A free module ( F ) with basis ( {e_i} ) means every element is a unique finite linear combination ( \sum r_i e_i ). Over commutative rings, the rank of a free module is well-defined if the ring has IBN (invariant basis number) — all fields, ( \mathbb{Z} ), and commutative rings have IBN.
Define addition pointwise: ( (f+g)(m) = f(m)+g(m) ). Define scalar multiplication: ( (rf)(m) = r f(m) ). Check module axioms. Dummit And Foote Solutions Chapter 10.zip
Use the relations: ( a \otimes b = a \otimes (b \bmod \gcd(m,n)) ). The result is isomorphic to ( \mathbb{Z}/\gcd(m,n)\mathbb{Z} ). The trick is to show that ( m(a\otimes b) = a\otimes (mb) = a\otimes 0 = 0 ), and similarly ( n ). Hence the tensor product is annihilated by ( \gcd(m,n) ). 11. Projective and Injective Modules (introduction) Definition: ( P ) is projective iff every surjection ( M \to P ) splits. Equivalently, ( \text{Hom}(P,-) ) is exact. A free module ( F ) with basis
The exercises in Chapter 10 are notoriously dense. They test not just computation, but conceptual understanding of exact sequences, direct sums, free modules, and the relationship between ( R )-modules and abelian groups. This essay provides a meta-solution : strategies for attacking each major problem type, with key lemmas and warnings. 1. Verifying Module Axioms Typical Problem: Show that an abelian group ( M ) with a ring ( R ) action is an ( R )-module. Define scalar multiplication: ( (rf)(m) = r f(m) )