Over a non-domain (e.g., ( \mathbb{Z}/6\mathbb{Z} )), torsion elements don’t form a submodule in general because the annihilator of a sum may be trivial. Part VI: Advanced Exercises (61–75) 10. Tensor Products (if covered in your edition) Typical Problem: Compute ( \mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z} ).
Define addition pointwise: ( (f+g)(m) = f(m)+g(m) ). Define scalar multiplication: ( (rf)(m) = r f(m) ). Check module axioms.
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.
A module homomorphism from a free ( R )-module ( F ) with basis ( {e_i} ) to any ( R )-module ( M ) is uniquely determined by choosing images of the basis arbitrarily in ( M ).
It is impossible for me to provide a complete, line-by-line solution set for an entire chapter (e.g., Chapter 10 on Module Theory) of Abstract Algebra by Dummit and Foote in a single response. Such a document would be dozens of pages long and exceed output limits.
Forgetting to check that ( 1_R ) acts as identity. This fails for rings without unity (though Dummit assumes unital rings for modules). 2. Submodules and Quotients Typical Problem: Given an ( R )-module ( M ), decide if a subset ( N \subset M ) is a submodule.
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.