By Eric Jespers

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**Example text**

AnnR (a) is maximal. Let r ∈ R. Suppose ra = 0. Then there exists an n ∈ N so that (ra)n = 0 and (ra)n+1 = 0. annR (ara). annR (ara), a proper inclusion. However, since ara ∈ L the maximality condition yields that ara ∈ B(R). 54 CHAPTER 4. PRIME AND PRIMITIVE RINGS Hence we have shown that aRa ⊆ B(R). Since B(R) is a semiprime ideal we get that a ∈ B(R), a contradiction. Let N be a maximal nilpotent ideal in R. Then N ⊆ B(R) and R/N does not have nonzero nilpotent ideals. So R/N is semiprime and N = B(R) is nilpotent.

5 (Jacobson) Let N be a nilpotent ideal in a ring R and assume R/N is semisimple. If S is a nil weakly closed subset of R, then there exists n so that S n = {0}. Proof. Clearly N = J(R) and k R/N ∼ = Mni (Di ) i=1 for some skew fields Di . Let Si denote the natural image of S in Mni (Di ) and n = max{n1 , · · · , nk }. Then Sin = {0} for all 1 ≤ i ≤ k. It follows that S n ⊆ N , as desired. ✷ The result shows the potential interest in finding a criterion for a ring to be contained in an Artinian ring.

We now show that I always is an ideal √ and we will write I as an intersection of prime ideals. 1. 4 Let I be an ideal in a ring R. Then √ I is the intersection of all prime ideals that contain I. In particular, I is an ideal of R. √ Proof. Let s ∈ I and let P be a prime ideal containing I. Then R get that s ∈ P . Hence √ \ P is an m-system. As P ∩ (S ∩ P ) = ∅ we √ I ⊆ ∩I⊆P, P prime P . Conversely, suppose s ∈ I. Hence there exists an m-system S with s ∈ S and I ∩ S = ∅. Because of Zorn’s Lemma there exists an ideal P of R, with I ⊆ P , and maximal with respect to the condition P ∩ S = ∅.