# Noetherian Rings = Generalization of PIDs

When I was first introduced to Noetherian rings, I didn't understand why my professor made such a big hoopla over these things. What makes Noetherian rings so special? Today's post is just a little intuition to stash in The Back Pocket, for anyone hearing the word "Noetherian" for the first time.

A ring is said to be Noetherian if every ideal in the ring is finitely generated. Right away we see that every principal ideal domain is a Noetherian ring since every ideal is generated by one element. Well, in short,

"Noetherian-ness" is a property which generalizes "PID-ness".

As K. Conrad so nicely puts it

The property of all ideals being singly generated is often not preserved under common ring-theoretic constructions (e.g. $\mathbb{Z}$ is a PID but $\mathbb{Z}[x]$ is not), but the property of all ideals being finitely generated does remain valid under many constructions of new rings from old rings. For example... every quadratic ring $\mathbb{Z}[\sqrt{d}]$ is Noetherian, even though many of these rings are not PIDs." (italics added)

And there you have it! We like rings with finitely generated ideals because it keeps the math (relatively) nice. For instance, one might ask, "Given a Noetherian ring $R$, can I build a new ring such that it, too, is Noetherian?"* Hilbert's Basis Theorem says the answer is yes! You can construct the polynomial ring $R[x]$ and it will be Noetherian whenever $R$ is (and in fact so will $R[x_1,\ldots,x_n]$).

As a final remark, there are three equivalent conditions associated with the Noetherian property. So when testing a ring for "Noetherian-ness," remember that one condition may be easier to invoke than another. They are as follows:

Proposition: Let $R$ be a commutative ring with 1. The following are equivalent:
1. Every ascending chain of ideals $I_1\subset I_2\subset\cdots$ in $R$ is stationary (i.e. there exists $N\in \mathbb{N}$ s.t. $I_n=I_N$ for all $n\geq N$.)
2. Every nonempty set of ideals in $R$ has a maximal element.**
3. Every ideal $I\triangleleft R$ is finitely generated.

Sketch of proof:

• (1 $\Rightarrow$ 2) Let $X$ be a nonempty set of ideals in $R$. Pick $I_1\in X$. If $I_1$ is maximal, then done. Otherwise there is $I_2\in X$ such that $I_1\subset I_2$. If $I_2$ is maximal, then done. Otherwise there is $I_3\in X$ such that $I_1\subset I_2\subset I_3$. Continuing in this fashion, we see that if $X$ does not contain a maximal ideal, then we can produce an ascending chain of ideals which never stabilizes, contradicting 1.

• (2 $\Rightarrow$ 3) Let $I\triangleleft R$ be an ideal and $X=\{J\triangleleft R:J\subset I \text{ and$J$is finitely generated}\}$ be the set of all ideals contained in $I$ which are finitely generated. Since $X\neq \emptyset$ (as $I$ contains the zero ideal), $X$ contains a maximal element, say $J$. We claim $J=I$. Suppose not. Then we can find an element $a\in I\smallsetminus J$ and can form the ideal $J+(a)$. This ideal is finitely generated and so is an element in $X$. But $J\subset J+(a)$ and this contradicts the maximality of $J$. Hence $J=I$ and so $I$ is finitely generated.

• (3 $\Rightarrow$ 1)
• Let $I_1\subset I_2\subset \cdots$ be an ascending chain and define $I=\bigcup_{n=1}^{\infty}I_n$. This is an ideal and so by assumption, $I=(a_1,a_2,\ldots,a_m)$ for some $a_i\in R$. Thus we can find $N$ such that $a_1,a_2,\ldots,a_m\in I_N$. Hence $I\subset I_N$ which implies $I_n=I_N$ for all $n\geq N$ and so the chain stabilizes.

Footnotes:

*This is a standard question a mathematician might ask. How can I build new things from existing ones?

**Don't forget, "maximal" is not the same thing as "a maximum"! - we've seen this before.

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