Noetherian Rings

Section 10.4 Noetherian Rings

“All good things must come to an end.”
―Geoffrey Chaucer

Most rings that commutative algebraists naturally want to study are noetherian. Noetherian rings are named after Emmy Noether, who is in many ways the mother of modern commutative algebra.

Definition 10.31. Ascending Chain Condition.

A commutative ring \(R\) has the ascending chain condition (on ideals) if given any chain of ideals in \(R\) of the form

\begin{equation*} I_1 \subseteq I_2 \subseteq \cdots \end{equation*}

there is an \(N\) such that

\begin{equation*} I_N = I_{N+1} = \cdots. \end{equation*}

Definition 10.32. Noetherian Ring.

Suppose \(R\) is a commutative ring. Then \(R\) is called a noetherian ring if \(R\) satisfies the ascending chain condition on ideals.

Theorem 10.33. PIDs are Noetherian.

If \(R\) is a PID then \(R\) is noetherian.

Proof.

Consider and ascending chain of ideals of \(R\text{;}\) it must have the form

\begin{equation*} (a_1)\subseteq (a_2) \subseteq (a_3) \subseteq \cdots \subseteq (a_i)\subseteq (a_{i+1})\subseteq \cdots \end{equation*}

Consider \(I=\bigcup_{i\geq 1} (a_i)\) which is an ideal of \(R\) by the argument given in [provisional cross-reference: cite]. Since \(R\) is a PID, \(I=(b)\) for some \(b\in R\text{.}\) Since \(b\in I=\bigcup_{i\geq 1} (a_i)\text{,}\) we must have \(b\in (a_n)\) for some \(n\text{.}\) Then we see that \(I=(b)\subseteq (a_n)\subseteq (a_{n+1})\subseteq \cdots (a_j) \subseteq \cdots \subseteq I\) for all \(j\geq n\text{,}\) thus \(I=(a_j)\) for \(j\geq n\) and the chain stabilizes as desired.

Corollary 10.34. Fields are Noetherian.

Every field is noetherian.

Exercise 10.35. Quotient Rings Noetherian in Noetherian Rings.

Let \(R\) be a commutative ring and \(I\) an ideal of \(R\text{.}\) Show that if \(R\) is noetherian then \(R/I\) is also noetherian.

Theorem 10.36. Factorization in Noetherian Domains.

If \(R\) is a noetherian integral domain, then every non-zero, not-unit element factors into a finite product of irreducible elements.

Proof.

Pick \(x \in R\) with \(x \ne 0\) and \(x \notin R^\times\text{.}\) If \(x\) is irreducible, there is nothing to prove. Otherwise, we have \(x = x_1 x_2\) for non-units \(x_1, x_2\text{.}\) If both \(x_1, x_2\) are irreducible, the proof is complete. Otherwise, one or both of them factors non-trivially. We may express this conveniently by saying that \(x_1 = x_3x_4\) and \(x_2 = x_5x_6\) such that either \(x_3\) and \(x_4\) are both non-units or \(x_5\) and \(x_6\) are both non-units. (E.g., if \(x_2\) is irreducible, we could set \(x_5 = x_2, x_6 = 1\text{.}\)) Continuing in the this manner, we form a binary tree with \(x\) at the top, \(x_1, x_2\) one level down, \(x_3,x_4,x_5,x_6\) one level below that, etc.

We halt the process of building the tree if at some stage all the leaves of the tree are irreducible elements, at which point we will have proven that \(x\) factors in to a product of the irreducible elements given by these leaves.

We need to rule out the possibility that the process never terminates. If it never terminates, we will have built an infinite binary tree with the property that some route downward through the tree consists of an infinite list of irreducible elements \(y_1, y_2, y_3,\dots\) such that \(x = y_1z_1\) for a non-unit \(z_1\) and, for each \(i \geq 1\text{,}\) \(y_{i} = y_{i+1} z_{i+1}\) for a non-unit \(z_{i+1}\text{.}\) Since \(R\) is an integral domain, we have \((x) \subsetneq (y_1)\) and \((y_i) \subsetneq (y_{i+1})\) for all \(i \geq 1\text{.}\) (E.g., if \((x) = (y_1)\) then \(y_1 = xv\) and hence \(x = xvz_1\text{,}\) so that \(vz_1 = 1\text{,}\) contrary to \(z_1\) being a non-unit.)

But then we have arrived at an infinite ascending chain of ideals in \(R\text{,}\)

\begin{equation*} (x) \subsetneq (y_1) \subsetneq (y_2) \subsetneq (y_3) \subsetneq \cdots, \end{equation*}

which is not possible in a Noetherian ring.

Remark 10.37.

For those mathematicians who refute the Axiom of Choice (though they are few and far between), noetherian rings provide a haven of sorts within the world of algebra. This is because the existence of a maximal ideal is guarenteed in noetherian rings, given that every ascending chain of ideals stabilizes.

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