GW], 10.7. Let r ∈ R and x ∈ S such that xr = 0 Then, since we have ACC on
right ideals of R, there exists an n ∈ N such that r.ann(xn ) = r.ann(xn+1 ). By the Ore
condition on S, there exists s ∈ R and y ∈ S such that ry = xn s. Then xn+1 s = xry = 0
implies s ∈ r.ann(xn+1 ) = r.ann(xn ) and hence ry = xn s = 0.
18
4.2
Goldie’s Theorem
The set of regular elements of a ring, denoted C(0), is a multiplicatively closed set and
it is often desirable to localise with respect to this set i.e. to form as large as possible a
ring of fractions for R. The question: Under what conditions does this localisation exist? was answered fully by Alfred Goldie, in around 1959. We say the ring is classically
localisable when this is the case. Goldie’s Theorem, and the standard results leading
to its proof, are essential tools when it comes to developing theory in connection with
localisation. Therefore, here we state some important results that we shall require; the
proofs 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. Then
the 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 be
a semiprime right Goldie ring and I a right ideal of R. Then I is essential in RR if and
only 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 R
with J ∩ I = 0, then (RJ)I ⊆ R(JI) ⊆ R(J ∩ I) = 0. But RJ is an ideal of R, hence
0 = RJ = J and I is essential as a right ideal of R. Hence, by Goldie’s Regular Element
Lemma 4.10 above, I contains a regular element of R (since a prime right Noetherian
ring is semiprime right Goldie).
Theorem 4.12 (Goldie’s Theorem, [GW], 6.15). A ring R has a semisimple classical right quotient ring if and only if R is a semiprime right Goldie ring.
Proof. For a proof see [GW], page 118.
19
If A is a right R-module, where R is a semiprime Noetherian ring and C(0) the set
of regular elements of R, then we define the torsion submodule of A, written t(A), to be
tC(0) (A) (since Goldie’s Theorem says that C(0) is a right Ore set). If R is not semiprime
Noetherian then we can’t generally define t(A), since it won’t be a submodule of A but
it 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 several
different 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 that
there is a link from P to Q, written P Ã Q, if there exists an ideal A of R such that
P ∩ Q ) A ⊃ P Q and (P ∩ Q)/A is a (R/P )-(R/Q)-bimodule, which is torsionfree on
both sides. If P Ã Q and A is as above, then (P ∩ Q)/A is called a linking bimodule
between P and Q.
Let R be a Noetherian ring and Spec(R) the set of all prime ideals of R. Then we
may thing of the elements of Spec(R) as the vertices in a graph, with a directed edge
between vertices P and Q if and only if P Ã Q. The connected components of this
graph are called the cliques of Spec(R), and we denote the connected component to
which a vertex P belongs as Cl(P ). Moreover, we define a right link-closed subset X
of 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 become
apparent 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 ideals
of R. Let Z be the centre of T (R). Then we say that P and Q are trace linked (“trlinked”) if there exists prime ideals P 0 , Q0 of T (R) such that P 0 ∩ R = P , Q0 ∩ R = Q
and P 0 ∩ Z = Q0 ∩ Z. If P and Q are tr-linked then we write this as P Ãtr Q, or
equivalently as Q Ãtr P .
We shall show later that, for prime P.I. rings, the two types of link share many
important 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 of
R such that S ⊆ C(Q) and P Ã Q. Then S ⊆ C(P )
20
In order to prove the above lemma, we need to make the following observation (which
is 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-torsion
modules so the result holds. Therefore we consider the case S ∩ J = ∅. Then S/J
is a right Ore set in the Noetherian ring R/J. So, without loss of generality, we may
assume that J = 0. We assume that the left S-torsion subset of I is nonzero (it is
not a submodule since S is right Ore only), and show that this implies that the right
S-torsion submodule is nonzero. So let c ∈ S and 0 6= x ∈ I such that cx = 0. Then
r.ann(c) 6= 0 and we get the ascending chain r.ann(c) ⊆ r.ann(c2 ) ⊆ . . . . Since R is
Noetherian, there exists an n ∈ N such that r.ann(cn ) = r.ann(cn+1 ). Since S is a right
Ore set, there exists y ∈ R and d ∈ C such that xd = cn y. Then cn+1 y = cxd = 0 and
hence y ∈ r.ann(cn+1 ). But r.ann(cn+1 ) = r.ann(cn ), so 0 = cn y = xd and x 6= 0 is in
the 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 torsionfree
both as a left R/P and right R/Q-module. Then S ⊆ C(Q) implies that P ∩ Q/A is
S-torsionfree on the right as an R-module. By Lemma 4.16, P ∩ Q/A is S-torsionfree on
the left too. I claim that for each c ∈ S we have r.annR/P (c + P ) = 0. Assume not, then
there exists a c ∈ S, and r ∈ R\P such that cr ∈ P . P ∩Q/A is a faithful (R/P )-module
so rP ∩ Q 6⊆ A. But cr is in P and P ∩ Q ⊆ Q means that crP ∩ Q ⊆ P Q ⊆ A. This is a
contradiction 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. Hence
S ⊆ C(P ).
Lemma 4.17. Let R be a Noetherian ring, P a prime ideal of R and S a right Ore
set of R. Then either S ⊆ C(P ) and R/P is S-torsionfree or S 6⊆ C(P ) and R/P is
S-torsion.
Proof. If S ⊆ C(P ), then by definition of C(P ), R/P must be S-torsionfree. So assume
S 6⊆ C(P ). Then I = tS (R/P ) 6= 0. I is finitely generated as a left R/P -module, so
I = (R/P )x1 + . . . (R/P )xn . There exists a c1 ∈ S such that x1 c1 = 0. x2 c1 ∈ I so
there exists a c2 ∈ S such that x2 c1 c2 = 0. By induction, we get c = c1 . . . cn in S such
that xi c = 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.
21
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 c2 r = p ∈ P or rc2 = p ∈ P . If c2 r = p then we use the
right reversibility of S2 /P in R/P to get rc02 = p0 ∈ P . Hence, without loss of generality, rc2 = p ∈ P . RS1 = RS2 implies that there exists x ∈ R, c1 ∈ S1 such that
φ(c2 )(φ(x)φ(c1 )−1 ) = 1 (where φ : R → RS1 is the localisation map with respect to
S1 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.
rc1 s = pxs ∈ P . Therefore S1 6⊆ C(P ).
Compare Proposition 4.15 above with the following Proposition. This Proposition
is Lemma 4 of [BW], though the proof we give here is different to the one given in the
paper.
Proposition 4.19 ([BW] 4). Let R be a prime Noetherian P.I. ring and S a right Ore
set in R. If P is a prime ideal of R such that S ⊆ C(P ), and P Ãtr Q, for some prime
ideal Q of R, then S ⊆ C(Q).
Proof. P Ãtr Q implies that there exists primes P 0 and Q0 of T (R) such that P 0 ∩ R =
P, Q0 ∩ R = Q and P 0 ∩ Z = Q0 ∩ 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 ) implies
that R/P is S-torsionfree so there exists x + P 6= P in R/P which is S-torsionfree.
P 0 ∩ R = P and x ∈ T (R) implies x + P 0 6= P 0 is S-torsionfree in T (R)/P 0 . Therefore
T (R)/P 0 can’t be S-torsion, thus, by Lemma 4.17, T (R)/P 0 is S-torsionfree and S ⊆
C(P 0 ). We note that, since every nonzero element of Z is invertible in Q, the simple
quotient ring of R (and T (R)), every nonzero element of Z is regular in T (R). Therefore
C(P 0 ) ∩ Z = Z\(Z ∩ P 0 ) = Z\(Z ∩ Q0 ) = C(Q0 ) ∩ Z. By Corollary 3.10, there is a two
sided Ore set S 0 ⊆ Z such that T (R)S = T (R)S 0 . Then, by Lemma 4.18 above, we have
S 0 ⊆ C(P 0 ) ∩ Z and hence S 0 ⊆ C(Q0 ). Applying Lemma 4.18 again, we get S ⊆ C(Q0 ).
Finally, S ⊆ R and Q0 ∩ 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 exists
a 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 next
section, these give a way of comparing rings, and it turns out that these dimensions are
equivalent for F.B.N. rings.
22
Definition 5.1. A ring R is right bounded if every essential right ideal of R contains a
two-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/P
is 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 a
left 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/P
is 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 (if
0 6= J Cr Q, then 0 6= J ∩ R and J, IQ have nonzero intersection in R). Since Q is a
central simple algebra and hence semisimple Artinian, IQ must be a direct summand of
Q, so IQ = Q. Therefore there exists a nonzero, regular, central element c in I. cR/P
is then a nonzero ideal of R/P contained in I. Since c is regular, by Lemma 4.9, cR/P
is 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. Set
T
N = 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 ideal
of 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 that
J ⊆ M for all maximal M .
To show the opposite inclusion, we show that every maximal right ideal M of R contains a
maximal twosided ideal of R and hence also N . Since J is the intersection of all maximal
right ideals, the result will then follow. Therefore let M be a maximal right ideal of
R and set P = r.annR (R/M ). Then P is a prime ideal because R/M is a simple right
R-module. If M/P were an essential right ideal of R/P , then there would exist an ideal
P ) I ⊇ M (since R/P is a right bounded ring). But we’d have (R/M )I = 0, which
contradicts P = r.annR (R/M ), hence M/P isn’t an essential right ideal. Therefore there
exists a nonzero right ideal J such that M ∩ J = 0. Since M is maximal, M ⊕ J = R
and J must be a minimal right ideal of R. This shows that S = soc((R/P )) 6= 0. But
in a semiprime right Noetherian ring, S is an ideal of R and R = S ⊕ T , for some ideal
23
T 1 . In our case R/P is prime so S = R/P , and R/P is semisimple (and hence simple
Artinian). 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 linked
primes and Ore sets. In particular, if P is a prime in a Noetherian ring R and we wish
to localise with respect to some Ore set S ⊆ C(P ), so that P is as “big” as possible in
the resulting ring, then we must have S ⊆ C(X), T
where X is the set of primes of R in
the smallest right link closure of P , and C(X) = Q∈X C(Q). Fortunately, it has been
shown (by B. J. Müller and A. V. Jategaonkar), that for sufficiently nice rings 2 and
if X is finite, C(X) is itself a right Ore set and hence right localisation with respect to
C(N ) exists.
This result has been extended to the case where X is infinite and we state the theorem
below. In order that the localisation formed is not “trivial” (see [BW], page 333, for
an 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 the
intersection 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 in
R. Then X is right localisable if and only if X is right link closed and satisfies the
intersection 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 weaker
condition: 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 that
R 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 all
P ∈ X, we have J ∩ C(X) 6= ∅. Equivalently, if J ∩ C(X) = ∅, then there exists P ∈ X
such that J ∩ C(P ) = ∅. Now let I be a right ideal of R such that I ∩ C(X) = ∅ and
set J = r.annR (R/I). Then J ⊆ I so J ∩ C(X) = 0 as well. Hence there exists a prime
ideal P ∈ X such that J ∩ C(P ) = ∅. Therefore (R/I)c 6= 0 for all c ∈ C(P ), and hence
I ∩ C(P ) = ∅.
1
If the reader is unfamiliar with this result, see [GW], page 131
The theorem holds for Noetherian rings satisfying the right second layer condition, which is a much
weaker condition than being a F.B.N. ring.
2
24
The result mentioned above when X is a finite set now follows from Theorem 5.5
and the following Lemma.
Lemma 5.7 ([GW], 7.5). Let N be a proper semiprime ideal in a Noetherian ring
R, and let P1 , . . . Pn be the prime ideals of R minimal over N . Then P1 , . . . Pn are
precisely 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 have
Pi (P1 . . . Pi−1 Pi+1 . . . Pn ) = 0. Therefore Pi = r.annR (Ii ), for some nonzero ideal Ii of
R. Now let x be a regular element of R with r ∈ R such that xr ∈ Pi . Then xrIi = 0
implies that rIi = 0, since x is regular, hence r ∈ Pi . Similarly, Pi = l.annR (Ji ), for
some nonzero ideal Ji of R. Following the same argument, x regular and r ∈ R such
that rx ∈ Pi implies that r ∈ Pi . Therefore x ∈ C(Pi ). Thus we have shown that
none of the Pi contains a regular element of R and C(N ) ⊆ C(P1 ) ∩ · · · ∩ C(Pn ). If
x ∈ C(P1 ) ∩ · · · ∩ C(Pn ), then l.annR (x), r.annR (x) ⊆ P1 ∩ · · · ∩ Pn = 0. Therefore x is
regular.
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 right
link 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 a
finite set. By Lemma 5.7, we have C(X) = C(N ). Therefore, in order to apply Theorem
5.5, it suffices to check that X satisfies the intersection condition described above. Let
I be a right ideal of R such that I ∩ C(Pi ) 6= ∅, for 1 ≤ i ≤ n. We prove by induction on
n, so assume there exists a c ∈ I ∩ C(P1 ) ∩ · · · ∩ C(Pn−1 ). By Goldie’s Regular Element
Lemma 4.10 (GREL), I ∩ C(Pn ) 6= ∅ implies that (I + Pn )/Pn is an essential right ideal
of R/Pn . Since the Pi are incomparable, (Y + Pn )/Pn 6= 0, where Y = P1 . . . Pn−1 . I
claim that (IY + Pn )/Pn is also essential as a right ideal. If Pn ⊆ J is a right ideal of
R 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, there
exists d ∈ IY ∩ C(Pn ), d = c + y, where y ∈ IY . Then d ∈ C(P1 ) ∩ · · · ∩ C(Pn ). To see
this - let r ∈ R with (c + y)r = n ∈ N . Since y, n ∈ Y , we have r ∈ Y . But dr ∈ N
means 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 way
of doing things. Theorem 5.8 can be (and normally is) proved directly without the use
25
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 right
link 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 is
a natural correspondence between ideals of R and RC(X) , as we shall see in the Lemma
below (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 of
R 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 the
localisation map R → RC(X) . Moreover, if J is prime then so too is φ−1 (J).
P
Proof. Let x ∈ IRC(X) , then x = ki=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 =
Pk
( i=1 φ(xi ai ))φ(s)−1 , for some s ∈ C(X) and ai ∈ R. But each xi belongs to I,
P
therefore x = φ(x0 )φ(s)−1 , where ki=1 xi ai = x0 ∈ I. Now let φ(r)φ(t)−1 be an element
of RC(X) , then φ(x0 )φ(s)−1 φ(r)φ(t)−1 = φ(x0 )φ(r0 )(φ(s0 )φ(t))−1 , where r0 ∈ R, s0 ∈ C(X)
(this follows from the Ore condition applied to s and r). x0 r0 ∈ I and s0 t ∈ C(X) so
φ(x0 )φ(s)−1 φ(r)φ(t)−1 ∈ IRC(X) .
Now consider φ(r)φ(t)−1 φ(x0 )φ(s)−1 : applying the right Ore condition to t, x0 , we get
x0 t0 = ty ∈ I, for some t0 ∈ C(X), y ∈ R. As in the proof of Lemma 4.16, this means
that there exists a z ∈ C(X) such that yz ∈ I. This gives φ(t)−1 φ(x0 ) = φ(y)φ(t0 )−1 =
φ(y)(φ(z)φ(z)−1 )φ(t0 )−1 = φ(yz)φ(t0 z)−1 . Therefore φ(r)φ(t)−1 φ(x0 )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(x1 s01 + x2 s02 )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. Clearly
φ(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 as
required. Now assume that J is a prime ideal of RC(X) and let A, B be ideals of R such
that 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) ⊆ J
or 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 the
form P RC(X) , where P ∈ X.
The next result is similar to those above and will also be required later.
26
Lemma 5.10. Let R be a right Noetherian ring, S a right Ore set and P a prime ideal
of R such that S ⊆ C(P ). Then C(P RS ) = 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), such
that (φ(x)φ(s)−1 + P RS )(φ(y)φ(c)−1 + P RS ) = 0. So φ(x)φ(s)−1 φ(y)φ(c)−1 ∈ P RS ,
φ(x)φ(s)−1 φ(y)φ(c)−1 = φ(p)φ(d)−1 where p ∈ P and d ∈ S (it was shown in the proof
of Lemma 5.9 that an element of P RS has this form). Using the right Ore condition,
φ(s)−1 φ(y)φ(c)−1 = φ(y 0 )φ(s0 )−1 φ(c)−1 with y 0 ∈ R, s0 ∈ S, and we choose a common
denominator so φ(x)φ(y 0 )φ(z1 )φ(e)−1 = φ(p)φ(z2 )φ(e)−1 , where φ(z1 ), φ(z2 ) are units
in RS and e ∈ S. Then xy 0 z1 − pz2 ∈ tS (R). Therefore there exists a d ∈ S such that
(xy 0 z1 − pz2 )d = 0. That is, x(y 0 z1 d) ∈ P . But x ∈ C(P ) implies that (y 0 z1 d) ∈ P .
Therefore φ(y 0 ) = φ(y 0 z1 d)φ(z1 d)−1 ∈ P RS , i.e. φ(s)−1 φ(y) ∈ P RS and since P RS is an
ideal we get φ(y)φ(c)−1 ∈ P RS . Hence φ(x)φ(s)−1 ∈ C(P RS ) and C(P )S −1 ⊆ C(P RS ).
Now let φ(x)φ(s)−1 ∈ C(P RS ) and y ∈ R such that xy ∈ P . Then (φ(x)φ(s)−1 +
P RS )(φ(sy)1−1 + P RS ) = 0 and hence φ(sy)1−1 ∈ P RS 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 Ore
condition, there exists q ∈ P, d ∈ S such that φ(s)−1 φ(p) = φ(q)φ(d)−1 . Therefore
φ(y)1−1 = φ(q)φ(d)−1 ∈ P RS and yd − q ∈ tS (R). As before, there exists an e ∈ S such
that (yd − q)e = 0, that is yde ∈ P . But de ∈ S ⊆ C(P ) implies that y ∈ P and so
φ(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 that
this is the case will take some work.
6.1
Classical Krull dimension
Classical Krull dimension is defined by transfinite induction and hence will involve ordinal 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 prime
ideals of R for each ordinal α. Firstly we let X−1 be the empty set. Next, consider an
ordinal α ≥ 0, if Xβ has been defined for all ordinals β < α, let Xα be theSset of those
primes 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
27
of R exists, and that this dimension (written Cl.K.dim(R)) equals γ, the smallest such
γ.
Note The following results are not stated in full generality, in many cases we assume
that R is Noetherian or F.B.N. in order to simplify the proof and because we do not
require the full result. In order to find the fully generality in which these results can be
proved, 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 dimension of R exists.
[GW] 14.1. Let the Xα ’s be the sets of prime ideals of R as defined above. The cardinality of these sets is bounded by the cardinality of P(R), the power set of R. Therefore
the transfinite chain X−1 ⊆ X0 ⊆ X1 ⊆ . . . cannot properly increase forever. So there
exists 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, there
exists a prime P maximal with respect to not being in Xγ . But then all primes properly
containing P are in Xγ which implies that P ∈ Xγ+1 . Xγ+1 = Xγ gives us the required
contradiction.
Proposition 6.3 ([GW] 14.2). Let R be a ring with Cl.K.dim(R) = γ. If α is any
nonnegative ordinal strictly less than γ, then there exists a prime P of R such that
Cl.K.dim(R/P ) = α. Moreover, if R is right or left Noetherian, then there is a minimal
prime 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 a
prime 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 a
contradiction. This shows that Cl.K.dim(R) is the supremum of Cl.K.dim(R/P ) as P
varies over all prime ideals of R. This will be greatest when P is minimal, so in fact it is
the supremum as P varies over all minimal primes of R. If R is right or left Noetherian
then R has finitely many minimal primes, one of which will have Cl.K.dim(R/P ) = γ
as required.
Theorem 6.4. Let R and S be Noetherian rings, such that R is F.B.N., and S MR a
bimodule 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 series
for M and won’t be given here. For a proof see [GW], page 237 or [Jat] page 231, noting
that a F.B.N. satisfies the second layer condition.
28
Corollary 6.5. Let R and S be F.B.N. rings, and S MR a bimodule which is finitely
generated 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 commutative rings. For noncommutative rings a more general definition which can be applied
to modules as well as rings is needed. Therefore we introduce the Krull dimension of a
module. 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 transfinite
induction. 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 class
of R-modules, such that for every countable
S decending chain M0 ⊇ M1 ⊇ M2 . . . of
submodules 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 dimension
of M and is denoted K.dim(M ). If M does not belong to any Yα then we say that M
has no Krull dimension.
The main aim of this section is to show that, for F.B.N. rings, and hence prime
P.I. rings, the Classical Krull dimension of R and the Krull dimension of RR are the
same. We denote the Krull dimension of RR by r.K.dim(R). If I is an ideal of R then
r.K.dim(R/I) will denote the Krull dimension of (R/I)(R/I) , whereas K.dim(R/I) will
generally 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 N
is a chain of submodules of M and any chain of submodules of M/N can be written as
the image of a chain of submodules of M under the natural quotient map. We prove
the 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/N
and 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 hence
max{K.dim(Ai /Ai+1 ), K.dim(Bi /Bi+1 )} < α. The induction hypothesis implies that
29
K.dim(Mi /Mi+1 ) ≤ max{K.dim(Ai /Ai+1 ), K.dim(Bi /Bi+1 )} < α, for all but finitely
many i. But this is the same as saying that K.dim(M ) ≤ α, and the induction is
complete.
Lemma 6.8 ([GW], 15.5). Let R be a right Noetherian ring, then there exists a
minimal 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 is
right Noetherian, there exist minimal prime ideals P1 , . . . Pn of R, who’s product is zero.
Then for each i = 1, . . . , n, we have
K.dim((P1 P2 . . . Pi−1 )/(P1 P2 . . . Pi )R ) = K.dim((P1 P2 . . . Pi−1 )/(P1 P2 . . . Pi )(R/Pi ) )
which is ≤ r.K.dim(R/Pi ). Therefore, by repeatedly applying Lemma 6.7 above, we
get r.K.dim(R) = K.dim(R/(P1 P2 . . . 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 a
monomorphism 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 submodules
M ⊇ 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 is
maximal.
Proof. The Artin-Wedderburn Theorem tells us that if R is a prime right Artinian ring
then it is simple. So assume that R has a nonmaximal prime ideal P . Then R/P is
a prime right Artinian ring which is not simple. This contradiction shows that P is
maximal.
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 with
I ) P , then K.dim(R/I) < K.dim(R/P ) as right R-modules. Since R/P is a prime
right Noetherian ring, Corollary 4.11 says that I/P contains a regular element c. So we
30
define a right R-homomorphism φ : R/P → R/P by x 7→ cx. This is a monomorphism
because c is regular. Hence, by Lemma 6.9,
µ
¶
(R/P )
r.K.dim(R/P ) > K.dim
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
µ
¶
(R/P )
r.K.dim(R/I) ≤ K.dim
< r.K.dim(R/P )
φ(R/P )
Now we prove the inequality by induction on α = r.K.dim(R). If α = 0, then R is
a 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 ideal
P 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 prime
ideal Q of R, Q 6= 0, such that Cl.K.dim(R/Q) = Cl.K.dim(R) − 1. By the first
part 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. ring
then the above inequality becomes an equality (this is not true for a general right
Noetherian ring - see [GW], page 259 for a counterexample). The proof of Theorem 6.13
is 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 descending
chain of right ideals of R. Then there exists a right ideal N of R and n ∈ N such that
Mn+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 any
essential right ideal of R. Therefore we may assume Mi 6= 0 for all i. We assume the
result isn’t true and show that this gives us a contradiction. So for each i, let Ni be a right
ideal of R such that Mi ⊕ Ni ≤e R. Moreover, we demand that Ni ⊇ Ni−1 (if not then
set Ni0 = Ni + Ni−1 ). Then there exists a k(i) > i such that Mk(i) ⊕ Ni is not essential in
R. Therefore we can create a subsequence i0 = 0, i1 = k(i0 ), . . . in = k(in−1 ) . . . , which
has the property that Min ⊕ Nin ≤e R and Min+1 ⊕ Nin 6≤e R. Then Min+1 ⊕ Nin+1 ≤e R
3
This can be rewritten simply as saying that R has finite rank since it is Noetherian. Hence there
exists 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.
31
and Min+1 ⊕ Nin 6≤e R implies that Nin+1 ) Nin . This gives us an ascending chain of
right 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). Therefore it suffices to show that Cl.K.dim(R) ≥ r.K.dim(R). We do this by induction
on α = r.K.dim(R), if α = −1 then R = 0 and contains no prime ideals, hence
Cl.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 Proposition
6.10, 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 that
r.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 a
descending 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. By
Lemma 6.12, there exists a right ideal N of R, and n ∈ N, such that Mn+k ⊕ N ≤e R
for all k ≥ 0. So let i ≥ n. Then, since Mi ∩ N = Mi+1 ∩ N = 0, we have
Mi + N ∼ Mi
=
Mi+1 + N
Mi+1
Because R is a prime F.B.N. ring and Mi+1 + N an essential right ideal of R, there
exists 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. Therefore
K.dim(Mi /Mi+1 ) ≤ r.K.dim(R/Ii ). It was shown in the proof of Lemma 6.11 that
Ii 6= 0 implies that r.K.dim(R/Ii ) < r.K.dim(R). Therefore we can apply the inductive
hypothesis to r.K.dim(R/Ii ) to conclude that r.K.dim(R/Ii ) = Cl.K.dim(R/Ii ). But
Cl.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 R BS a bimodule which is finitely generated 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 gives
r.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. Then
r.K.dim(R/P ) = r.K.dim(R/Q).
32
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 unfaithful as a left R/P -module, then I = annR/P (P ∩ Q/A) is a nonzero, twosided ideal
of R/P . Since R/P is a prime right Noetherian ring, Corollary 4.11 says that I contains
a 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 on
the 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 ring
and Z the centre of T (R). If P is a prime ideal of R, Q a prime ideal of T (R) such that
Q ∩ 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 says
that T (R) is a finite central extension of R and hence Noetherian on both sides. This
allows 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). If
Cl.K.dim(Z/p) = 0, then p is a maximal prime of Z. By Proposition 3.4, T (R) satisfies
LO, 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 the
result holds for all β < α. By Proposition 6.3, there exists a prime ideal p0 of Z such
that p ( p0 and Cl.K.dim(Z/p0 ) = α − 1. LO impies that there is a prime Q0 of
T (R) such that Q0 ∩ Z = p0 , GU then says that Q ( Q0 . Finally, GU and INC imply
that there is no prime ideal of T (R) lying between Q and Q0 so Cl.K.dim(T (R)/Q0 ) =
Cl.K.dim(T (R)/Q) − 1. The inductive hypothesis tells us that Cl.K.dim(T (R)/Q0 ) =
α − 1, therefore Cl.K.dim(T (R)/Q) = α = Cl.K.dim(Z/p). Then Theorem 6.13 says
that 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 closed
subset of Spec(R) containing P . Moreover, this set is finite.
We shall first prove that this set is finite.
33
Lemma 7.2. 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. Let X be the set of all prime ideals Q of R such that
P Ãtr Q, then X is finite.
Proof. Let p = P ∩ Z and Y the set of all prime ideals of R minimal over pR. Since
R is a Noetherian ring, this set is finite. I claim that X ⊆ Y , i.e. the primes of X are
minimal over pR. Choose Q ∈ X, then p ⊆ Q implies that pR ⊆ Q. Since p ⊆ Z, pR
is an ideal of R. Let Q0 be a prime ideal of R with pR ⊆ Q0 ⊆ Q. Intersecting with Z
gives p ⊆ pR ∩ Z ⊆ Q0 ∩ Z ⊆ Q ∩ Z = p, hence Q0 ∩ Z = p. Proposition 3.4 says that
INC holds for R over Z, therefore Q = Q0 is minimal over pR. So we get X ⊆ Y and X
must 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, since
Z\p ⊆ C(P ) is a right Ore set, Proposition 4.15 implies that Z\p ⊆ C(Q). Therefore
Q ∩ Z ⊆ p. But we can apply the same argument to Z\(Z ∩ Q), which is a left Ore
set, to get p ⊆ Z ∩ Q. Similarly, if Q0 is a prime ideal such that P Ã Q0 , repeating
the argument gives Q0 ∩ Z = p. Hence Cl(P ) is a subset of the tr-link closure of
P in Spec(R). Moreover, Cl(P ) is finite by Lemma 7.2. Now let Y be the smallest
right link-closed subset of Spec(R) containing P . Then
Tn Y ⊆ Cl(P ) implies that Y is
a finite set. If Y = {P1 , P2 , . . . Pn = P } and N = i=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 can
form the localisation RN . Lemma 5.9 shows that the maximal ideals of RN are precisely
those of the form Pi RN , for 1 ≤ i ≤ n. Corollary 3.10 shows that there is a twosided
Ore set S ⊆ Z such that RN = RS . Therefore if Q is any other prime ideal of R such
that Q ∩ Z = p, then S ∩ Q = ∅ and QRN 6= RN is a proper ideal of RN . So QRN
is contained in a maximal ideal Pi RN 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 shown
above 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-link
closed subset of Spec(R) containing P , we have shown X ⊆ Y ⊆ Cl(P ) ⊆ X, which is
finite.
The next Proposition is the key to our main result because it will allow us to connect
the link structure of R with the link structure of T (R). The Proposition we state and
prove below is a weaker result than Proposition 5 of [BW], however it is sufficient for
our needs and its proof is considerably simpler. We break the proof into two parts in
order to make it easier to understand.
Lemma 7.3. Let R be a Noetherian and Artinian P.I. ring and S a finite central
extension of R. Let P Ã Q be prime ideals of R. Then there exists prime ideals P 0 , Q0
34
of S such that P 0 ∩ R = P , Q0 ∩ R = Q and there exists a simple, faithful (S/P 0 )-(S/Q0 )bimodule M 4 .
The main idea of this proof (to apply the Jordan-Holder Theorem to R ⊗ Rop and
S ⊗ S op series) comes from [BW], page 329. If the reader is unfamiliar with the JordanHolder Theorem and the existence of composition series for modules which are both
Artinian and Noetherian then they may wish to read Chapter 4 of [GW].
Proof. Since P Ã Q, there exists an ideal P Q ⊆ A ( P ∩ Q with P ∩ Q/A a torsionfree (R/P )-(R/Q)-bimodule. Therefore we have a sequence of left R ⊗ Rop -modules
0 ⊆ P Q ⊆ A ( P ∩ Q ⊆ R ⊆ S, and we let 0 = C0 ⊆ C1 ⊆ · · · ⊆ Cn = S be a composition series of S as a left S ⊗ S op -module. By applying the Jordan-Holder Theorem to
refinements of the two sequences of R⊗Rop -modules defined above, we can conclude that
there exists an ideal A ⊆ B of R, and R ⊗ Rop -modules Ci ⊆ D1 ⊆ D2 ⊆ Ci+1 such that
B/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 0 = l.annS (Ci+1 /Ci ),
Q0 = r.annS (Ci+1 /Ci ), then they are prime ideals of S because Ci+1 /Ci is a simple left
S ⊗ S op -module. Ci ⊆ D1 ⊆ D2 ⊆ Ci+1 and P = l.annR (D2 /D1 ), Q = r.annR (D2 /D1 )
imply that P 0 ∩ R ⊆ P , Q0 ∩ R ⊆ Q. But P 0 ∩ R, Q0 ∩ R are prime ideals of R and, since
R is Artinian, every prime ideal of R is maximal (Proposition 6.10), so it follows that
P 0 ∩ R = P , Q0 ∩ 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 of
R. Then there exist prime ideals P 0 , Q0 of T (R) such that P 0 ∩ R = P, Q0 ∩ R = Q and
P 0 Ãtr Q0 .
Proof. Since T (R) is a finite central extension of R, P QT (R) is an ideal of T (R). We
let R0 = R/P Q and S = T (R)/P QT (R) so that S is a finite central extension of R0 .
e be the images of P , Q in R0 . Then Pe à Q
e in R0 because P Q ⊆ A. Since
Let Pe, Q
0
e
e
P and Q are precisely the minimal primes of R and Corollary 6.15 says that A Ã B
e must be a right link closed set. Let
implies r.K.dim(R0 /A) = r.K.dim(R0 /B), {Pe, Q}
e
e
C = C(P ∩ Q) = C(0) be the set of regular elements of R0 . Then Theorem 5.8 says that C
is a right Ore set and we can form the localisation RC0 . Since S is a central extension of
R0 , C is also a right Ore set in S and SC is a finite central extension of RC . The argument
e 0 . Also, since PeR0 , QR
e 0
given in the proof of Lemma 7.9 below shows that PeRC0 Ã QR
C
C
C
0
are precisely the maximal ideals of RC and their intersection is 0, the argument given in
the proof of Lemma 7.8 below shows that RC0 (and hence SC ) is Artinian.
e0
Therefore we can apply Lemma 7.3 and conclude that there exists prime ideals Pe0 , Q
0
0
0
0
0
0
e
e
e
e
in SC such that P ∩ RC = P RC and Q ∩ RC = QRC . We also know that there exists a
4
This result says that there exists a “bond” between P 0 and Q0 , see [Jat], page 123 for a definition
of a bond between prime ideals.
35
e 0 )-bimodule M . Let φ : T (R) → SC be the natural map,
simple, faithful (SC /Pe0 )-(SC /Q
who’s restriction to R is the natural map R → RC0 . Then we set P 0 = φ−1 (Pe0 ) and
e 0 ). M becomes a (T (R)/P 0 )-(T (R)/Q0 )-bimodule by defining (x + P 0 )m =
Q0 = φ−1 (Q
0
(φ(x) + Pe )m and similarly for m(x + Q0 ). From the way it is defined, M is faithful on
both sides. We use this fact to show that P 0 ∩ Z = Q0 ∩ Z, where Z is the centre of
T (R). Let p = P 0 ∩ Z, then for c ∈ Z\p, (c + P 0 )M 6= 0. But c belongs to the centre
of T (R), which means that (c + P 0 )M = M (c + Q0 ), and hence P 0 ∩ Z ⊆ Q0 ∩ Z. The
same 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 prime
ideals 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 P Q ⊆ A ⊂ P ∩ Q such that P ∩ Q/A is
a (S/P )-(S/Q)-bimodule which is finitely generated and torsionfree on both sides. Let
P 0 = P ∩ R and Q0 = Q ∩ R, then P 0 (P ∩ Q/A) = (P ∩ Q/A)Q0 = 0 imply that P ∩ Q/A
is also a (R/P 0 )-(R/Q)-bimodule. Clearly it is torsionfree on both sides and, since S is
finitely 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 0 ) = r.K.dim(R/Q0 ). Now A ∩ R ⊆ P 0 , Q0 , hence
r.K.dim(R/P 0 ) = r.K.dim(R/Q0 ) implies that A ∩ R = P 0 ⇔ A ∩ R = Q0 . Therefore,
either P 0 = Q0 or A ∩ R is a proper subset of both P 0 and Q0 . If the latter is the case,
then (P 0 ∩ Q0 )/(A ∩ R) is a nonzero finitely generated (R/P 0 )-(R/Q0 )-bimodule. Since it
is isomorphic to P ∩ Q ∩ R/A, it is torsionfree on both sides. Hence P 0 Ã Q0 as required.
Corollary 7.6. Let R be a prime Noetherian P.I. ring and P, Q prime ideals of R such
that 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 X
a 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.
36
Proof. By Corollary 7.6, if X has any of properties 1. 2. or 3. then the elements of X
are all trace linked. Therefore it suffices to show that if P, Q are any tracelinked prime
ideals of R, then there exists primes P1 , . . . Pn such that P à P1 à . . . à Pn à Q (a
similar 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 0 , Q0 of T (R), such that P 0 ∩ Z = Q0 ∩ Z
and P 0 ∩ R = P , Q0 ∩ R = Q, where Z is the centre of T (R). Proposition 7.1 says
that P 0 is in the right link closure of Q0 . Therefore there exists prime ideals P10 , . . . Pn0
of T (R) such that P 0 Ã P10 Ã . . . Ã Pn0 Ã Q0 . Let Pi = Pi0 ∩ R, then by Lemma 7.5,
either Pi = Pi+1 or Pi à Pi+1 , for each 0 ≤ i ≤ n. So, after removing repetitions, we
get P Ã P1 Ã . . . Ã Pn0 Ã Q as required.
We note here that, though the cliques of T (R) are all finite, and there is a clear
connection between linked primes in T (R) and R, it does not follow that the cliques in
R are finite. In [BM], Bruno J. Müller 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 again
below. This next lemma will enable us to show that every localisation in a F.B.N. ring
can 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 proof
below.
Lemma 7.8. Let R be a F.B.N. ring and X the set of all maximal ideals of R. Then
C(X) is precisely the set of units of R.
T
Proof. Clearly every unit of R is contained in C(X). Set J = M ∈X M , then, by
Proposition 5.4, J is the Jacobson radical of R. It suffices to show that elements of
C(X) are invertible in R/J. To see this, let x ∈ R such that x + J is invertible, then
there exists a y in R such that (1 − xy) ∈ J. This means that 1 − (1 − xy) is right
invertible in R, that is, (xy)z = 1 for some z ∈ R. Therefore yz is a right inverse for x
in 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 minimal
prime P of R/J such that Cl.K.dim(R/J) = Cl.K.dim((R/J)/P ). But every prime
ideal of R/J is maximal, so 0 = Cl.K.dim((R/J)/P ) = Cl.K.dim(R/J). Since R/J is
a F.B.N., ring we have Cl.K.dim(R/J) = r.K.dim(R/J) = 0 (by Theorem 6.13), and
hence R/J is a right Artinian ring. Therefore all regular elements of R/J are invertible
(consider the ascending chain xR ⊇ x2 R ⊇ · · · ⊇ xn R ⊇ . . . , for each regular x ∈ R/J).
But the regular elements of R/J are C(X) + J, as required.
37
Lemma 7.9. Let R be a F.B.N. ring, S a right Ore set and φ : R → RS the localisation
map of R with respect to S. Let Q0 be a maximal ideal of RS , Q = φ−1 (Q0 ) and P a
prime ideal of R such that P Ã Q. Then P RS is a maximal ideal of RS .
Proof. We note that Lemma 5.9 shows that Q is indeed a prime ideal of R. Also, since
Q0 is a maximal ideal of RS , S ⊆ C(Q) and, by Proposition 4.15, S ⊆ C(P ). P Ã Q
means that there exists an ideal P Q ⊆ 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 be
the corresponding (RS /P RS )-(RS /QRS )-bimodule. Then, since Lemma 5.10 says that
C(P RS ) = C(P )S −1 and C(QRS ) = C(Q)S −1 , it is clear that (P ∩Q/A)S −1 is torsionfree
on 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 /P RS ) =
Cl.K.dim(RS /QRS ). Since QRS is maximal, it follows that P RS 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) and
RC(X) = RS .
The proof of Proposition 7.10 is a combination of the proof given in [BW], page 334
and Lemmas 7.8 and 7.9 above.
Proof. Let X 0 be the set of all maximal ideals in RS and X = {φ−1 (M )|M ∈ X 0 } (where
φ : R → RS is the localisation map). Then Lemma 5.9 says that X consists of prime
ideals and Lemma 7.9 shows that it is right link closed. Since φ(C(X)) = C(X 0 ), Lemma
7.8 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 such
that I ∩ C(X) = ∅, then φ(I)RS is a proper ideal of RS and hence is contained in some
maximal ideal of RS . Therefore I ⊆ P for some P ∈ X and I ∩ C(P ) = ∅ i.e. X satisfies
the weaker, symmetric, intersection condition (and hence, by Lemma 5.6, the standard
intersection condition). Theorem 5.5 says that C(X) is right Ore and we can form the
localisation 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, and
S a right (left) Ore set in R, then there exists a two sided Ore set S 0 such that S ⊆ S 0
and the right (left) localisation of R by S equals the two-sided localisation of R by S 0 .
[BW], Theorem B. By Proposition 7.10, there exists a set X of right localisable prime
ideals of R such that RC(X) = RS . We know that X is right link closed and satisfies
the weaker, symmetric, intersection condition. Theorem 7.7 says that X is also a left
link closed set, so we apply Theorem 5.5 again to conclude that C(X) is also a left Ore
set.
38
The main theorem tells us that, if R is a prime Noetherian P.I. ring and S a left or
right Ore set, then in the localisation RS any element can be written both as φ(r)φ(s)−1
or φ(s0 )−1 φ(r0 ), for some r, r0 ∈ R and s, s0 ∈ S.
References
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Mathematics 67, 233-245 (1976).
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[BW] A. Braun and R.B. Warfield, Jr, Symmetry and Localization in Noetherian
Prime PI Rings, Journal of Algebra 118, 322-335 (1988).
[CH] A.W. Chatters and C.R. Hajarnavis, Rings with chain conditions, Research
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