ABSTRACT. The paper concerns some cases of ring extensions R C S, where S is finitely generated as a right i?-module and R is right Noetherian.In §1 it is shown that if R is a Jacobson ring, then so is S, with the converse true in the PI case. In §2 we show that if S is semiprime PI, R must also be left (as well as right) Noetherian and S is finitely generated as a left ñ-module. §3 contains a result on ¿J-rings.In this paper we collect some results concerning the relationship between rings R C S, where S is finitely generated as a right .R-module. For the special cases in which the finite extension 5 of R is normalizing or centralizing, many theorems have been proved. In this paper we obtain some results of a more general nature. Jacobsonproperty.We call a ring R a Jacobson ring if every prime ideal of R is an intersection of primitive ideals. Let J(R) denote the Jacobson radical of R and N(R) the lower nilradical, that is, the intersection of all the prime ideals of R. With this notation, the alternative formulations of the Jacobson property are:J(R/P) = 0 for all prime ideals P of R.J(R) = N(R) for every homomorphic image R of R. If R is PI or right Noetherian, then every nil ideal is contained in N(R), and the last condition is equivalent to: J(R) is nil for every homomorphic image R of R.THEOREM l. Let R be a right Noetherian subring of a ring S, such that S is finitely generated as a right R-module. Then, if R is Jacobson, so must S be Jacobson.PROOF. Given any prime ideal P of S, we want to show that J(S/P) = 0. Since the hypothesis of the theorem holds for the ring embedding R/P n R C S/P, the problem reduces to the case where 5 is a prime ring. We therefore want to show that, if S is prime, then J(S) -0.Suppose J(S) ^ 0. By a standard result on Goldie rings, J(S) must contain a regular element, say a. Since Sr is finitely generated and R is right Noetherian, there is a positive integer n, such that the elements l,a,a2,... , a™ are integrally dependent over R. That is, for some rn_i,...,ri, ro G R, we have a" +an~1rn-1 + ■ ■ ■ + ari + ro = 0. If n is minimal, then, since a is regular, ro ^ 0. But then r0 G J(S) n R. Hence J(S) flfi/0.We claim that J(S)nR C J(R). Let x E J(S)nR and suppose x & J(R). Then there is an element r G R for which 1 -rx is not invertible in R. But 1 -rx is
ABSTRACT. The paper concerns some cases of ring extensions R C S, where S is finitely generated as a right i?-module and R is right Noetherian.In §1 it is shown that if R is a Jacobson ring, then so is S, with the converse true in the PI case. In §2 we show that if S is semiprime PI, R must also be left (as well as right) Noetherian and S is finitely generated as a left ñ-module. §3 contains a result on ¿J-rings.In this paper we collect some results concerning the relationship between rings R C S, where S is finitely generated as a right .R-module. For the special cases in which the finite extension 5 of R is normalizing or centralizing, many theorems have been proved. In this paper we obtain some results of a more general nature. Jacobsonproperty.We call a ring R a Jacobson ring if every prime ideal of R is an intersection of primitive ideals. Let J(R) denote the Jacobson radical of R and N(R) the lower nilradical, that is, the intersection of all the prime ideals of R. With this notation, the alternative formulations of the Jacobson property are:J(R/P) = 0 for all prime ideals P of R.J(R) = N(R) for every homomorphic image R of R. If R is PI or right Noetherian, then every nil ideal is contained in N(R), and the last condition is equivalent to: J(R) is nil for every homomorphic image R of R.THEOREM l. Let R be a right Noetherian subring of a ring S, such that S is finitely generated as a right R-module. Then, if R is Jacobson, so must S be Jacobson.PROOF. Given any prime ideal P of S, we want to show that J(S/P) = 0. Since the hypothesis of the theorem holds for the ring embedding R/P n R C S/P, the problem reduces to the case where 5 is a prime ring. We therefore want to show that, if S is prime, then J(S) -0.Suppose J(S) ^ 0. By a standard result on Goldie rings, J(S) must contain a regular element, say a. Since Sr is finitely generated and R is right Noetherian, there is a positive integer n, such that the elements l,a,a2,... , a™ are integrally dependent over R. That is, for some rn_i,...,ri, ro G R, we have a" +an~1rn-1 + ■ ■ ■ + ari + ro = 0. If n is minimal, then, since a is regular, ro ^ 0. But then r0 G J(S) n R. Hence J(S) flfi/0.We claim that J(S)nR C J(R). Let x E J(S)nR and suppose x & J(R). Then there is an element r G R for which 1 -rx is not invertible in R. But 1 -rx is
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