Associated Primes under Noncommutative Localization (Preliminary Version)

Associated Primes under Noncommutative Localization (Preliminary Version) Scott Annin Nicholas J. Werner September 19, 2012 Abstract We investigate the effect of noncommutative localization on associated prime ideals of a ring. Given a (possibly noncommutative) ring and a right denominator set S one may form the right ring of fractions Q := S 1 of with respect to S. When is commutative and M Q is a Q-module, it is known that the associated primes of M Q are extended from the associated primes of M. We prove that an analogous result is true for noncommutative rings provided that either Q is (right) Noetherian or S consists of central elements. Keywords: Associated prime, prime module, right ring of fractions, right denominator set, extension and contraction of ideals MSC Primary 16D10, Secondary 16U20 1 Introduction ing theorists have studied broadly the ways in which ring-theoretic and moduletheoretic concepts behave under change of rings, such as subrings, polynomial rings, and matrix rings. An introduction to this line of inquiry can be found in [8, 6F], and a good example of it in the context of this paper s theme, associated primes, can be found in a paper of C. Faith s, [6]. Faith showed how the associated primes of a module M over a commutative ring behave upon extension to the polynomial module M[x] over the polynomial ring [x]. Later, S. Annin generalized these results to the noncommutative setting, including Ore extension rings, in [1] and [2]. In commutative ring theory, an associated prime of a module is defined to be a prime ideal that is the annihilator of a module element. Associated Department of Mathematics, California State University Fullerton, Fullerton, CA 92834 sannin@fullerton.edu Department of Mathematics, University of Evansville, Evansville, IN 47722 nw89@evansville.edu 1

primes and localization have already been explored together in some contexts of commutative algebra; see, for example, [5, Theorem 3.1(c)] or [10, pp. 38 39]. Both topics are important for a variety of reasons that are well-documented in the literature. Associated primes, for example, hold a prominent place in the theory of primary decomposition and in the study of indecomposable injective modules. Many sources are appropriate to consult for more details, such as [5], [7], [8], or [10]. Continuing the study of associated primes under change of rings, this paper studies how the associated prime ideals of a module M over a noncommutative ring behave under localization. The theory of associated primes for modules over (possibly) noncommutative rings can be found in a variety of sources, including [7] or [8]. For convenience, Definitions 1.1 and 1.2 below recall the basic terminology we need. Throughout this paper, will denote a (possibly noncommutative) ring, and M will stand for a right -module. Definition 1.1. We say that an -module N is prime if N 0 and ann(n ) = ann(n ) for every nonzero submodule N N. It is easy to see that if N is prime, then ann(n ) is a (two-sided) prime ideal 1 of. The converse of this is false, however, and it is easy to construct a counterexample (cf. [2]). Definition 1.2. Let M be an -module. A two-sided ideal p of is called an associated prime of M if there exists a prime submodule N M such that p = ann(n ). The set of associated primes of M is denoted by Ass(M ). An easy exercise shows that, if is a commutative ring, this definition agrees with the standard one given above, and various well-known properties of associated primes over a commutative ring continue to hold in this noncommutative setting as well. However, if the generalization of the associated primes of a module M to noncommutative ring theory is relatively straightforward and predictable, localization is decidedly not! Noncommutative localization is fraught with difficulties. To save space here, we shall not go into a full discussion of the issues, but instead refer the reader to [8] for a detailed treatment. We will highlight only the most crucial definitions; this will also enable us simultaneously to introduce the notation that will appear throughout the paper. We will always write S for a multiplicative set in a ring, which means that S S S, 1 S, and 0 / S. Definition 1.3. A (possibly noncommutative) ring is said to be a right ring of fractions 2 of (with respect to S ) if there is a given ring homomorphism φ : such that 1. φ(s) U( ) (the units of ) 1 ecall that a two-sided ideal p in a (possibly noncommutative) ring is said to be a prime ideal if p and for any a, b, ab p implies that either a p or b p. 2 Alternatively, right ring of quotients 2

2. every element of has the form φ(a)φ(s) 1 for some a and some s S 3. ker(φ) = {r rs = 0 for some s S} In order for such a right ring of fractions to exist, it is well known that the multiplicative set S must satisfy two fundamental properties: (I) for every a and s S, as s (S is right permutable) (II) for every a, if s a = 0 for some s S, then as = 0 for some s S (S is right reversible) Multiplicative sets S satisfying (I) and (II) are called right denominator sets. emarkably, these conditions also turn out to be sufficient for the existence of a right ring of fractions. We summarize this in the following nontrivial theorem. Theorem 1.4 ([8], Thm. 10.6). The ring has a right ring of fractions with respect to S if and only if S is a right denominator set. Given a right denominator set S, it is well known that the right ring of fractions for is uniquely determined up to isomorphism. Therefore, we speak of the right ring of fractions of with respect to S and denote it by Q := S 1. Elements of Q are often written as as 1 or a/s (where a and s S). Two fractions a/s and a /s are equal in Q if there exist b, b such that sb = s b S and ab = a b (see [8] for details). In particular, if we let b = 1 above, then a/s = (ab)/(sb) as long as sb S. We will use this latter property several times in what follows. The right permutability of S can be used to bring any two fractions to a common denominator in order to add them. To multiply two fractions a 1 /s 1 and a 2 /s 2, we use right permutability to find c and d S such that s 1 c = a 2 d. Then, we define (a 1 /s 1 ) (a 2 /s 2 ) := (a 1 c)/(s 2 d). This will be freely used many times in what follows. It is also possible to localize a module over the (possibly noncommutative) ring. Specifically, given a module M, we can form a new module MS 1 Q consisting of elements of the form ms 1 or m/s (where m M, s S). There is a natural Q-action on MS 1 that conforms with the rule for multiplying two fractions in Q, and we leave this generalization to the reader (cf. [8, Exer. 10.18]). Naturally, the relationship between the ideals of and those of S 1 will be key to our work on associated primes. In this preliminary discussion, we will work with right ideals, although later we shall only apply these basic facts to two-sided ideals. From now on, we shall denote by φ : S 1 the natural map a a/1 (a ). Given a right ideal A of, we define A e (the extension of A) to be φ(a) Q, the right ideal generated by φ(a) in Q. Similarly, given a right ideal B of Q, we define B c (the contraction of B) to be φ 1 (B). The following proposition summarizes the relationships we need. 3

Proposition 1.5 ([8], Props. 10.32, 10.33). Let A and B be as above. Then, 1. B ce = B 2. A e = {a/s a A, s S} 3. A ec = {r rs A for some s S} 4. If is right Noetherian, then so is Q 5. Assume that Q is right Noetherian. If A is a two-sided ideal of, then A e is a two-sided ideal of Q. We are now ready to present our results concerning the behavior of associated primes under passage to right rings of fractions. 2 Main esults Throughout this section, we assume that is a ring, S is a right denominator set, and Q := S 1 is the right ring of fractions of with respect to S. Our two main results (Theorem 2.1 and Corollary 2.8) essentially provide a noncommutative analogue for Theorem 6.2 from [10] on the behavior of associated primes of modules upon passage to and from a right ring of fractions. For both results, we will assume that and Q satisfy the following condition: whenever N is an -submodule of the Q-module M Q, the extended right ideal ann(n ) e of Q is actually a two-sided ideal of Q (1) By Proposition 1.5, (1) holds under the fairly mild assumption that either or Q is right Noetherian. Hence, our main results will apply in either case. At the end of the paper, we will give a second condition (2) that implies (1) and that can hold independent of or Q being right Noetherian. Our first main theorem is the following: Theorem 2.1. Assume that (1) holds, and let M be a right Q-module. Then, Ass(M Q ) = {p e p Ass(M ) and p S = } = {I I c Ass(M ) and I c S = } Before proving Theorem 2.1, we will prove several lemmas on the relationships between annihilators and prime modules over and Q. Condition (1) is needed for some, but not all, of these results; we will point out where it comes into play. Given any -submodule N of any Q-module M Q, let N e := N Q Q denote the Q-submodule of M Q generated by N. Lemma 2.2. Let M Q be a right Q-module. Then, 4

1. ann(m ) = ann(m Q ) c 2. Let N be an -submodule of the Q-module M Q. Then, ann(n e Q ) ann(n ) e, and if (1) holds, we have equality. Proof. The first part is trivial. For the second part, let x/y ann(nq e ), where x and y S. Then, for each n N, we have n(x/1) = n(x/y)(y/1) = 0, since x/y ann(nq e ). Thus, nx = 0 and x ann(n ). By Proposition 1.5 x/y ann(n ) e. Assume now that (1) holds. Then, ann(n ) e is a two-sided ideal of Q. Let a/s ann(n ) e and r/t Q. By the right permutability of S, we can find c and u S be such that tc = au. Then, au ann(n ), and since ann(n ) e is two-sided, we have c = (1/t)au ann(n ) e. Since c is also in, we must have c ann(n ). Then, when n(r/t) N e, we have n(r/t)(a/s) = n(rc/su) = 0, so a/s ann(nq e ). Lemma 2.3. Let M Q be a right Q-module, and assume that (1) holds. If M Q is prime, them M is prime. Proof. Let M be a nonzero -submodule of M. We prove that ann(m ) = ann(m ). By Lemma 2.2 part (2), we have ann(m )e = ann((m Q )e ), and ann((m Q )e ) = ann(m Q ) because M Q is prime. So, ann(m ) = ann(m ) e = ann(m Q ) = ann(m ). Hence, ann(m ) = ann(m ) and M is prime. We will also prove a corresponding lemma about the primeness of extensions of -submodules of M Q, but we first need two results about right denominator sets. Essentially, the following proposition says that there is no harm in taking S to be as large as possible, and subsequently, Lemma 2.5 says that we can always clear a finite number of denominators in Q with a single element of S. Proposition 2.4. Given a right denominator set S and Q = S 1, let S = {u u U(Q)}. Then, 1. S is a right denominator set 2. S 1 = S 1 Thus, we may assume without loss of generality that S = S. Proof. For (1), we prove that S is right permutable and right reversible. For right permutability, let a and s S. Then, 1/s Q. Write 1/s = b/t, where b and t S. Let c and u S be such that tc = au. Then, (b/t)au = (b/t)(a/1)u = (bc/u)u = bc. Moreover, we have s(bc) = s(b/t)au = au. Thus, as s. For right reversibility, assume that a and sa = 0 for some s S. Then, sa = 0 in Q. Since 1/s Q, we get a = 0 in Q. Thus, a is in the kernel 5

of the localization homomorphism φ. By definition, ker(φ) = {r rt = 0 for some t S}. Hence, there exists t S such that at = 0. For (2), since S S, we have S 1 S 1. However, since S 1 Q, S 1 Q. Thus, equality holds. In light of the above proposition, we will assume for the remainder of this paper that S = S. Lemma 2.5. Let a 1 /s 1, a 2 /s 2,..., a k /s k Q. Then, there exists t S such that (a i /s i )t for each 1 i k. Proof. We use induction on k. If k = 1, then we can take t = s 1. Assume that k > 1 and that the lemma holds for k 1, and let t k 1 S be such that (a i /s i )t k 1 for all 1 i k 1. By right permutability of S, let c and u S be such that t k 1 c = s k u. Then, in Q, c = (1/t k 1 )s k u U(Q), so c S. Taking t = t k 1 c S, we have (a i /s i )t for all 1 i k, as required. We can now prove another result about prime modules. Lemma 2.6. Let N be an -submodule of the Q-module M Q. Assume that (1) holds. If N is prime, then NQ e is prime. Proof. Assume that N is prime; then, NQ e {0}. Let N Q be a nonzero Q- submodule of NQ e. We claim that N N {0}. Indeed, let n N Q \ {0} and write n = k i=1 n i(a i /s i ), where n i N, a i, and s i S. By Lemma 2.5, there exists t S such that (a i /s i )t for each i. Hence, n t N N, and n t 0 because n 0 and t S. Now, since N is prime, we have ann(n ) = ann(n N ) ann(n ). Let a/s ann(n Q ); then, a ann(n ), so a ann(n ). Since (1) holds, by Lemma 2.2 we have a/s ann(nq e ). Therefore ann(n Q ) = ann(n Q e ) and N Q e is prime. We are now prepared to prove Theorem 2.1. Proof of Theorem 2.1. Consider the first equality. For, let I Ass(M Q ). By definition, there exists {0} N Q M Q with N Q prime and I = ann(n Q ). Set p := I c. Since Q I = I ce = p e by Proposition 1.5, we must have p S = ; otherwise, 1 = a/a p e whenever a p S. It remains to prove that p Ass(M ). To do this, in turn, it suffices to prove that p = ann(n ) and N is prime. The first property follows from Lemma 2.2 part (1) and the second property from Lemma 2.3. To prove, let p Ass(M ) with p S =. Find a prime (nonzero) submodule N M with p = ann(n ). By Lemma 2.2 part (2), we have ann(nq e ) = pe, and furthermore NQ e is prime by Lemma 2.6. Hence, pe Ass(M Q ), which completes the proof of the first equality. For the second equality, is true by Proposition 1.5 and follows from the proof of for the first equality. 6

Our second main result extends Theorem 2.1 to localization of modules (as mentioned prior to Proposition 1.5; see [8, Exer. 10.18]). We will use the following lemma. Lemma 2.7. Let M be a right -module, and assume that M is S-torsionfree 3. Then, Ass(M ) = Ass(MS 1 ). Proof. Since M is S-torsion-free, there is an embedding M MS 1 ; it follows that Ass(M ) Ass(MS 1 ). For the other inclusion, let p Ass(MS 1 ) and let N be a prime submodule of MS 1 such that p = ann(n ). We will exhibit a prime submodule M of M with p = ann(m ). Let m/s N \ {0}, where m M and s S; then, m 0. Consider the nonzero cyclic submodule (m/s) of N. Since s, we have m = (m/s) s (m/s). Let M = m. Then, M is nonzero and M (m/s) N. Since N is prime, so is M. Moreover, ann(m ) = ann(n ) = p. Finally, since m M we get M M. Hence, p Ass(M ), as required. Corollary 2.8. Let M be a right -module, and assume that M is S-torsionfree. Assume that (1) holds. Then, Ass(MS 1 Q ) = {pe p Ass(M ) and p S = } Proof. By Theorem 2.1, we know that = {I I c Ass(M ) and I c S = } Ass(MS 1 Q ) = {pe p Ass(MS 1 ) and p S = } = {I I c Ass(MS 1 ) and Ic S = }. The result now follows from Lemma 2.7, since Ass(MS 1 ) = Ass(M ). As mentioned at the beginning of Section 2, there is an alternate condition that implies (1) and that can be true when neither nor Q is right Noetherian. This condition is: for each t S, there exists t and q in the center of Q such that 1/t = t q (2) Proposition 2.9. Assume (2) holds for S. If A is a two-sided ideal of, then A e is a two-sided ideal of Q. In particular, (1) holds for and Q. Proof. Let A be a two-sided ideal of, let a/s A e and r/t Q. It suffices to show that A e is closed under multiplication on the left. Let c and u S be such that tc = au. Using (2), in Q we have c = (1/t)au = t qau = t auq, for some t and q in the center of Q. So, (r/t)(a/s) = rc/su = rt auq/su = (rt a)(uq/su) = (rt aq)/s A e, since q is in the center of Q. 3 ecall that M is S-torsion-free if, whenever m M and s S with ms = 0, then m = 0. 7

Note that (2) is satisfied if S is a central multiplicative subset of. Thus, Theorem 2.1 and Corollary 2.8 are true whenever S is central. As a topic for further investigation, we note that a theory dual to that of a module s associated primes has also been developed. In 1973, I.G. Macdonald introduced the concept of the attached prime ideals of a module M over a commutative ring, written Att(M ), and a theory of secondary representation that is dual to that of primary decomposition; see [9] for details. Later, Annin generalized this theory to the noncommutative setting in [3] and investigated how the attached prime ideals behave under passage to skew polynomial extensions in [4]. It is natural to ask how the attached prime ideals of a module behave under localization. This question remains open even in the commutative setting. eferences [1] S. Annin, Associated primes over skew polynomial rings, Comm. Algebra 30(5) (2002) 2511 2528. [2] S. Annin, Associated primes over Ore extension rings, J. Algebra Appl. 3 (2004) no. 2 193-205. [3] S. Annin, Attached primes over noncommutative rings, J. Pure Appl. Algebra 212 (2008) 510 521. [4] S. Annin, Attached primes under skew polynomial extensions, J. Algebra Appl. 10 (2011) no. 3 537 547. [5] D. Eisenbud, Commutative Algebra with a View Towards Algebraic Geometry, Graduate Texts in Mathematics 150, Springer-Verlag, New York, 1995. [6] C. Faith, Associated primes in commutative polynomial rings, Comm. Algebra 28(8) (2000) 3983 3986. [7] K. Goodearl and. Warfield Jr., An Introduction to Noncommutative Noetherian ings, London Mathematical Soceity, Student Texts 16, Cambridge University Press, Cambridge, 1989. [8] T. Y. Lam, Lectures on Modules and ings, Graduate Texts in Mathematics 189, Springer-Verlag, New York, 1999. [9] I. G. Macdonald, Secondary representation of modules over a commutative ring, Sumpos. Math. 11 (1973) 23 43. [10] H. Matusumura, Commutative ing Theory, Cambridge Studies in Advanced Mathematics 8, Cambridge University Press, Cambridge, 1986. 8