A.4 · Rings, ideals and modules

For the domain, ideal and quotient examples below, use a commutative ring with identity 1 distinct from 0. A source problem may specify a general ring instead; do not assume its multiplication commutes unless stated or proved. A unit has a multiplicative inverse. A nonzero zero divisor multiplies some nonzero element to zero. An integral domain has no such zero divisors; cancellation of a nonzero factor then works. A field is a domain in which every nonzero element is a unit. Z is a domain but not a field; Z/6Z is neither.

An ideal I is an additive subgroup that absorbs multiplication by every ring element. This is stronger than being a subring. In Z, nZ is an ideal; quotient elements are integer residue classes modulo n. In a commutative ring, R/I is a field exactly when I is maximal, and it is a domain exactly when I is prime. The ideal must be proper in both statements.

A module allows scalars from a ring instead of requiring a field. Every abelian group is a Z-module by repeated addition, but it need not have a vector-space basis. In Z/6Z as a Z-module, 6 times the nonzero residue 1 is zero; this is torsion. For a vector space over a field, a nonzero scalar is invertible and cannot annihilate a nonzero vector.

A submodule is closed under addition and all permitted scalar actions. A linear map of modules preserves both. The kernel and image are submodules, and the quotient by the kernel is isomorphic to the image. Do not apply finite-dimensional rank-nullity to an arbitrary module without establishing an appropriate free-module setting; integer row operations and field row operations permit different divisions. In a general Boolean ring, every a satisfies a²=a. Do not assume commutativity to prove it: idempotence of a+a gives 4a=2a, hence 2a=0. Expanding (a+b)²=a+b gives ab+ba=0, and characteristic two makes −ba=ba; therefore ab=ba. Idempotence does not imply nilpotence: in F2, the nonzero element 1 satisfies 1^n=1 for every positive n.

In Z/6Z, 2·3=0, so 2 and 3 are zero divisors. The units are 1 and 5 because their gcd with 6 is 1. In Z, the ideal 5Z gives the field Z/5Z, while 6Z gives a quotient with zero divisors. As Z-modules, the map from Z to Z/6Z has kernel 6Z; the quotient identifies integers differing by a multiple of 6, rather than producing a real vector space.

A set can be a submodule without being a vector space over the rationals. Scalar division is valid only when the scalar inverse belongs to the structure.

In $\mathbb Z/12\mathbb Z$, list the units and nonzero zero divisors. Is the ideal $(4)$ prime or maximal?

Worked solution

Units are the residues coprime to 12: $1,5,7,11$. The remaining nonzero residues $2,3,4,6,8,9,10$ are zero divisors. For each such a, $a\cdot(12/\gcd(a,12))=0$ modulo 12, with the second factor nonzero. The ideal $(4)=\{0,4,8\}$ gives quotient isomorphic to $\mathbb Z/4\mathbb Z$. This is not a domain, since $2\cdot2=0$, and is not a field. Hence the ideal is neither prime nor maximal.

As a $\mathbb Z$-module, can $\mathbb Z/6\mathbb Z$ be free? Compare the ideals $(2)$ and $(x)$ in $\mathbb Z$ and $\mathbb Q[x]$, respectively, using their quotients.

Check after attempting

A nonzero free $\mathbb Z$-module has no nonzero vector annihilated by a nonzero integer: examine each integer coordinate. But $6[1]=0$ in $\mathbb Z/6\mathbb Z$. It is not free; a field-style basis argument is invalid. $\mathbb Z/(2)\cong\mathbb F_2$ is a field, so $(2)$ is maximal and prime. Evaluation at zero maps $\mathbb Q[x]$ onto $\mathbb Q$ with kernel $(x)$; its quotient is a field too. Both arguments use proper ideals in commutative rings with identity.