Modules: Linear Algebra Over a Ring

Here R is a commutative ring with identity. An R-module has abelian-group addition and scalar multiplication satisfying r(m + n) = rm + rn, (r + s)m = rm + sm, (rs)m = r(sm), and 1m = m. A vector space is a module whose scalar ring is a field. Ring scalars need not all have inverses, but fields and invertible ring elements are allowed.

A basis spans the module and is independent: only zero coefficients give a zero finite combination. For a nonzero commutative ring R with identity, R^k has k standard basis vectors. Coordinates prove spanning, and a zero tuple forces every coefficient to be zero. Finite bases over such a ring have the same size, called rank. For example, Z^3 has three standard basis vectors. These rank claims exclude the zero ring.

Over a domain such as Z, a nonzero element killed by a nonzero scalar is torsion and cannot belong to an independent set. In the Z-module Z/n for n > 1, n kills every element. Thus no nonempty independent set exists; the independent empty set does not span this nonzero module. Do not use this test to reject freeness over a ring with zero divisors: Z4 over itself has basis {1}, although 2 times 2 = 0.

For a principal ideal domain R, every finitely generated R-module is a finite direct sum of a free part R^r and cyclic torsion parts R/(p^e), with prime elements p and positive exponents e. The parts are unique up to order and associated prime choices. They need not be finite sets. Over Z, these quotients are finite cyclic groups of prime-power order. A free Z-coordinate contributes no nonzero torsion element.

A finite direct sum has coordinate tuples, with each coordinate chosen independently. Cyclic groups C_a and C_b therefore give a times b elements, not a + b. For example, C3 plus C5 has 15 pairs. A pair's order is the lcm of its coordinate orders. A finite group is cyclic only if it has an element whose order equals its size. You can also recover an unknown factor size by dividing the total size by the known size.

For coprime positive a and b, send a residue m modulo ab to (m modulo a, m modulo b). Equal pairs imply a and b both divide the input difference, hence ab divides it. The map is therefore injective; equal finite sizes make it bijective. It respects addition and integer scaling, giving Z/(ab) = Z/a plus Z/b as modules, up to isomorphism. For the separate example Z10, use moduli 2 and 5.

For an n by n complex matrix A, evaluate p(A) by replacing x^j with A^j and constants c with cI. Let p(x)v = p(A)v on C^n. Cayley-Hamilton says the characteristic polynomial chi_A satisfies chi_A(A) = 0, so this finitely generated module is torsion. C[x] is a PID, and complex polynomials split into linear factors. Summands C[x]/((x - lambda)^k) give Jordan blocks J_k(lambda). Over another field, this requires the characteristic polynomial to split.

A Jordan block J_k(lambda) has lambda on the diagonal, 1 directly above it and zeros elsewhere. Each block contributes one independent eigenvector direction. For a single eigenvalue, block sizes sum to the matrix dimension and the number of blocks equals its eigenspace dimension. For a 3 by 3 complex matrix with only eigenvalue 2 and eigenspace dimension 2, the sizes are 2 and 1. Polynomial powers (x - lambda)^k encode these sizes.