Noetherian Rings and the Hilbert Basis Theorem

Three faces of one condition

In a first course you proved PIDs are wonderfully tame: every ideal needs just one generator. That is too strong to ask of, say, k[x,y], where the ideal (x,y) genuinely needs two. The right generalization keeps the spirit — ideals don't sprawl out of control — without demanding a single generator. A commutative ring R is a Noetherian ring if every ideal of R is finitely generated.

What makes this hypothesis so usable is that three apparently different statements are equivalent. Get fluent switching between them; in practice you pick whichever is easiest to verify.

1. Finite generation. Every ideal I ⊆ R is generated by finitely many elements.
2. Ascending chain condition (ACC). Every chain I_1 ⊆ I_2 ⊆ I_3 ⊆ … of ideals eventually stabilizes: I_n = I_{n+1} = … for some n.
3. Maximal condition. Every nonempty set of ideals has a maximal element under inclusion.

The equivalence, worked

The cycle (1) ⇒ (2) ⇒ (3) ⇒ (1) is short and worth doing once by hand, because the same union/chain trick recurs everywhere in the subject. The only subtle step uses the axiom of choice in the form of the maximal condition.

(1) => (2): Given a chain I_1 c I_2 c ... , let I = union of all I_n.
  I is an ideal (a chain-union of ideals always is).
  By (1), I = (a_1, ..., a_k) for finitely many generators.
  Each a_j lies in some I_{n_j}; take N = max(n_1, ..., n_k).
  Then all a_j in I_N, so I = I_N, and the chain stops at N.

(2) => (3): Suppose a nonempty set S of ideals has NO maximal element.
  Pick I_1 in S. Not maximal => exists I_2 in S with I_1 (strictly) c I_2.
  Repeat forever => a strictly ascending chain that never stabilizes,
  contradicting (2).  [uses dependent choice]

(3) => (1): Let I be any ideal. Let S = { finitely generated ideals c I }.
  S is nonempty (0 is in it). By (3) it has a maximal element J = (a_1,...,a_m).
  If J != I, pick x in I \ J. Then J + (x) is finitely generated, c I,
  and strictly bigger than J -- contradicting maximality. So J = I.

Hilbert's basis theorem: Noetherian is contagious

The reason the condition is omnipresent is the Hilbert basis theorem: if R is Noetherian, so is the [[polynomial-ring|polynomial ring]] R[x]. Iterating, R[x_1,…,x_n] is Noetherian over any field or over Z. Since every coordinate ring of an affine variety is a quotient of such a ring — and quotients of Noetherian rings are Noetherian — essentially every ring in algebraic geometry is Noetherian for free.

The proof is a leading-coefficient argument. Be warned: it is one of those proofs that feels like sleight of hand the first time. The trick is to organize an arbitrary ideal of R[x] by the degrees of its members and lean on R being Noetherian at each level.

Claim: R Noetherian => R[x] Noetherian.
Let J be an ideal of R[x]. For each d >= 0 set
  L_d = { leading coeffs of degree-d elements of J } u {0}  c R.
Each L_d is an ideal of R, and L_0 c L_1 c L_2 c ...  (multiply by x).
By ACC in R this chain stabilizes at some N: L_d = L_N for all d >= N.

Each L_d (0 <= d <= N) is f.g.: L_d = (a_{d,1}, ..., a_{d,k_d}).
Choose f_{d,i} in J of degree d with leading coeff a_{d,i}.
Claim: these finitely many f_{d,i} generate J.

Given g in J of degree d, subtract an R[x]-combination of the f_{*,i}
matching its leading term (using L_d, or L_N times x^{d-N} if d > N).
This lowers deg g. Induct on degree => g lands in the ideal generated
by the f_{d,i}. Hence J is finitely generated.  QED