Advanced Algebra II

Advanced Algebra II

Noetherian rings and primary decomposition

We will assume R is commutative with identity in the following. One can define the similar notion of Noetherian rings and Artinian rings as:

Definition 0.1. A ring R is said to be Noetherian (resp. Artinian) if R is a Noetherian (resp. Artinian) R-module.

Even though the definition looks quite similar to modules. We would like to remark that there is some subtlety here.

Example 0.2. We know that a submodule of a Noetherian module is Noetherian. But a subring S ⊂ R of a Noetherian ring R is not neces- sarily a Noetherian ring. For example, let S = k[x₁, ...] be a polynomial ring with infinitely many indetermenates and R be its quotient field. R is a Noetherian ring but S is not.

We will first examine some properties of Noetherian ring. It might be surprising at the first glance that an Artinian ring is Noetherian.

Proposition 0.3. If R is Noetherian ring and I C R is an ideal, then R/I is Noetherian.

Proof. Since ideals of R/I is of the form J/I for some J C R. An ascending chain of ideal in R/I corresponds to an ascending chain of

ideals in R hence stationary. ¤

Theorem 0.4 (Hilbert Basis Theorem). If R is Noetherian, then the polynomial ring R[x] is Noetherian.

Proof. Let I C R be an ideal. We define

I₀ := {a ∈ R|f = axⁿ+ l.o.t ∈ I}.

It’s clear that I₀ C R. Let b₁, ..., b_r be a set of generator of I₀. And let fi = bixⁿⁱ + l.o.t be a polynomial with leading coefficient bi. Let n = max{ni}. If f ∈ I has degree ≥ n with leading coefficient b , then b = P

sibi for some si ∈ R. If follows that we can ”divide” f by fi

f =X

s_ixⁿ⁻ⁿⁱf_i+ g.

Where g ∈ I and deg(g) < deg(f ). Inductively, one can write f = Pg_if_i+ h with deg(h) < n.

Let M = R+Rx+...+Rxⁿ⁻¹which is a Noetherian R-module. Then I ∩ M is a Noetherian R-module. Let h₁, ..., h_t be a set of generators of I ∩ M as R-module. Then we have seen that I is generated by f1, ..., fr, h1, ..., ht as R[x]-module. Thus R[x] is Noetherian. ¤ Corollary 0.5. Let R be a Noetherian ring. Then any finitely gener- ated R-algebra is Noetherian.

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Proof. If S is a finitely generated R-algebra, say generated by s1, .., sn, then there is a surjective ring homomorphism

ϕ : R[x₁, ..., x_n] → S = R[s₁, ..., s_n] f (x₁, ..., x_n) 7→ f (s₁, ..., s_n).

By Hilbert Basis Theorem, R[x1, ..., xn] is a Noetherian ring. And hence the homomorphic image S ∼= R[x1, ..., xn]/Kerϕ is a Noetherian

ring. ¤

Example 0.6. Let k be a field. We consider Aⁿ_k := {(a₁, ..., a_n)|a_i ∈ k}

the affine space. For a subset S ⊂ Aⁿ, one can consider I(S) := {f ∈ k[x₁, ..., x_n]|f (x) = 0, ∀x ∈ S}.

It’s clear that I(S)Ck[x1, ..., xn] and hence finitely generated. The ring A(S) := k[x1, ..., xn]/I(S) represents ”polynomial functions on S” is a Noetherian ring.

By the similar argument as in the proof of Hilbert Basis Theorem, one can show that: (while leading term now is the term of smallest degree)

Proposition 0.7. Let R be a Noetherian ring. Then R[[x]] is a Noe- therian ring.

An important consequence is the following:

Corollary 0.8. Let R be an Noetherian ring and S be a finitely gen- erated R-algebra, then S is finitely generated.

Another important feature of Noetherian ring is that it allows us to perform the primary decomposition. As we’ll see later in this section that there is a nice correspondence between algebra (polynomial rings) and geometry (algebraic sets). The primary decomposition in the al- gebraic side coincide with the decomposition of an algebraic set into union of irreducible components.

We first work on the primary decomposition for Noetherian rings.

Definition 0.9. An ideal q C R is said to be primary if xy ∈ q then either x ∈ q or yⁿ ∈ q for some n.

Proposition 0.10. If q is primary, then √

q is a prime ideal. More- over, every prime ideal containing q contains √

q.

Let p :=√

q, then we say q is p-primary.

Proof. Let p := √

q. We first show that p is prime. To this end, if xy ∈ p, then xⁿyⁿ= (xy)ⁿ∈ q for some n. Since q is primary, one has either xⁿ ∈ q or (yⁿ)^m ∈ q for some m. In any case, one has either x ∈ p or y ∈ p.

Let p⁰ be a prime ideal containing q. If x ∈ p =√

q, then xⁿ ∈ q ⊂ p⁰ for some n. It follows that x ∈ p⁰. Hence we have p ⊂ p⁰. ¤

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Definition 0.11. Let a C R be an ideal. We say that a is irreducible if there is no non-trivial decomposition, i.e. for any a = b ∩ c, then a = b or a = c.

Proposition 0.12. If R is Noetherian, then every ideal can be written as intersection of finitely many irreducible ideals.

Proof. Let aCR be an ideal. Suppose that a is irreducible, then nothing to prove. It a is not irreducible, then a = a₁ ∩ a⁰. If both a₁ and a⁰ are irreducible, then we are done. Otherwise, we may assume that a₁ is reducible. Note that a ⊂ a₁. By continuing this process, we get a sequence of ideals

a ⊂ a₁ ⊂ a₂ ⊂ ...

Since R is Noetherian, this process must terminates and hence we are done.

Another way to put it is: Let Σ be the set of ideals which cannot be written as decomposition of irreducible ideals. We would like to prove that Σ is empty. If Σ 6= ∅, by the maximal condition, there is a maximal element a ∈ Σ. a = b ∩ c since a is not irreducible. By the maximality of a, one has that both b and c have finite decomposition, hence so is a. This is the required contradiction. ¤ Proposition 0.13. If R is a Noetherian ring, then irreducible ideal is primary.

Proof. Let a C R be an irreducible ideal. We need to show that a is primary. Let’s pass to the ring ¯R := R/a. If suffices to show that if xy = 0 ∈ ¯R, then either x = 0 or yⁿ= 0 for some n.

Claim. 0 = (x) ∩ (yⁿ).

Grant this claim, then by the irreducibility of a, one has that 0 C ¯R is irreducible. Hence 0 = (x) or 0 = (yⁿ) and we are done.

To prove the claim, we consider the ascending chain Ann(y) ⊂ Ann(y²) ⊂ ...

By the a.c.c., there is n such that Ann(yⁿ) = Ann(yⁿ⁺¹). If a ∈ (x) ∩ (yⁿ), then a = bx = cyⁿ for some b, c. ay = 0 since (bx)y = b(xy) = 0.

Hence cyⁿy = cyⁿ⁺¹ = 0. So c ∈ Ann(yⁿ⁺¹) = Ann(yⁿ). Therefore,

a = cyⁿ = 0. ¤

Combining all these, we have

Proposition 0.14. Let R be a Noetherian ring, then every ideal in R has a primary decomposition, i.e. can be written as a finite union of primary ideals.

Exercise 0.15. Let I C R be a radical ideal in a Noetherian ring R, then I = ∩^r_i=1pi for some primes ideal pi.