# Exercises in Classical Ring Theory, Second Edition (Problem by T.Y. Lam

By T.Y. Lam

Best mathematics books

Mathematik für Physiker 2: Basiswissen für das Grundstudium der Experimentalphysik

Die für Studienanfanger geschriebene „Mathematik für Physiker'' wird in Zukunftvom Springer-Verlag betreut. Erhalten bleibt dabei die Verbindung einesakademischen Lehrbuches mit einer detaillierten Studienunterstützung. DieseKombination hat bereits vielen Studienanfangern geholfen, sich die Inhalte desLehrbuches selbständig zu erarbeiten.

Additional resources for Exercises in Classical Ring Theory, Second Edition (Problem Books in Mathematics)

Sample text

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.