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I'm currently working my way through the foundations of ring theory (Dummit & Foote), and I am somewhat confused on how to think about the notion of a generating set for a ring.

For a group $(\mathcal{G},+,-,0)$ the notion is clear; a subset $G\subseteq\mathcal{G}$ is a generating set for $\mathcal{G}$ iff every element of $\mathcal{G}$ can be expressed as a finite composition of the members of $G$ under addition and negation.

This notion still exists canonically in a ring $(\mathcal{R},+,-,\times,0,1)$, however it seems that we may want to consider a second notion as well which specifically leverages the existence of multiplication when trying to simplify the structure of a ring. Specifically, it seems like we can define a generating set for $\mathcal{R}$ as a subset $R\subseteq\mathcal{R}$ such that every element of $\mathcal{R}$ can be expressed as a finite composition of members of $R$ under addition, multiplication and negation. My question is this:

Is the latter definition of a generating set for a ring well-known, and if so where can I find some literature on it?

In general, the second definition yields a smaller set in order type or sometimes cardinality, and seems like it should still contain much of the information we care about for generating sets. This is easier to see when considering an example:

Let $n$ be a natural number, and consider the Grothendieck ring of the limit ordinal $\omega^{\omega^n}$ under natural addition and multiplication, denoted $\mathfrak{G}(\omega^{\omega^n})$, as constructed here [if we set $n=0$ we obtain $\mathfrak{G}(\omega)=\mathbb{Z}$]. Then $$G=\{\omega^\beta\}_{\beta<\omega^n}$$ is a countable generating set for $\mathfrak{G}(\omega^{\omega^n})$ if we only use the group operations $+$ and $-$, however $$R=\{1\}\cup\{\omega^{\omega^\beta}\}_{\beta<n}$$ is a finite generating set for $\mathfrak{G}(\omega^{\omega^n})$ if we allow the usage of $\times$. In the simplest non-trivial case where $n=1$, we have $$G=\{\omega^m\}_{m<\omega},$$ $$R=\{1,\omega\}$$ as the generating sets for $\mathfrak{G}(\omega^\omega)$. When $n=2$ we have $$G=\{\omega^m\}_{m<\omega^2},$$ $$R=\{1,\omega,\omega^\omega\},$$ etc. In each situation it seems that $R$ is much simpler than $G$ and obviously recaptures its members under multiplication, since $$\omega^n\times\omega^m=\omega^{n+m},$$ $$\omega^{m\omega}\times\omega^{n\omega}=\omega^{n\omega+m\omega}=\omega^{(n+m)\omega},$$ and in general the algebraic expressions factor in the expected fashion. This seems to imply that, so long as we know that we have access to a multiplicative structure to work with, the second notion of a generating set is easier to work with than the first.

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    $\begingroup$ I've never seen a usage of "the subring generated by" other than the one you describe. In general, in any algebraic structure, the substructure generated by a subset is the smallest substructure containing that subset, which will happen to be the set of all elements generated using the subset's elements and any of the operations. | Also, I don't get why you think the subring generated by a subset will be smaller than the additive subgroup generated by it. Surely the subring generated will strictly contain the additive subgroup generated? $\endgroup$
    – anon
    Commented Jan 6, 2018 at 2:02
  • $\begingroup$ @anon Thanks for the clarification, and my apologies; I meant to say that $R$ is smaller in cardinality/order type than $G$ as a generating set, not that the objects they generate differ in cardinality/order type. I believe we could actually generate many different groups of a smaller order type than $\mathfrak{G}(\omega^\omega)$ by stopping at $\gamma$-numbers between $\omega$ and $\omega^\omega$ in the generating set. $\endgroup$
    – Alec Rhea
    Commented Jan 6, 2018 at 3:47

2 Answers 2

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This is a nice question, and one I had myself at one point. As anon notes in the comments, your description of the subring generated by a subset of a particular ring $R$ is one I've seen. Perhaps the following reformulation of things may help.

If we think about your description of elements of the subring generated by some subset $S$ of a ring $R$, what sorts of combinations can we get? We can multiply powers of elements by each other, and add or subtract integer multiples of them from each other. Hence, what we're really describing are integer polynomials in the elements of $S$. Perhaps more precisely, we could make the following definition: Let $S \subset R$. Then we say that the subring generated by $S$ is the image of the ring homomorphism $\mathbb{Z}[\{X_{s}\}_{s \in S}] \to R$ which sends $X_{s}$ to $s$, where $\mathbb{Z}[\{X_{s}\}_{s \in S}]$ denotes the polynomial ring with coefficients in $\mathbb{Z}$ whose indeterminates are indexed by the elements of $S$. We could further say that $R$ is generated by a subset $S$ if this morphism is surjective.

This admits a kind of nice categorical interpretation, if you're into that sort of thing. One can equivalently observe that if $G$ is a group and $S$ is a subset of $G$, the subgroup of $G$ generated by $S$ is the image of the canonical morphism $F(S) \to G$ induced by the set-theoretic inclusion $S \to G$, where $F(S)$ denotes the free group on the alphabet $S$. In both cases, we are looking at the image of the right morphism out of the free object in that category on our generating set; in the category of rings, the polynomial algebra (in however many indeterminates) plays the role of the free object. I imagine that the notion of "object generated by a subset" can be given a more categorical interpretation, though I'm a neophyte in these matters. (In particular, in the language I use above, I am restricting my attention to sets with additional algebraic structure.) I would welcome comments from those more knowledgeable about what's really going on here.

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  • $\begingroup$ Thanks for this; I'm more of a model than category theorist, but this is a nice way to view the matter at hand. $\endgroup$
    – Alec Rhea
    Commented Jan 6, 2018 at 3:51
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As others have said, this is indeed the standard definition of what it means to generate a ring, and it generalizes to other kinds of sets equipped with operations such as vector spaces and modules (where now the operations include multiplication by a fixed scalar).

There is an interesting subtlety here. Namely, you can imagine defining the subring generated by a subset $S$ of a ring $R$ via the following procedure:

  1. First take the closure under addition.
  2. Then take the closure under multiplication.
  3. Then take the closure under addition again.
  4. Then take the closure under multiplication again.

And so on. Now there's a question of at what point this construction terminates: a priori constructions of this form need not terminate at any finite step, and we may need to continue them transfinitely. However, in fact it terminates at step 3, and also step 1 is redundant. That is, it suffices to

  1. First take the closure under multiplication.
  2. Then take the closure under addition.

The result is now closed under both addition and multiplication. To show that the result of this procedure is closed under multiplication we need to make crucial use of the distributivity of multiplication over addition.

By contrast, if we drop distributivity then this construction needs to be repeated $\omega$ times!

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  • $\begingroup$ May I ask, in the above process, how one get to inverses for units? Maybe this is silly, but is it possible you could answer this question: math.stackexchange.com/questions/4916590/…? Thanks a lot $\endgroup$
    – user760
    Commented May 15 at 6:09

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