# Get Exercises in Classical Ring Theory, Second Edition (Problem PDF By T.Y. Lam

ISBN-10: 0387005005

ISBN-13: 9780387005003

ISBN-10: 0387217711

ISBN-13: 9780387217710

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Xn ], where R is any ring. (1) If I · g = 0 for some nonzero g ∈ A, show that I · r = 0 for some nonzero r ∈ R. 22 Chapter 1. Wedderburn-Artin Theory (2) Assume R is a commutative ring or a reduced ring, and let f ∈ A. If f · g = 0 for some nonzero g ∈ A, show that f · r = 0 for some nonzero r ∈ R. (3) Show that the conclusion of (2) need not hold if R is an arbitrary ring. Solution. (1) We induct on n. First assume n = 1 and write A = R[x]. Let g ∈ A\{0} be of minimal degree such that I · g = 0, say g = bxd + · · · , b = 0.

Quotienting out z w 0 0 the ideal , we get the eigenring ER (A) ∼ = Mr (k). z w Ex. 28. Do the same for the right ideal A = xR in the free k-ring R = k x, y , and for the right ideal A = xR where x = i + j + k in the ring R = Z ⊕ Zi ⊕ Zj ⊕ Zk of quaternions with integer coeﬃcients. Solution. (1) Let R = k x, y and A = xR. For r ∈ R, we have r ∈ IR (A) iﬀ r · x ∈ xR. Writing r = r0 + r1 where r0 is the constant term of r, we see that r · x = r0 x + r1 x ∈ xR iﬀ r1 ∈ xR. This shows that IR (A) = k + xR, from which we get ER (A) = (k + xR)/xR ∼ = k.

Therefore, n · ε = 0 ∈ E, as claimed. Ex. 11. If R V is a simple module over a commutative ring R, show that any direct product of copies of V is semisimple. Does this hold over an arbitrary ring R? Solution. Representing V in the form R/m where m is a maximal ideal of R, we can solve the ﬁrst part as in the solution to Ex. 9. However, the same conclusion does not hold (in general) without the commutativity assumption. To produce a counterexample, let V be a right vector space 32 Chapter 1. Wedderburn-Artin Theory over a division ring D with a basis {e1 , e2 , .