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:
- 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 $n\geq N$.)
- Every nonempty set of ideals in $R$ has a maximal element.**
- 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.