A RESULT ON THE INCOMPARABILITY OF LINKED PRIME IDEALS

It is shown that linked prime ideals in certain fully semiprimary Noetherian ring are incomparable. Mathematics Subject Classification (2010): 16D20, 16D25, 16D80, 16P40


Introduction
If P and Q are prime ideals of a two-sided Noetherian ring R, then there is a link from Q to P , denoted Q P , provided there exists an ideal A such that QP ⊆ A ⊂ Q ∩ P and (Q ∩ P )/A is torsion-free as a left R/Q-module and as a right R/P -module.A long-standing conjecture in noncommutative ring theory is that no link can exist when P and Q are comparable.From [3,Theorem 1.8] this is true if R satisfies the second layer condition.More recently, Vyas in [8] showed that this is true if R is a Noetherian ring with global dimension 1.In this note, we use a result of Vyas' from [8] to prove that this result holds if R is a fully semiprimary

Notation and definitions
Throughout, R will always be a Noetherian ring.We will write M R to indicate that M is a right R-module.Similarly, S M and S M R indicate that M is a left S-module and an S-R bimodule respectively.A Noetherian bimodule is a nonzero bimodule S M R where both S M and M R are Noetherian.We use For a nonempty subset See [1], [2] and [4] for the definition of the second layer condition and its role in the structure of Noetherian rings.

Preliminaries
In this section, we collect the results that we need to prove the main result of the paper.(

A Noetherian bimodule
In this situation, it is easy to see that C is the unique largest cell in B.
The (1) B is right uneven if and only if P = Q and C is not an essential submodule of B R .
(2) If C ≤ e B R , then C = ann B (P ).
(3) B is right FSN iff no subfactor bimodule of B is right uneven.
Corollary 3.5.Let R be a Noetherian ring and let Q be a prime ideal.Then no Proof.Note that Q P if and only if P Q in R op , the opposite ring of R.
It then follows from Proposition 3.4 that no P ⊃ Q has Q P if and only if

Incomparability and ( †)
Let R and S be Noetherian rings.A biuniform Noetherian bimodule S B R is said to satisfy ( ‡) r provided there exists a cell C ⊂ B such that B/C is a cell and

Proposition 3 . 1 . ( 1 )( 2 )
S C R is called a right cell provided r R (C) is a prime ideal, C is torsion-free as a right R/r R (C)-module and for all 0 = C ≤ S C R , C/C is torsion over R/r R (C).A left cell is defined likewise and, of course, cell means right and left cell.The next result follows from [2, Proposition 2.1] and its proof.Every Noetherian bimodule contains a (right) cell.If S B R is a biuniform Noetherian bimodule, then the sum of all the (right) cells in B is the unique largest (right) cell in B. Definition 3.2.[2, page 383].A biuniform bimodule S B R is called right uneven provided there exists a cell C ⊂ B such that B/C is a cell and the following statements are true: (1) r R (B/C) ⊂ r R (C).
first two statements in the next result are parts of [2, Proposition 2.2] and [2, Proposition 2.4] respectively.The third statement is a slight restatement of [2, Theorem 3.2].The last statement is contained in the proof of [5, Proposition 3.1].Proposition 3.3.Let S B R be a biuniform Noetherian bimodule with unique largest cell C such that B/C is a cell.Set P = r R (C) and Q = r R (B/C).

( 4 )
If P is a maximal associated prime ideal of B R and ann B (P ) ≤ e B R , then any associated prime ideal of B/ ann B (P ) is linked to P .If R is a Noetherian prime ring, then the torsion submodule of a right R-module M is denoted by T(M R ) and similarly for left modules.If M is an S-R bimodule, then T(M R ) is a subbimodule of M .When M is a Noetherian S-R bimodule, then T(M R ) is annihilated by a regular element of R cf. [2, Lemma 1.1].The next result is the special case of [8, Corollary 3.8] for a Noetherian ring R. Proposition 3.4.Let Q be a prime ideal of a Noetherian ring R. No prime ideal