Why are ideals more important than subrings?

Question

I have read that subgroups, subrings, submodules, etc. are substructures.

But if you look at the definition of the Noetherian rings and Noetherian modules, Noetherian rings are defined with ideals and Noetherian modules are defined with submodules. Isn't it awkward? Why does submodule correspond to ideal, not subring? Is there any definition of Noetherian with subrings?

As I'm studying commutative algebra, it looks like ideals are more important than subrings. But why is it ideal, not subring (which seems to correspond to all other substructures)? Though I am not very familiar with pseudo-rings, is it true that ideal is a sub-pseudo-ring (or sub-rng) and thus we can view ideal as a kind of substructure?

Answer 1

Any ring $R$ is a module over itself, in the obvious manner: $R$ is an abelian group, and we define $a\cdot b$ for $a\in R$ and $b\in R$ to be $ab$.

A submodule of $R$-as-an-$R$-module is precisely an ideal of $R$ (work out the relevant definitions to see this). Thus, the definition of a Noetherian ring is really saying that it is a Noetherian module over itself. Indeed, one of the important points about commutative algebra that I learned from Atiyah-Macdonald is that given a ring $R$, we are interested in both ideals $I\subset R$ and quotient rings $R/I$ of $R$, and introducing the concept of an $R$-module - which $I$, $R$, and $R/I$ are all examples of - allow us to treat everything on a roughly equal footing.

I don't believe that rings satisfying the ascending chain condition for subrings, which we might call "subring-Noetherian", are studied or have nice properties, but I could be wrong about this.

I think it's fair to say that ideals are more important than subrings, but subrings are still integral (ha ha) for commutative algebra. You are correct that if we don't require our rings to have a multiplicative identity, ideals are subrings, i.e. "sub-rngs".

Answer 2

The "right" notion of a substructure of an algebraic gadget is the kernel of a homomorphism. For abelian groups, and more generally modules, these are subgroups, respectively submodules. For groups, we need normal subgroups. For rings, we need ideals.

Answer 3

Both ideals and subrings are important in algebra. But generally there is no close relationship between congruences on $\rm\,A\,$ and subalgebras of $\rm\,A.\,$ Instead, generally congruences are related to subalgebras of the square $\rm\,A^2,\,$ e.g. see here. Rings (and groups) are special in that their congruences are determined by a single congruence class - which has the effect of collapsing the relationship of congruences with subalgebras from $\rm\,A^2\,$ down to $\rm\,A.\,$

One way to better understand the importance of ideals is to study other algebras whose congruences are determined by a single congruence class - so-called ideal determined varieties. They are characterized by two properties of congruences, being $\,0$-regular and permutable, e.g. see below for an entry point into literature on such topics.

ON SUBTRACTIVE VARIETIES IV: DEFINABILITY OF PRINCIPAL IDEALS

Paolo Agliano and Aldo Ursini

$0.\ \ $ Foreword

We have been asked the following questions:

(a) $\ $ What are ideals in universal algebra good for?
(b) $\ $ What are subtractive varieties good for?
(c) $\ $ Is there a reason to study definability of principal ideals?

Being in the middle of a project in subtractive varieties, this seems the right place to address them.

To (a). The notion of ideal in general algebra [13], [17], [22] aims at recapturing some essential properties of the congruence classes of $0$, for some given constant $0$. It encompasses: normal subgroups, ideals in rings or operator groups, filters in Boolean or Heyting algebras, ideals in Banach algebra, in l-groups and in many more classical settings. In a sense it is a luxury, if one is satisfied with the notion of "congruence class of $0$". Thus in part this question might become: Why ideals in rings? Why normal subgroups in groups? Why filters in Boolean algebras?, and many more. We do not feel like attempting any answer to those questions. In another sense, question (a) suggests similar questions: What are subalgebras in universal algebra good for? and many more. Possibly, the whole enterprise called "universal algebra" is there to answer such questions?

Having said that, it is clear that the most proper setting for a theory of ideals is that of ideal determined classes (namely, when mapping a congruence E to its $0$-class $\,0/E\,$ establishes a lattice isomorphism between the congruence lattice and the ideal lattice). The first paper in this direction [22] bore that in its title.

It comes out that -- for a variety $V$ -- being ideal determined is the conjunction of two independent features:

$1$. $V$ has $0$-regular congruences, namely for any congruences $E,E'$ of any member of $V,\,$ from $\,0/E = 0/E'$ it follows $E = E'$.

$2$. $V$ has $0$-permutable congruences, namely for any congruences $E,E'$ of any member of $V,\,$ if $\ 0\, E\, y\, E'\, x,\,$ then for some $z,\ 0\, E'\, z\, E\, x$.

Answer 4

This answer is (by intent) very similar to Zev Chonoles' answer, but I have tried to put it in terminology a bit more similar to that of the question.

When one wants to study a ring $R$, there are (at least) two avenues of approach. The more naive approach is to study $R$ as a ring structure. In this sense, you are correct: a substructure of $R$ is precisely a subring of $R$. However, while the study of ring structures is undoubtedly important, it turns out that it is even more useful to study structures of $R$-modules. [One way to obtain a language of $R$-modules is to take the language of abelian groups and add a 1-ary function symbol for every element of $R$.] When one considers $R$ as having the structure of an $R$-module rather than a ring, then its substructures are precisely its ideals.

Why is it more useful to study $R$-modules than rings? It's hard to give a precise answer to that question (especially since the study of ring structures, subrings, etc. is in fact very useful and important), but one way of thinking about this is that when you are looking at $R$-modules, there is a nice notion of "quotient." On the other hand, if you have $S$ as a subring of $R$, then in particular $1 \in S$; so if you try to take a "quotient ring" $R/S$, you have to set $1=0$ in the quotient ring, i.e., you get the zero ring.