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Communications in Algebra® , 34: 3751–3759, 2006 Copyright © Taylor & Francis Group, LLC ISSN: 0092-7872 print/1532-4125 online DOI: 10.1080/00927870600862573

FINITE NORMALIZING EXTENSIONS OF FSN RINGS Karl A. Kosler Department of Mathematics, University of Wisconsin–Waukesha, Waukesha, Wisconsin, USA Fully semiprimary Noetherian bimodules and their bimodule extensions are examined. In the presence of incomparability of the link graph of prime ideals, certain bimodule extensions preserve the fully semiprimary property. In particular, a finite normalizing extension ring of a fully semiprimary Noetherian ring is also fully semiprimary as a bimodule over the base ring. It is shown that the extension ring is itself a fully semiprimary ring. An application to crossed products over finite groups is given. Key Words:

Bimodule; Finite extension; Fully semiprimary; Noetherian; Normalizing extension.

Mathematics Subject Classification:

Primary 16P40, 16S20; Secondary 16D20, 16S35.

1. INTRODUCTION A Noetherian bimodule S BR is called right primary provided Ass!BR " is a singleton. A Noetherian bimodule S BR is called right semiprimary provided there is a bimodule monomorphism B → B1 ⊕ · · · ⊕ Bn where, for each i, Bi is a right primary S-R bimodule. Finally, a bimodule S BR is called right fully semiprimary Noetherian, or right FSN for short, if S BR is Noetherian and every bimodule factor of B is right semiprimary. The left-hand versions of these concepts are defined likewise. When used to describe a ring R, these definitions refer to the bimodule R RR . According to Jategaonkar (1986, Theorem 8.3.9), examples of right FSN rings include Noetherian rings that satisfy the right second layer condition. However, from Kosler (2003, §4) the rings in Stafford (1985, Theorem 2.3) are right FSN rings for which the right second layer condition fails. In part, this article is motivated by the desire to construct a large class of FSN rings that do not satisfy the second layer condition. The aim of this article is to determine when a finite normalizing extension of a right FSN ring is a right FSN ring. A finite normalizing extension of a ring R is a ring S ⊇ R such that, S = a1 R + · · · + an R where ai R = Rai for all i. If, for all i, there exists #i ∈ Aut!R" such that ai r = #i !r"ai for all r ∈ R, then S is called a finite automorphic extension of R. In the special case that #i = 1 for all i, ai r = rai for all r ∈ R and then S is called a centralizing extension of R. If G is a finite group, Received May 17, 2005; Revised July 14, 2006. Communicated by R. Wisbauer. Address correspondence to Karl A. Kosler, Department of Mathematics, University of Wisconsin–Waukesha, 1500 University Drive, Waukesha, Wisconsin 53188-2799, USA; E-mail: [email protected] 3751

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then by McConnell and Robson (1987, 10.1.4 Example (ii)), a crossed product R ∗ G is a finite automorphic extension of R. From McConnell and Robson (1987, 10.1.4 Example (iv)), there exist finite normalizing extensions that have no finite automorphic generating set and so are not of the form R ∗ G for any finite group G. The main result of this paper states that if R is a right FSN ring for which the link graph of Spec!R" satisfies incomparability, then a finite normalizing extension of R is also a right FSN ring. In particular, if G is a finite group, then the ordinary group ring RG is a right FSN ring. Furthermore, thanks to a result of Bell’s (1987), RG does not satisfy the right second layer condition if R does not satisfy the right second layer condition. 2. NOTATIONS AND DEFINITIONS All rings considered have an identity and modules are unitary. Properties and adjectives, when not accompanied by one-handed modifiers are meant to hold on both sides. However, all modules will be right R-modules unless otherwise specified. The notation N ≤ MR , or N ≤ M when R is understood, will be used to indicate that N is a (right) R-submodule of M. If N ≤ M is essential in M, then we will write N ≤e M. In case M is an S-R bimodule, we will write S MR and use N ≤ S MR to indicate that N is an S-R sub-bimodule of M. If N ≤ S MR and N ∩ N $ %= 0 for all 0 %= N $ ≤ S MR , then N is called biessential in M and we will write N ≤bi M. For a nonempty subset X of an R-module M, rR !X" = #r ∈ R ' xr = 0 for all x ∈ X$. In case X is a nonempty subset of a left R-module, then lR !X" = #r ∈ R ' rx = 0 for all x ∈ X$. If X is a nonempty subset of R, then annM !X" = #m ∈ M ' mx = 0 for all x ∈ X$. A module M is said to be finitely annihilated if there exist finitely many elements m1 % & & & % mk ∈ M such that rR !M" = rR !m1 % & & & % mk ". Note that if M is any finitely annihilated module, then there is a 1-1 map R/rR !M" → M !n" for some n. The collection of all prime ideals of R is denoted by Spec!R". If U is a uniform R-module, and R is right Noetherian, then ass!U" = #r ∈ R ' Vr = 0 for some 0 %= V ≤ U$ ∈ Spec!R" and is called the associated prime ideal of U . For an arbitrary module M, the associated prime ideals of M is the set of all prime ideals ass!U" where U is a uniform submodule of M. The set of maximal members of Ass!M" will be denoted by max Ass!M". 3. BIMODULES AND LINKED PRIMES This section is devoted to showing that the right FSN property is preserved under certain bimodule extensions. For example, if S is a finite normalizing extension of a right FSN ring R and the link graph of Spec!R" satisfies incomparability, then R S/XR is right FSN for every R-R sub-bimodule X ⊆ S. This result is used in Section 5 to show that the ring S is right FSN. We begin by collecting a few preliminary results. In the following, a bimodule B = % 0 is called biuniform if for every nonzero sub-bimodule D of B, D ≤bi B. S R Proposition 3.1. Let R be a right Noetherian ring, S any ring and S BR %= 0 a Noetherian bimodule. (1) Every right R-submodule of B is finitely annihilated.

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(2) Let P ∈ Ass!BR ". Then annB !P" is a torsion-free right R/P-module iff P ∈ max Ass!BR ". (3) B is right FSN iff every biuniform factor of B is primary. Proof. (1) Pick a 0 %= U ≤ BR . As a left S-module, SU is Noetherian and so SU = Su1 + · · · + Sum for some u1 % & & & % um ∈ U . It follows that rR !U" ⊆ rR !u1 % & & & % um " ⊆ rR !U". Therefore, rR !U" = rR !u1 % & & & % um ". (2) If annB !P" is a torsion-free right R/P-module and P ⊂ Q ∈ Ass!BR ", then 0 %= annB !Q" ⊆ annB !P". Since !annB !Q""!Q/P" = 0, annB !Q" is a torsion R/Psubmodule of annB !P", a contradiction. Suppose that P ∈ max Ass!BR ". Let Z %= 0 be the torsion R/P-submodule of annB !P". Let Q ∈ Ass!ZR ". By Proposition 3.1(1), Z is finitely annihilated. It follows that P ⊂ rR !Z" ⊆ Q contradicting our choice of P.

(3) If B is right FSN and X ≤ S BR , then there is a bimodule monomorphism f ' B/X → B1 ⊕ · · · ⊕ Bn for some right primary S-R bimodules Bi % i = 1% & & & % n. If, in addition, B/X is biuniform, then one of the maps (i f is a monomorphism where (i denotes the canonical projection onto Bi . For the converse, let X ≤ S BR . Set B$ = B/X and choose ! a maximal independent family #Bi ' i = 1% & & & % n$ of sub-bimodules of B$ . Then ≤bi B$ . i Bi ! $ For each j = 1% & & & % n, let Dj be a maximal biessential extension in B of #Bi ' $ i %= j$. Each factor B$ /Dj is biuniform and D1 ∩ · · · ∩ Dn = 0. Thus, each B /D j is ! right primary and there is a bimodule monomorphism B$ → i B/Di . !

A right cell is a nonzero Noetherian bimodule S CR such that rR !C" is a prime ideal, C is torsion-free over R/rR !C" and for all 0 %= C $ ≤ S CR , C/C $ is torsion over R/rR !C". By Jategaonkar (1981, Proposition 1.2), every nonzero Noetherian bimodule contains a right cell. Furthermore, by Jategaonkar (1981, Proposition 1.3) every biuniform Noetherian bimodule contains a unique largest right cell. Proposition 3.2. Let R be a right Noetherian ring. If S BR is a biuniform Noetherian bimodule, then max Ass!BR " = #P$ where P = rR !C", C the unique largest right cell in B.

Proof. Let P = rR !C". We claim that P ∈ max Ass!BR ". If not, then for some Q ∈ Ass!B", P ⊂ Q. Since Q ∈ Ass!B", there exists Q$ ∈ max Ass!B" such that Q ⊆ Q$ . By Proposition 3.1(2), annB !Q$ " is torsion-free over R/Q$ . Let D ⊆ annB !Q$ " be a right cell with, say, Q$$ = rR !D". Clearly, Q$ ⊆ Q$$ . Since D is torsion-free over R/Q$ , Q$ = Q$$ . In particular, since C ∩ D %= 0, Q$ = rR !C ∩ D". However, P = rR !C ∩ D" ⊂ Q ⊆ Q$ which is impossible. Thus, P ∈ max Ass!BR ". Let P $ ∈ max Ass!B". Then H = annB !P" ∩ annB !P $ " %= 0 and, by Proposition 3.1(2), H is torsion-free over both R/P and R/P $ . In particular, P ⊆ rR !H" ! forces P = rR !H". Likewise, P $ = rR !H". Therefore, P = P $ . Let R be a Noetherian ring and let P% Q ∈ Spec!R". There is a link from Q to P, denoted Q ∼→ P provided there exists an ideal A of R with QP ⊆ A ⊂ Q ∩ P such that Q ∩ P/A is torsion-free as a left R/Q-module and as a right R/P-module.

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Proposition 3.3 (Kosler, 2003, Proposition 3.1). Let R be a Noetherian ring, S any ring and S BR %= 0 a Noetherian bimodule. If P ∈ max Ass!BR " and Q ∈ Ass!B/annB !P"R ", then Q ∈ Ass!BR " or Q ∼→ P. If B and B$ are right FSN bimodules, then it is natural to ask if the direct sum B ⊕ B$ is also right FSN. Give the nature of the definition of fully semiprimary, it seems reasonable to conjecture a positive answer. However, we have been able to prove this only in case members of the right cliques of Spec!R" (defined below) are incomparable. If R is a Noetherian ring and P ∈ Spec!R", then the right clique of P, denoted )!P", is the set of all Q ∈ Spec!R" such that for some P1 & & & Pn ∈ Spec!R" Q ∼→ P1 · · · ∼→ Pn ∼→ P& A Noetherian ring R satisfies incomparability if for all P ∈ Spec!R", any two members of )!P" are incomparable. It is well known that a Noetherian ring that satisfies the right second layer condition also satisfies incomparability (Goodearl and Warfield, 1989, Corollary 12.6 or Jategaonkar, 1982, Theorem 1.8). It is an open question if a right FSN ring satisfies incomparability. In fact, there is no known example of a Noetherian ring which does not satisfy incomparability. Theorem 3.4. Let R be a Noetherian ring that satisfies incomparability and let %= 0 be a Noetherian bimodule. If B = B1 + · · · + Bn where each Bi is a right FSN sub-bimodule of B, then B is right FSN.

S BR

Proof. Assume, by way of Noetherian induction, that the result is true for all proper bimodule factors of B. By Proposition 3.1(1), It suffices to show that every biuniform factor of B is right primary. If B/X is such a factor with X %= 0, then B/X = *i !Bi + X"/X and each Bi right FSN yields, by hypothesis, that B/X is right FSN. In particular, B/X is primary. Thus, it suffices to assume that B itself is biuniform and show that B is primary. Note then that each Bi is biuniform. Let C be the unique largest right cell contained in B and let P = rR !C". By Proposition 3.2, max Ass!B" = #P$. We claim that annB !P" ≤e BR . If not, then for some prime, uniform U ≤ BR , U ∩ annB !P" = 0. Let Q = rR !U" = ass!U". Then Q ⊆ P. If Q = P, then UP = 0 whence U ⊆ annB !P", a contradiction. Thus, Q ⊂ P. Since Bi is right FSN and biuniform, Ass!Bi " = #P $ $ for some prime P $ . Also, $ P ∈ Ass!B" forces P $ ⊆ P. Now, Bi ∩ annB !P" %= 0 and so Ass!Bi ∩ annB !P"" = #P $ $; but !Bi ∩ annB !P""P = 0 whence P ⊆ P $ . Thus, for all i, Ass!Bi " = #P$. In particular, as right R-modules, ! annBi !P" ≤e Bi . Let V = i Bi with the canonical S-R bimodule structure. Then annV !P" = ! ann !P" and consequently, annV !P" ≤! Bi e V . Passing to the quotient yields an i S-R bimodule isomorphism V/annV !P" - i !Bi /annBi !P"". It follows that there is a bimodule epimorphism V/annV !P" → B/annB !P". Also, since U ∩ annB !P" = 0, there is a 1-1 map of right R-modules U → B/annB !P". Let 0 = Y0 ⊂ · · · ⊂ Ym = V/annV !P" be a right affiliated series with corresponding prime ideals Qi = rR !Yi /Yi−1 " for i = 1% & & & % m. By Jategaonkar (1986, Theorem 8.3.1) Qi ∈ )!Ass!V/annV !P""" for all i. Furthermore, !V/annV !P"" !Qm · · · · · Q1 " = 0, and as a consequence of the epimorphism established above,

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U!Qm & & & Q1 " = 0. Thus, Q1 · · · · · Qm ⊆ Q and so Qj ⊆ Q for some j. Since Q ⊂ P, Qj ⊂ P. There exists a chain P ⊃ Qj ∼→ P1 ∼→ · · · ∼→ Pn ∈ Ass!V/annV !P""& From Proposition 3.3, either Pn ∈ Ass!V" or Pn ∼→ P. From the definition of V , it’s easy to see that in the first case, Pn = P. Thus, in either case, Qj ∈ )!P" and incomparability is violated. ! If S ⊇ R is an extension ring, then an element a ∈ S normalizes R provided aR = Ra. It is easy to see that if a ∈ S normalizes R, then rR !a" and lR !a" are twosided ideals of R. In the following, we use Bim!B" to denote the lattice of all R-R sub-bimodules of a bimodule R BR ordered under inclusion. Lemma 3.5. Let R be a right FSN ring and let S ⊇ R be an extension ring. If a ∈ S normalizes R and X ≤ R SR with a 1 X, then aR + X/X is a right FSN R-R bimodule. Proof. Let Y/X ⊆ aR + X/X be a sub-bimodule such that aR + X/Y is biuniform. Since X ⊆ Y , aR + X/Y = aR + Y/Y . Further, as right R-modules, aR + Y/Y - R/H where H = rR !a + Y". Since a normalizes R, it is easy to see that H is an ideal of R. Consider the map + ' Bim!R/H" → Bim!aR + Y/Y" given by +!B/H" = aB + Y/Y . It is routine to verify that + is a 1-1 lattice map. Consequently, R/H is a biuniform R-R bimodule. Since R is a right FSN ring, Ass!R R/H" is a singleton by 3.1(3). The right R-isomorphism aR + Y/Y - R/H yields that Ass!aR + Y/Y"R is a singleton. It follows from 3.1(3) that aR + X/X is a right FSN R-R bimodule. ! Corollary 3.6. Let S be a finite normalizing extension of a ring R with generators a1 % & & & % an . If R is a right FSN ring that satisfies incomparability and X ≤ R SR , then S/X is a right FSN R-R bimodule. In particular, S is a right FSN R-R bimodule. Proof. Since S/X = *i !ai R + X/X", the result follows from Theorem 3.4 and Lemma 3.5 ! 4. THE ZERO BOUNDED SUB-BIMODULES OF S Throughout, S will be a finite normalizing extension of a Noetherian ring R with generators a1 % & & & % an i.e., S = a1 R + · · · + an R where ai R = Rai for all i. The purpose of this section is to examine the associated prime ideals of SR in case S SS is biuniform. It is shown that when R is right FSN, there is a ring isomorphism R/P - R/Q for all P,Q ∈ Ass!SR ". This forms a crucial step in the proof of the main result of this paper. We will make use of the notion of the bound of a sub-bimodule as given in McConnell and Robson (1987): For every Noetherian bimodule S MS and N ≤ R MR , the largest S-S sub-bimodule of M contained in N is called the bound of N and is denoted by B!N". Since R SR is Noetherian, any Noetherian S-S bimodule is also Noetherian as an R-R bimodule. It follows that a Noetherian bimodule S MS contains an R-R sub-bimodule N maximal with respect to B!N" = 0.

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Lemma 4.1. Let S MS be Noetherian and biuniform. If N ≤ R MR is maximal with respect to B!N" = 0, then R M/NR is biuniform. Proof. Suppose that 0 %= C/N , D/N ≤ R M/NR and C/N ∩ D/N = 0. By the maximality of N , B!C"% B!D" %= 0. It follows that 0 %= B!C" ∩ B!D" ⊆ N contradicting that B!N" = 0. ! For V ≤ R SR and a ∈ S that normalizes R, we define Va−1 = #x ∈ S ' xa ∈ V$

a−1 V = #x ∈ S ' ax ∈ V$& Note that both Va−1 and a−1 V are R-R sub-bimodules of S. Moreover, if b ∈ S normalizes R, then b−1 !Va−1 " = !b−1 V"a−1 and so there is no harm in using b−1 Va−1 to denote this sub-bimodule. −1 The sub-bimodules a−1 i Vaj provide a useful description of the bound of V . Lemma 4.2. If V ≤ R SR , then B!V" =

"

−1 a−1 i Vaj &

i%j

Proof. Let W denote the bimodule on the right-hand side of the equality. If −1 x ∈ B!V", then for all i and j, ai xaj ∈ B!V" ⊆ V and hence x ∈ a−1 i Vaj . Thus, B!V" ⊆ W . On the other hand, if x ∈ W , then for all i and j, ai xaj ∈ V . It easily follows that for all s, s$ ∈ S, sxs$ ∈ V whence SWS ⊆ V . From the definition of B!V", W ⊆ SWS ⊆ B!V". ! Corollary 4.3. If V ≤ R SR is maximal with respect to B!V" = 0, then there is an R-R bimodule monomorphism S→

#

−1 S/a−1 i Vaj &

i%j

It is clear from Corollary 4.3 that every associated prime ideal of SR is an −1 associated prime ideal of at least one of the quotients S/a−1 i Vaj . To take advantage of this relationship, we show that each quotient is biuniform and right FSN in case S SS is biuniform and R is right FSN. Proposition 4.4. Suppose that S SS is biuniform. Let V ≤ R SR be maximal with respect to B!V" = 0. Then: 1. 2. 3. 4.

S/V is biuniform as an R-R bimodule; For all i, S/a−1 i V is biuniform as an R-R bimodule; is biuniform as an R-R bimodule; For all i, S/Va−1 i −1 Va For all i,j, S/a−1 i j is biuniform as an R-R bimodule.

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Proof. The first statement follows directly from Lemma 4.1. For the second −1 statement, define a function + ' Bim!S/a−1 i V" → Bim!S/V" by +!X/ai V" = !ai X + $ $ V"/V . (Note that if r ∈ R, then for some r ∈ R, rai X = ai r X ⊆ X while ai Xr ⊆ X.) It is routine to verify that + is a 1-1 lattice map. It immediately follows that S/a−1 i V is biuniform. The third statement is the left-hand version of the second. The last statement is biuniform and then, as is obtained by using the third to conclude that S/Va−1 j −1 −1 above, applying the lattice map Bim!S/a−1 Va " → Bim!S/Va ! i j j ". Lemma 4.5 (McConnell and Robson, 1987, Lemma 10.2.2). 1. If a ∈ S normalizes R, then there is a ring isomorphism R/rR !a" → R/lR !a" given by r¯ → r¯$ where ar = r $ a. 2. If S is a finite normalizing extension of R and I ⊆ S is an ideal, then S/I is a finite normalizing extension of R/!I ∩ R". Proposition 4.6. Suppose that S SS is biuniform and R is a right FSN ring that satisfies incomparability. Let V ≤ R SR be maximal with respect to B!V" = 0. Then R S/VR is biuniform and right FSN with Ass!S/VR " = #P$ for some prime ideal P. Furthermore: −1 1. For each i with a−1 i V %= S, S/ai V is a biuniform right FSN R-R bimodule with −1 Ass!S/ai V" = #P$; −1 is a biuniform right FSN R-R bimodule with 2. For each i with Va−1 i %= S, S/Vai −1 Ass!S/Vai " = #Q$ where R/Q - R/P as rings; −1 −1 −1 is a biuniform right FSN R-R bimodule 3. For all i,j, with a−1 i Vaj %= S, S/ai Vaj −1 −1 with Ass!S/ai Vaj " = #Q$ where R/Q - R/P as rings.

Proof. In the body of the proposition and in each of the three statements, the corresponding bimodule is right FSN and biuniform by Corollary 3.6 and Proposition 4.4, respectively. By Proposition 3.1, each bimodule is right primary. −1 Let Ass!S/V" = #P$. For any i with a−1 i V %= S, the map S/ai V → S/V given by −1 x + ai V → ai x + V is a right R-monomorphism. The last part of (1) immediately follows. −1 For (2), let Ass!S/Va−1 i " = #Q$ where Vai %= S. We first show that ai P = −1 Qai . Since Vai %= S, Sai " V and, since ai is normal, Sai + V/V is a nonzero right R-submodule of S/V . Pick r ∈ P. There exists X ≤ SR such that Xai + V/V %= 0 with !Xai + V/V" r = 0 i.e., Xai r ⊆ V . Now, ai r = r $ ai for some r $ ∈ R and hence Xr $ ai ⊆ V . It follows −1 −1 −1 $ that Xr $ ⊆ Va−1 i . Also, Xai " V forces X " Vai . Since !X + Vai /Vai "r = 0, $ $ r ∈ Q. It follows that ai r = r ai ∈ Qai . −1 −1 Let r ∈ Q. For some right R-submodule 0 %= X/Va−1 i ≤ S/Vai , Xr ⊆ Vai and $ $ $ −1 so Xrai ⊆ V . Again, rai = ai r for some r ∈ R. Thus, Xai r ⊆ V . Also, Vai %= X forces Xai " V whence Xai + V/V is a nonzero right R-submodule of S/V with !Xai + V/V"r $ = 0. Therefore, r $ ∈ P and hence rai = ai r $ ∈ ai P. From above, 0 %= Sai + V/V ≤ S/VR . It follows that rR !ai " ⊆ P. Since −1 S/Va−1 i %= 0 and SlR !ai " ⊆ Vai , lR !ai " ⊆ Q. Let + ' R/rR !ai " → R/lR !ai " be the ring isomorphism supplied by Lemma 4.5(1). If r ∈ P, then ai r ∈ ai P = Qai and so ai r = r $ ai for some r $ ∈ Q. By definition, +!¯r " = r¯$ . Conversely, if r $ ∈ Q, then

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r $ ai = ai r for some r ∈ P and hence +!¯r " = r¯$ . Therefore, +!P/rR !ai " = Q/lR !ai ". Upon passing to the quotient, + induces an isomorphism R/P → R/Q. −1 −1 −1 −1 For (3), note that if Va−1 j = S, then ai Vaj = ai S = S. Thus, S/Vaj %= 0. −1 By (2), Ass!S/Vaj " = #Q$ where R/Q - R/P as rings. As in the proof of (1), −1 −1 −1 −1 the map S/a−1 given by x + a−1 is a right Ri Vaj → S/Vaj i Vaj → ai x + Vaj monomorphism. The result follows. ! Corollary 4.7. Suppose that S SS is biuniform and R is a right FSN ring that satisfies incomparability. Then for all P% Q ∈ Ass!SR ", R/P - R/Q as rings. Proof. Pick V ≤ R SR maximal with respect to B!V" = 0. By Corollary 4.3, there is an R-R bimodule monomorphism S→

#

−1 S/a−1 i Vaj &

i%j

−1 −1 −1 It follows that P ∈ Ass!S/a−1 i Vaj " and Q ∈ Ass!S/ah Vak " for some i, j, h and k. The conclusion now follows from Proposition 4.6. !

5. FINITE NORMALIZING EXTENSIONS ARE FSN As before, we will assume that S is a finite normalizing extension of a Noetherian ring R with generators a1 % & & & % an . A prime ideal Q of S is said to lie over a prime ideal P of R provided Q ∩ R ⊆ P and P/!Q ∩ R" is a minimal prime ideal of R/!Q ∩ R". Lemma 5.1. Let S XS %= 0 be Noetherian. If Q ∈ Ass!XS " and Q lies over P ∈ Spec!R", then P ∈ Ass!XR ". Proof. Since Q ∈ Ass!XS ", Q = rS !Y" for some 0 %= Y ≤ XS . By Proposition 3.1(1), Q = rS !Y" = rS !y1 % & & & % yn " for some y1 % & & & % yn ∈ Y . Thus, Q ∩ R = rR !Y" = rR !y1 % & & & % yn " and there is a monomorphism R/!Q ∩ R" → Y n . According to Goodearl and Warfield (1989, Theorem 7.27), Q ∩ R is a semiprime ideal. Since P is minimal over Q ∩ R, P ∈ Ass!R/!Q ∩ R"R " ⊆ Ass!YR " ⊆ Ass!XR " ! Theorem 5.2. If R is a right FSN ring that satisfies incomparability, then S is a right FSN ring. Proof. Proceed by Noetherian induction: Assume that for every ideal 0 %= I ⊆ R, any finite normalizing extension of R/I is a right FSN ring. To show that S is a right FSN ring, it suffices, by Proposition 3.1(3), to show that every biuniform S-S bimodule factor of S is right primary. Let H be an ideal of S for which S S/HS is biuniform. If H ∩ R %= 0, then by Lemma 4.5(2), S/H is a finite normalizing extension of R/!H ∩ R". By the induction hypothesis, S/H is a right FSN ring. In particular, S/H is right primary. If H ∩ R = 0, then S/H is a finite normalizing extension of R. Thus, without loss of generality, we may assume that S is a finite normalizing extension of R such that S SS is biuniform.

FINITE NORMALIZING EXTENSIONS OF FSN RINGS

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By Proposition 3.2, max Ass!SS " consists of a single prime ideal, say Q0 . Thus, if Q ∈ Ass!SS ", then Q ⊆ Q0 . Suppose that Q ⊂ Q0 . Pick P, P0 ∈ Spec!R" with Q lying over P and Q0 lying over P0 . By Lemma 5.1, P% P0 ∈ Ass!SR " and by Corollary 4.7, R/P - R/P0 as rings. Now, Q ∩ R ⊆ Q0 ∩ R ⊂ P0 and hence, there is a minimal prime ideal P $ /Q ∩ R ⊆ R/Q ∩ R with P $ /Q ∩ R ⊆ P0 /Q ∩ R i.e., Q lies over P $ . Thus, by Lemma 5.1, P $ ∈ Ass!SR ". Reasoning as above, R/P $ - R/P0 as rings. This excludes the possibility that P $ ⊂ P0 (for example, R/P $ and R/P0 have equal Krull dimension.) Thus, P $ = P0 and so Q lies over P0 . However, Q ⊂ Q0 and Q0 lies over P0 contradicting Letzter (1990, Corollary 2.4) or Goodearl and Warfield (1989, Theorem 10.6). It follows that Ass!SS " = #Q0 $. Therefore, S SS is right primary. ! Corollary 5.3. If R is a right FSN ring that satisfies incomparability and G is a finite group, then R ∗ G is a right FSN ring. Furthermore, R ∗ G satisfies the right second layer condtion iff R satisfies the right second layer condition. Proof. The first statement follows directly from Theorem 5.2. The last is a consequence of Bell (1987, Proposition 7.5). ! REFERENCES Bell, A. D. (1987). Localization and ideal theory in Noetherian strongly group-graded rings. J. Algebra 105:76–115. Goodearl, K. R., Warfield, R. B. (1989). An Introduction to Noncommutative Noetherian Rings. London Math. Soc. Student Texts. Vol. 16. Cambridge: Cambridge University Press. Jategaonkar, A. V. (1981). Noetherian bimodules, primary decomposition and Jacobson’s conjecture. J. Algebra 71:379–400. Jategaonkar, A. V. (1982). Solvable Lie algebras, polycyclic-by-finite groups and bimodule Krull dimension. Comm. in Algebra 10:19–69. Jategaonkar, A. V. (1986). Localization in Noetherian Rings. London Math. Soc. Lecture Note Series. Vol. 98. Cambridge: Cambridge University Press. Kosler, K. A. (2003). On symmetric radicals over fully semiprimary Noetherian rings. J. Algebra and Its Applications 2:351–364. Letzter, E. S. (1990). Prime ideals in finite extensions of Noetherian rings. J. Algebra 135:412–439. McConnell, J. C., Robson, J. C. (1987). Noncommutative Noetherian Rings. New York: WileyInterscience. Stafford, J. T. (1985). On the ideals of a Noetherian ring. Trans. AMS 289:381–392.