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No chain of ideals climbs for ever, no ideal needs infinitely many generators, every family of ideals has a biggest member: one finiteness condition in three forms, plus the Hilbert basis theorem.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the ascending chain condition and its two equivalent forms, decide whether a named ring is Noetherian, exhibit a chain that never stops in a ring that is not, quote the Hilbert basis theorem and its consequence for polynomial rings, say what the condition does and does not give, and prove that the chain condition forces every ideal to be finitely generated.
The last lesson said that every non-zero non-unit factors into irreducibles, and passed over why the stripping-off process stops. It stops because of a chain condition, and that condition turns out to be the single most important finiteness hypothesis in ring theory — strong enough to make arguments terminate, weak enough that almost every ring met in practice satisfies it.
A ring satisfies the ascending chain condition on ideals when every increasing chain $I_1 \subseteq I_2 \subseteq \cdots$ is eventually constant. A ring with that property is Noetherian, after Emmy Noether. An ideal is finitely generated when $I = (a_1, \ldots, a_k)$ for finitely many elements. The Hilbert basis theorem says $R$ Noetherian implies $R[x]$ Noetherian.
For a commutative ring $R$ the following are equivalent, and any of them may be taken as the definition of Noetherian:
Why they are equivalent. From 2 to 1: the union of a chain is an ideal, its finitely many generators lie in finitely many terms, so one term contains them all and the chain is constant from there. From 1 to 2: if $I$ were not finitely generated, one could choose $a_1, a_2, a_3, \ldots$ each outside the ideal generated by its predecessors, producing a chain that never stops. From 1 to 3 and back: a collection with no maximal member supplies an infinite strictly increasing chain, and conversely.
Each face is the convenient one somewhere. Chains for termination arguments, finite generation for computation, maximality for choosing a worst counterexample.
Which rings are Noetherian. Fields, trivially — two ideals. Principal ideal domains, since one generator is finitely many; so $\mathbb{Z}$ and $F[x]$. Quotients of Noetherian rings. And, by the Hilbert basis theorem, $R[x]$ whenever $R$ is, hence $\mathbb{Z}[x_1, \ldots, x_n]$ and $F[x_1, \ldots, x_n]$ for every finite $n$. Also every ring of algebraic integers in a number field, including $\mathbb{Z}[\sqrt{-5}]$.
Which are not. A polynomial ring in infinitely many variables: $(x_1) \subset (x_1, x_2) \subset \cdots$ never stops. The ring of all algebraic integers: $(\sqrt{2}) \subset (\sqrt[4]{2}) \subset (\sqrt[8]{2}) \subset \cdots$. The ring of continuous real functions on an interval. These have to be built on purpose.
What it buys, and what it does not. It buys termination — factorisations exist, inductions on ideals work, algorithms stop. It does not buy uniqueness: $\mathbb{Z}[\sqrt{-5}]$ is Noetherian and is not a UFD. Nor does it make ideals principal: $\mathbb{Z}[x]$ is Noetherian and $(2, x)$ needs two generators. Noetherian is about finiteness, not about factorisation.
Another way: picture
Imagine trying to build an ideal by adding generators one at a time, each genuinely new. In a Noetherian ring you always run out: after finitely many additions there is nothing left outside. That is all the condition says, and the reason it is everywhere is that almost every proof which constructs something step by step needs precisely the promise that the steps run out.
Another way: steps
To show a ring is Noetherian: 1. If it is a PID or a field, it is. 2. If it is $R[x]$ or a quotient of a Noetherian ring, it is. 3. Otherwise show every ideal has a finite generating set directly. To show it is not: 4. Exhibit one strictly increasing chain that never stops, or 5. Exhibit one ideal needing infinitely many generators — usually the same example seen two ways.
The statement. If $R$ is Noetherian then so is $R[x]$. By induction, $R[x_1, \ldots, x_n]$ is Noetherian for every finite $n$.
The proof, in outline. Suppose $I \subseteq R[x]$ is an ideal that is not finitely generated. Choose $f_1 \in I$ of least degree, then $f_2 \in I \setminus (f_1)$ of least degree, and so on. The leading coefficients generate an increasing chain of ideals of $R$, which must stop; at that point some leading coefficient is a combination of the earlier ones, and subtracting the corresponding multiple produces an element of $I$ outside the ideal so far but of smaller degree — contradicting the choice. So $I$ is finitely generated.
Why it was famous. Hilbert proved it in 1888 to settle a question in invariant theory that Gordan had attacked computationally for twenty years. The proof exhibits no generators at all; it shows a finite set must exist. Gordan's reported reaction — this is not mathematics, this is theology — is the standard anecdote about non-constructive proof, and he later accepted it.
What it means in practice. Any system of polynomial equations in finitely many variables is equivalent to a finite subsystem, however many equations were written down. That is the foundation on which algebraic geometry is built: every algebraic set is cut out by finitely many polynomials, so geometry over a field is a finite subject even when the ideal is described infinitely.
The failure, for contrast. In $F[x_1, x_2, x_3, \ldots]$ the ideal generated by all the variables needs all of them: no finite subset generates $x_{n}$ for large $n$. The chain $(x_1) \subset (x_1, x_2) \subset \cdots$ is the same fact in the other formulation. Finitely many variables is exactly the hypothesis the theorem needs.
*Reading Noetherian as finitely many ideals.* $\mathbb{Z}$ has infinitely many ideals and is Noetherian. The condition is on chains, not on the total.
*Reading it as every ideal is principal.* $\mathbb{Z}[x]$ is Noetherian and not a PID. Finitely generated is much weaker than singly generated.
Expecting unique factorisation. $\mathbb{Z}[\sqrt{-5}]$ is Noetherian and $6$ factors two ways. The chain condition gives existence, not uniqueness.
Assuming subrings inherit it. They do not. Quotients and polynomial extensions do; subrings need not.
Forgetting the chain must be increasing. The descending chain condition is a different and much stronger property — a ring with both is Artinian, and an Artinian domain is a field.
Thinking the union of a chain is always an ideal. It is for a chain, because any two elements lie in a common term. For an arbitrary family of ideals the union is usually not an ideal at all.
In $F[x_1, x_2, x_3, \ldots]$, consider $I_n = (x_1, \ldots, x_n)$.
One ideal per stage.
Each containment is strict: $x_{n+1} \notin I_n$, because every element of $I_n$ has every term divisible by one of $x_1, \ldots, x_n$.
Strictly increasing.
So the chain never becomes constant, and the ring is not Noetherian. The same fact in the other language: the ideal generated by all the variables has no finite generating set.
Not Noetherian.
Let $R$ be Noetherian and $I$ an ideal. Ideals of $R/I$ correspond exactly to ideals of $R$ containing $I$.
The correspondence theorem.
An increasing chain in $R/I$ pulls back to an increasing chain in $R$, which is eventually constant.
Pull the chain back.
Pushing forward again, the original chain is eventually constant too. So $R/I$ is Noetherian — and with Hilbert's theorem this makes every finitely generated algebra over a field Noetherian.
Inherited downwards.
Consider the elements $\sqrt{2}, \sqrt[4]{2}, \sqrt[8]{2}, \ldots$, all of them algebraic integers.
A sequence to try.
Each divides the one before: $\sqrt{2} = (\sqrt[4]{2})^{2}$, so $(\sqrt{2}) \subseteq (\sqrt[4]{2})$, and the containment is strict since $\sqrt[4]{2}$ is not a multiple of $\sqrt{2}$ there.
A strictly increasing chain.
The chain never stops, so the ring is not Noetherian — even though each individual $\mathbb{Z}[\sqrt[2^{k}]{2}]$ is. Being a union of Noetherian rings is not enough.
The Noetherian condition has three standard formulations. Match each name to what it says.
| Every increasing chain of ideals is eventually constant | Every ideal has a finite set of generators | Every non-empty family of ideals has a maximal member | The ring has only finitely many ideals | Every ideal is principal | |
|---|---|---|---|---|---|
| The ascending chain condition | |||||
| The finite generation condition | |||||
| The maximal condition | |||||
| A strictly stronger condition than these |
Is this ring Noetherian: $\mathbb{Z}[x]$?
Select every statement that is true of Noetherian rings.
This task has no paper form; do it on a device.
In $\mathbb{Z}$, start at the ideal $(2^{3})$ and climb: $(2^{3}) \subset (2^{2}) \subset \cdots \subset (2) \subset (1)$. How many strict inclusions are there?
Answer:
Put in order the steps showing that if every ideal is finitely generated then every increasing chain of ideals stops.
Number the steps in order (write the number in the box):
Build the proof that if every increasing chain of ideals stops, then every ideal is finitely generated.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In $\mathbb{Z}$, start at the ideal $(2^{6})$ and climb: $(2^{6}) \subset (2^{5}) \subset \cdots \subset (2) \subset (1)$. How many strict inclusions are there?
Answer:
You can test a ring against the ascending chain condition and move between the three equivalent statements of it. Say in your own words why Noetherian gives the existence of factorisations and not their uniqueness. Next: extensions of fields, where a degree replaces an order.
9. Your turn: is the ring of all algebraic integers Noetherian?, step 3