Read quotient structure from an ideal

Try

Compare Z/6Z with Z/7Z. The first has nonzero zero divisors, while the second is a field. Both arise by quotienting the same ring by a principal ideal. The different behavior can be expressed as a property of the ideal before constructing the quotient table.

A prime ideal P is proper and satisfies ab∈P only if a∈P or b∈P. A maximal ideal M is proper and has no ideal strictly between M and R. These definitions concern multiplication and containment respectively.

Predict

Translate “a+P is nonzero” back into membership in P. Then translate a zero product of two cosets. Predict that primality is exactly the condition saying the quotient has no nonzero zero divisors. For maximality, predict why adjoining any missing element a to M should make the generated ideal contain one.

Experiment

Use quotients to test several ideals of Z.

Ideal	Quotient	Prime?	Maximal?
(0)	Z	Yes	No
(5)	Field with five elements	Yes	Yes
(6)	Residues with 2·3=0	No	No

For the zero ideal, the strict chain (0)⊊(2)⊊Z proves nonmaximality directly. For (6), the product two times three belongs to the ideal while neither factor does, disproving primality without relying on a theorem name.

Understand

For a proper ideal P, the quotient R/P is nonzero. Its product (a+P)(b+P) is zero exactly when ab∈P. Requiring one factor coset to be zero is exactly the definition of P being prime. This proves both directions of the domain criterion.

For maximal M and a∉M, the ideal M+(a) strictly contains M and therefore equals R. Write 1=m+ra with m∈M. In the quotient, (r+M)(a+M)=1+M. Every nonzero coset has an inverse, so R/M is a field.

Conversely suppose R/M is a field and an ideal I strictly contains M. Choose a∈I outside M. Its coset has an inverse r+M, so ra−1∈M⊆I. Since ra∈I, subtraction gives 1∈I and hence I=R. Thus M is maximal. The proof explains why finding one invertible nonzero image forces an intermediate ideal to become the whole ring.

Every maximal ideal is prime because every field is a domain. The converse is false: (0) in Z is prime but not maximal. In a principal ideal domain, every nonzero prime ideal is maximal, a stronger statement requiring the principal-ideal hypothesis and exclusion of the zero ideal. Do not silently apply that conclusion to arbitrary rings.

For Z/nZ, ideals correspond to ideals of Z containing nZ. Maximal ideals therefore correspond to prime divisors of n. For n=12 there are two, generated by the residues two and three. Their quotients are fields of sizes two and three. Different prime divisors produce different maximal ideals even though both satisfy the same abstract definition.

Apply

Prove that (x) in F[x] is maximal by identifying its quotient with F. Then consider (x) in Z[x]; the quotient is Z, so the ideal is prime but not maximal. The notation for the ideal is the same, but the coefficient ring changes the conclusion.

When a proposed quotient is called a field, check both nonzero-ness and inverses. When an ideal is called prime, check properness separately from the product condition; the whole ring satisfies the latter vacuously but is excluded by definition.

Continue

Euclidean division next provides a constructive route to principal ideals and greatest common divisors. It will supply the Bézout identities that make several quotient and irreducibility arguments work.