A field is always noetherian

Question

I learned from this a recent posting here that a field is always notherian since a field has only $\{0\}$ and the field itself as ideals. However in a 3-year old posting here, it says $\mathbb Q$ as $\mathbb Z$-module is not notherian, see the third part of Rudy the Reindeer's response. I am sure I have messed up with something, therefore I would appreciate if anyone could shed me some light.

Thank you for your time and help.

---

POST SCRIPT: Tobias Kildetoft has responded to me below. Thank you. Let's say that I am pretty familiar with the definitions of $\mathbb Q$ as $\mathbb Z$-module and as $\mathbb Q$ as $\mathbb Q$-module, but on formal level only. Is there anyone care to explain to me the two's practical difference instead? Thanks again to Tobias.

Answer 1

Practically speaking, it is a lesson in understanding terminology.

The phrase "$R$ is a right Noetherian ring" hides the dependency of the Noetherian condition on a module structure. It is equivalent to "the right $R$ module $R_R$ is Noetherian."

An abelian group can be a module over many different rings, and whether or not the group is Noetherian as an $R$-module depends on the ring $R$. Changing the can completely alter what the submodules are.

Consider the abelian group $\Bbb C$ (the complex numbers) for example. Of course it is a $\Bbb Z$-module, $\Bbb Q$-module, $\Bbb R$-module, and $\Bbb C$-module all at the same time.

As a $\Bbb Q$-module, you can find uncountably many $\Bbb Q$-submodules inside whose direct sum is the whole module. As an $\Bbb R$-module you can do the same, but you will always need exactly two modules to get the direct sum, and no more. Certainly not infinitely many. As a $\Bbb C$-module, there are exactly two submodules of $\Bbb C$: the whole set and the zero submodule.

To make it even more concrete, consider the following:

$\Bbb Z$ is a $\Bbb Z$ submodule of $\Bbb C$ that isn't a $\Bbb Q$ submodule.

$\Bbb Q$ is a $\Bbb Q$ submodule of $\Bbb C$ that isn't a $\Bbb R$ submodule.

$\Bbb R$ is an $\Bbb R$ submodule of $\Bbb C$ that isn't a $\Bbb C$ submodule.

So you can see that the $R$-submodules of a module really do depend on what $R$ you are talking about. Different rings will radically change the submodules, and any statements you might make about them like "Noetherian" or "Artinian" or anything else.

Answer 2

Before answering your question, lets conceptualize a bit.

(All my rings are commutative.)

1. Given a ring $R$, we may speak of $R$-modules.
2. Given an $R$-module $X$, we may speak of the submodules of $X$.
3. The submodules of $X$ form a poset $\mathrm{Sub}(X).$

4. In order of decreasing strength, several conditions we can put on $X$ include:

   • C0. $\mathrm{Sub}(X)$ has precisely two elements
   • C1. $\mathrm{Sub}(X)$ has finitely-many elements
   • C2. $\mathrm{Sub}(X)$ is co-wellfounded (i.e. it satisfies the ascending chain condition)
5. We say that a module is simple iff it satisfies C0.
6. We say that a module is noetherian iff it satisfies C2.
7. Every ring $R$ has an underlying $R$-module, defined in the obvious way. Lets denote this $\overline{R}.$
8. An ideal of $R$ is, by definition, a submodule of $\overline{R}.$ Exercise. Check that this agrees with the more usual "explicit" definition of an ideal.
9. We say that $R$ is a field iff the module $\overline{R}$ is simple (i.e. it satisfies C0); equivalently, $R$ is a field iff its poset of ideals has precisely two elements.

10. We say that $R$ is noetherian iff the module $\overline{R}$ is noetherian (i.e. it satisfies C2); equivalently, $R$ is noetherian iff its poset of ideals is co-wellfounded.

11. Since every simple module is noetherian, hence every field is noetherian.

So being noetherian isn't really a property of rings, its a property of modules! But since it is often convenient to view $R$ as a module over itself, hence we have a lot of convenient terminology to refer to this situation, like "ideal" and "field" and "noetherian ring."

Therefore, the phrase "$\mathbb{Q}$ as a ring is noetherian" can be viewed as shorthand. The full phrase is "$\mathbb{Q}$ as a $\mathbb{Q}$-module is noetherian," which is clearly true, since $\mathbb{Q}$ is a field, which just means that "$\mathbb{Q}$ as a $\mathbb{Q}$-module is simple," which is a stronger condition!

How about $\mathbb{Q}$ viewed as a $\mathbb{Z}$-module? Well, a $\mathbb{Z}$-module is the same thing as an abelian group. Therefore submodules of $\mathbb{Q}$ are the same as (additive) subgroups of $\mathbb{Q}$. So a subset of $\mathbb{Q}$ is a submodule iff it is closed under addition and the taking of additive inverses. Hence, for each $q \in \mathbb{Q}$, the set $q\mathbb{Z}$ is a submodule of $\mathbb{Q}$. Now observe that the sequence of submodules $\mathbb{Z},(1/2)\mathbb{Z},(1/4)\mathbb{Z},(1/8)\mathbb{Z},\ldots$ contradicts the possibility of $\mathbb{Q}$ being noetherian.