By T.Y. Lam
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Additional resources for Exercises in Classical Ring Theory, Second Edition (Problem Books in Mathematics)
2) Here, we need only prove the inclusion “⊇”. For any element x ∈ (eR + A) ∩ (eR + B), we can write x = er + a = es + b, where r, s ∈ R, a ∈ A, and b ∈ B. Then ex = er + ea = x − a + ea, so x − ex ∈ A. Similarly, x − ex ∈ B. Thus, x = ex + (x − ex) ∈ eR + (A ∩ B), as desired. (3) To ﬁnd the desired counterexamples, take e = ring R = M2 (Q), and A=R· 1 0 1 0 = a b a b , B =R· 0 0 1 0 0 0 0 1 = in the matrix 0 0 c d . a 0 , we have Re ∩ A = Re ∩ B = (0). On the other b 0 hand, A + B = R so Re ∩ (A + B) = Re.
Comment. A similar construction shows that A is also isomorphic to the ring S of 2 × 2 upper triangular matrices over k. ) Ex. 19. Let R be a domain. If R has a minimal left ideal, show that R is a division ring. ) Solution. Let I ⊆ R be a minimal left ideal, and ﬁx an element a = 0 in I. Then I = Ra = Ra2 . In particular, a = ra2 for some r ∈ R. Cancelling a, we have 1 = ra ∈ I, so I = R. The minimality of I shows that R has no left ideals other than (0) and R, so R is a division ring (cf. 2).
Let F be the prime ﬁeld of D. We may view M as a D-vector space, so M is also an F -vector space. As such, M is isomorphic to a direct sum of copies of F . This gives the desired conclusion since we have either F ∼ = Zp for some prime p. = Q, or F ∼ Comment. The solution above oﬀers perhaps the best “conceptual view” of this exercise. However, a direct proof not using Schur’s Lemma is possible too. One observes that, if M has p-torsion for any prime p, then pM = 0; otherwise, argue that pM = M for any prime p so M is a torsionfree divisible group.