1. ## Rings with zero Jacobson radical.

Hi:
In McCoy, The Theory of Rings, 1964 I read:
7.29 EXERCISES
2. Show that if R is a ring such that J(R) = (0) and the d.c.c. for right ideals holds in R, then R is a regular ring. [J = Jacobson radical, d.c.c. = descending chain condition, R is an arbitrary ring i.e., not necessarily commutative or with unity except it is subject to the two specified conditions.]

The author's definition of a regular ring is
7.2 Definition. Let c be an element of an arbitrary ring R. If there exists an element c' of R such that c = cc'c, c is siad to be a regular element of R. The ring R is said to be a regular ring if each of its elements is regular.

Going to the proof, I find in the book
7.28 Theorem. If the d.c.c. for right ideals holds in the ring R, then N(R) = J(R) = B(R).

Nevermind what the radicals N or B are. But I also have this:
7.27 Theorem. If R is a ring with more than one element, then N(R) = (0) iff R is isomorphic to a subdirect sum of simple rings with unity.

So R, in the exercise, can be considered to be a subdirect sum of simple rings with unity and, to deal with the simplest case, that in which the sum consists of only one ring, let us say R is a simple ring with unity. It would be nice if simple rings with unity were always regular. Or simple ring + unity + dcc = regular. This is as far as I could go. Not a long way indeed. Thanks.

EDIT: searching the book, I found the Wedderburn-Artin therem (5.59), by which R is isomorphic to a direct sum of a finite number of rings, each of which is a complete matrix ring over some division ring. Also this proposition: For each positive integer n and each division ring D the ring D_n of all matrices of order n over D is a regular ring. This immediately gives R is regular. E.g. in R1 x R2, given (c,d) there exist c', d' such that c = cc'c, d = dd'd. So (c,d) = (cc'c, dd'd) = (c,d) (c',d') (c,d).

2. Originally Posted by ENRIQUESTEFANINI
Hi:
In McCoy, The Theory of Rings, 1964 I read:
7.29 EXERCISES
2. Show that if R is a ring such that J(R) = (0) and the d.c.c. for right ideals holds in R, then R is a regular ring. [J = Jacobson radical, d.c.c. = descending chain condition, R is an arbitrary ring i.e., not necessarily commutative or with unity except it is subject to the two specified conditions.]

The author's definition of a regular ring is
7.2 Definition. Let c be an element of an arbitrary ring R. If there exists an element c' of R such that c = cc'c, c is siad to be a regular element of R. The ring R is said to be a regular ring if each of its elements is regular.

Going to the proof, I find in the book
7.28 Theorem. If the d.c.c. for right ideals holds in the ring R, then N(R) = J(R) = B(R).

Nevermind what the radicals N or B are. But I also have this:
7.27 Theorem. If R is a ring with more than one element, then N(R) = (0) iff R is isomorphic to a subdirect sum of simple rings with unity.

So R, in the exercise, can be considered to be a subdirect sum of simple rings with unity and, to deal with the simplest case, that in which the sum consists of only one ring, let us say R is a simple ring with unity. It would be nice if simple rings with unity were always regular. Or simple ring + unity + dcc = regular. This is as far as I could go. Not a long way indeed. Thanks.
that's a trivial result of Wedderburn-Artin (theorem 5.59) and theorem 7.3.

3. Thank you. Please forgive my having edited the original post. I had not refreshed the page or otherwise seen your replay when I did. Farewell.