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17 Semiprime Rings. Our next goal is to characterize those rings whose classical ring of left quotients is semisimple. This requires that we degress briey to discuss the notion of a prime ideal in a non-commutative setting. This, in turn, leads to a generalization of the prime radical from commutative algebra. Let R be an arbitrary ring. An ideal P of R is prime in case for every pair a, b ∈ R,ifaRb P , then a ∈ P or b ∈ P . Of course, if R is commutative, this reduces to the usual denition of primeness. 17.1. Proposition. For an ideal in the ring R the following are equivalent: (a) P is prime; (b) For every pair I,J of left ideals of R, IJ P =⇒ I P or J P ; (c) For every pair I,J of right ideals of R, IJ P =⇒ I P or J P ; (d) For every pair I,J of ideals of R, IJ P =⇒ I P or J P . Proof. (a) =⇒ (d) Let IJ P and suppose that J 6 P . Then there is some b ∈ J \ P . But then aRb P for all a ∈ I, so by (a), a ∈ P for all a ∈ I. (d) =⇒ (c) If I,J are right ideals with IJ P , then RIRJ P ,soby(d)I RI P or J RJ P . (c) =⇒ (b) is similar. (b) =⇒ (a) If aRb P , then RaRb P ,soby(b)a ∈ R or b ∈ R. A ring R is prime if 0 is a prime ideal. Thus, an ideal P is prime i the ring R/P is a prime ring. More generally, an ideal I of R is semiprime if it is an intersection of prime ideals. A ring R is semiprime in case 0 is a semiprime ideal of R. So an ideal I is semiprime i R/I is a semiprime ring. The smallest of all the semiprime ideals, N(R)=∩{P : P is a prime ideal of R} is the prime radical of R.SoR is semiprime i N(R) = 0. Our immediate goal is to nd an internal characterization of N(R). An element a of a ring R is strongly nilpotent in case for every sequence a0,a1,... in R if a0 = a and an+1 ∈ anRan for all n 0, there must exist some n ∈ N with an = 0. That this really is a statement about the nilpotence of a may be made a bit more clear with the following characterization whose easy proof we leave to the reader. 106 Section 17 17.2. Lemma. An element a ∈ R is strongly nilpotent i for every sequence x1,x2,... in R, there exists some n ∈ N with a(x1a(x2a(a(xn1a(xn)axn1)a a)x2)ax1)a =0. 17.3. Theorem. For a ring R, its prime radical is N(R)={a ∈ R : a is strongly nilpotent }. Proof. Suppose that a 6∈ N(R), so there is some prime ideal P with a 6∈ P . Since P is prime, there is a function : R \ P → R \ P such that (x) ∈ xRx \ P for each x ∈ R \ P . Now dene a sequence a0,a1,... in R \ P recursively by a0 = a and an+1 = (an), for all n 0. Since an =6 0 for all n ∈ N, a is not strongly nilpotent. On the other hand suppose that a ∈ R is not strongly nilpotent. Then there is a sequence a = a0,a1,a2,... in R with 0 =6 an+1 ∈ anRan for all n 0. Then there is an ideal P maximal w.r.t. an 6∈ P for all n ∈ N. We claim that P is prime. Indeed, suppose that I,J are two ideals neither contained in P . Then by maximality, there must be some n ∈ N with an ∈ I + P and an ∈ J + P . But then an+1 ∈ anRan IJ + P .SoIJ 6 P , and P is prime. Of course, this means that a 6∈ N(R). 17.4. Corollary. For every ring R, its prime radical N(R) is a nil ideal. We can now use Theorem 17.3 to characterize semiprime rings. 17.5. Corollary. For a ring R the following are equivalent: (a) R is semiprime; (b) N(R)=0; (c) I2 =0implies I =0for every left (right/two-sided) ideal I of R; (d) aRa =0implies a =0; (e) R has no non-zero nilpotent left (right/two-sided) ideals; (f) IJ =0implies I ∩ J =0for every pair I,J of left (right/two-sided) ideals. Non-Commutative Rings Section 17 107 Proof. (a) ⇐⇒ (b) This is simply by denition. (f) =⇒ (c) =⇒ (d) and (c) ⇐⇒ (e) are all trivial. (b) =⇒ (f) If IJ = 0, then IJ P for every prime ideal P ,soI ∩ J P for every prime ideal P . Thus, by (b), I ∩ J =0. (d) =⇒ (b) By (d) there is a map : R \{0}→R \{0} such that (a) ∈ aRa for each a =0in6 R. Then for each a =0in6 R the sequence a, (a),2(a),...is never 0, so a is not strongly nilpotent. Thus, by Theorem 17.3, N(R)=0. In turn this gives us various characterizations of semiprime ideals. 17.6. Corollary. For an ideal H of R the following are equivalent: (a) H is semiprime; (b) I2 H implies I H for all left (right/two-sided) ideals I; (c) aRa H implies that a ∈ H. If R is left noetherian, then the prime radical N(R) has a particularly nice form. Indeed, since R is left noetherian, it has a maximal nilpotent left ideal, say I. But then clearly the ideal IR is also nilpotent, so by maximality, I is a maximal two-sided nilpotent ideal. We now see that this maximal nilpotent ideal is actually N(R). 17.7. Theorem. If R is a left noetherian ring, then N(R) is the unique largest nilpotent ideal of R. Proof. As we saw above, R has a maximal nilpotent ideal I.IfJ is another nilpotent ideal, then there is some n with In = J n =0.So(I + J)2n = 0 and then by maximality I + J = I,soJ I. That is, I is the unique largest nilpotent ideal. Of course, then, I N(R). But if J is an ideal nilpotent modulo I, then it must be nilpotent and hence in I. So by Corollary 17.6, the ideal I is semiprime, so N(R) I. Exercises 17. 17.1. For a ring R and an ideal P , (a) Prove that every maximal ideal P is prime. (b) Prove that every primitive ideal P is prime. (c) Deduce that N(R) is contained in the Jacobson radical J(R). 108 Section 17 (d) Show that N(R) need not equal J(R). (e) Prove that if R is artinian, then every prime ideal is maximal, so that N(R)=J(R). 17.2. For a ring R prove that it contains a unique largest nil ideal. This ideal U(R)istheupper nil radical of R. The prime radical N(R) is sometimes called the lower nil radical of R.So N(R) U(R). (a) Show that N(R) need not be U(R). (b) Prove that if R is left noetherian, then N(R)=U(R). 17.3. Let R be a ring. (a) Let (P)∈ be a chain of prime ideals in R. Prove that both ∪∈P and ∩∈P are prime ideals of R. (b) Prove that any non-empty set S of prime ideals of R contains a maximal element and a minimal element. (c) Prove that if P is a prime ideal in R, then P is contained in a maximal prime ideal and contains a minimal prime ideal. 17.4. (a) Prove that if R is left noetherian, then there exists a nite sequence P1,P2,...,Pn of prime ideals with P1P2 Pn = 0. [Hint: Suppose false. So R contains an ideal maximal w.r.t. I contains no nite product of primes. Then w.m.a. I = 0. So every non-zero ideal of R contains a nite product of primes. But R is prime.] (b) Find such a prime factorization of 0 in Z72. 17.5. Let R = C([0, 1], R) be the ring of all continuous real-valued functions on [0, 1]. For each f ∈ R, let Z(f)={x ∈ [0, 1] : f(x)=0}. So each Z(f) is a closed set. (a) Prove that for each x ∈ [0, 1] the set Mx = {f ∈ R : x ∈ Z(f)} is a maximal ideal of R. (b) For each x>0 and each y<1in[0, 1], set Px = {f ∈ R :(a, x) Z(f) for some 0 <a<x} and Qy = {f ∈ R :(y, b) Z(f) for some y<b<1}. Prove that each Px and Qy is a prime ideal. (c) Prove that for each x>0 and y<1, R/Px and R/Qy is a local ring. (d) Prove that for each open set U [0, 1], {f ∈ R : U Z(f)} is a semiprime ideal..
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