Correction : the Embedding of Radical Rings in Simple Radical Rings

As G. M. Bergman has pointed out, in the proof of the lemma on p. 187, we cannot conclude that S ¯ is universal in the sense stated. However, the proof can be completed as follows: Any element of S ¯ can be obtained as the first component of the solution u of a system (A−I)u+a = 0, (1) where A ɛ S n , a ɛ n S and A−I has an inverse over L . Since S is generated by R and k{s} , A can (by the last part of Lemma 3.2 of [1]) be taken to be linear in these arguments, say A = A 0 + sA1 , where A 0 ɛ R n , A 0 ɛ R n , A 1 ɛ K n . Multiplying by ( I−sA 1 ) −1 , we reduce this equation to the form (Σ S v B v −I)u +a=0, (2) with the same solution u as before, where B v \sɛ R n , s v ɛ k{s} 1 and a ɛ n S . Now consider the retraction S → k{s} (3) obtained by mapping R → 0. If we denote its effect by x ↦ x * , then (2) goes over into an equation − I.v + a * 0, (4) which clearly has a unique solution v in k{s} ; therefore the retraction (3) can be extended to a homomorphism S ¯ ↦ k{s} , again denoted by x ↦ x * , provided we can show that u 1 * does not depend on the equation (1) used to define it. This amounts to showing that if an equation (1), or equivalently (2), has the solution u 1 = 0, then after retraction we get v 1 = 0 in (4), i.e. a 1 * = 0. We shall use induction on n ; if u 1 = 0 in (2), then by leaving out the first row and column of the matrix on the left of (2), we have an equation for u 2 ,…, u n and by the induction hypothesis, their values after retraction are uniquely determined. Now from (2) we have∑ j = 2∑ v S v b 1vju j + a 1 = 0 ,where B = (b ij v ). Applying * and observing that b ij v ɛ R , we see that a 1 * = 0, as we wished to show. The proof still applies for n = 1, so we have a well‐defined mapping S ¯ → k{s} , which is a homomorphism. Now the proof of the lemma can be completed as before.