Exercises in Modules and Rings by LAM T. Y.

By LAM T. Y.

This quantity bargains a compendium of routines of various measure of trouble within the conception of modules and earrings. it's the significant other quantity to GTM 189. All workouts are solved in complete aspect. every one part starts off with an creation giving the final historical past and the theoretical foundation for the issues that stick with.

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34. ("Unimodular Column Lemma") Let F = EB~=l eiR and a= Ei ei ai E F be as in Exercise 33. (1) Show that E~=l Rai = R iff a· R is a direct summand ofF free on {a}. (2) In case E~=l Rai = R, show that a direct complement of a · R in F is free of rank n- 1 iff there exists a matrix (aij) E GLn (R) with ail = ai for all i. Solution. (1) follows by applying Exercise 33 to A = Ei Rai with the idempotent f = 1 E R. For (2), assume that Ei Rai = R. Note that the isomorphism type of a direct complement of a · R is uniquely determined (being isomorphic to F/a·R).

This example is drawn from the paper of M. Ojanguren and R. Sridharan, "Cancellation of Azumaya algebras," J. Algebra 18(1971), 501-505. In this paper, it is also shown that the stably free R-module P constructed above is not free, but 3 · P( = P EB P EB P) is free. g. projective right modules over Rn = D[x 1 , ... , Xn] was started by A. Suslin (see Trudy Mat. Inst. Steklov. g. projective module P of rank(*) > 1 over Rn is free. Later, in "Projective modules over polynomial extensions of polynomial rings," Invent.

Comment. The following important special cases of this exercise should be kept in mind. (1) If annr(a) = 0, thenaR is a freeR-module of rank 1, with a singleton basis {a}. (2) If a is itself an idempotent, then indeed aR is projective. In this case, annr(a) = eR for the complementary idempotent e: = 1- a. Rings R for which all principal right ideals are projective ("principal ::::} projective") are called right P P-rings, or right Rickart rings. For more information on these rings and examples of them, see LMR-§7D.

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