## reference request – Is there any structural characterization of the rings in which every element other than the identity is a (two-sided) zero divisor?

(I fear that I’m missing something obvious here, but I’ll dare to ask anyway.)

A division ring is a (unital, associative, non-zero) ring where every non-zero element is a unit. So, let an anti-division ring be a ring where every element other than the (multiplicative) identity is a zero divisor. Is there any classification whatsoever of anti-division rings?

There are many simple things one can actually say about these rings: They are closed under direct products; they all have characteristic $$2$$; every boolean ring is a member of the family; any such ring is, in fact, a subdirect product of domains (see Jose Brox’s answer to a related question on SME); the only nilpotent element in such a ring is zero; if a ring in the family is semilocal, then it’s a direct product of copies of the field $$F_2$$ with two elements (see znc’s answer from the same thread of Jose Brox’s); etc.

Question. Is it perhaps the case that a more precise characterization (say, than the one provided by Jose Brox for rings whose non-units are all zero divisors) is indeed possible?

If so, it’s clear to me that this must be very well known (at least in some large circles) and I would much appreciate it if you could offer a reference.

## ra.rings and algebras – Are corner rings of (semi)perfect rings (semi)perfect?

Part of this question (asked by someone else) for semiperfect rings has circulated a few weeks on math.se but without much attention. I think it might be above the threshold of difficulty to be on mathoverflow, but you can let me know if I should move it.

Let $$R$$ be a ring and $$e$$ be an idempotent element of $$R$$ (that is, $$e^2=e$$).

1. If $$R$$ is semiperfect, is $$eRe$$ necessarily semiperfect?

2. If $$R$$ is right perfect, is $$eRe$$ necessarily right perfect?

This wiki page has the definitions, if you are not familiar.. “$$eRe$$” is the so-called corner ring.

I was a bit surprised I could not turn up the answer for this quickly in the literature or through my own computations.

I’m familiar with the characterization of both conditions via projective covers, and also for the characterization of semiperfect rings using the existence of a finite decomposition of $$1$$ into orthogonal local idempotents, and the characterization of right perfect rings as satisfying the DCC on left principal ideals.

But they don’t seem to avail me since I don’t understand the relationships of $$eRe$$ modules to $$R$$ modules. I am not even very clear on what can be said about the relationship of right ideals of $$R$$ and $$eRe$$.

## ra.rings and algebras – rings isomorphism

ra.rings and algebras – rings isomorphism – MathOverflow

## nt.number theory – Simultaneous embeddings of ring of integers into product of rings

This question is about something mentioned in Katz’s “$$p$$-adic $$L$$-functions for CM fields” in section 2.0.

Let $$K$$ be a number field with $$(K:mathbb{Q}) = d$$ and ring of integers $$mathcal{O}_K$$. Let $$Sigma$$ be the set of embeddings $$sigma colon K to overline{mathbb{Q}}$$, and $$K^{gal}$$ the compositum of the fields $$sigma(K)$$ over all $$sigma in Sigma$$, the Galois closure of $$K$$. Then let $$mathcal{O}^{gal}$$ be the ring of integers in $$K^{gal}$$, and $$R$$ an algebra over $$mathcal{O}^{gal}$$. For each $$sigma in Sigma$$, we have a ring homomorphism $$sigma colon mathcal{O}_K otimes R to R$$ given on pure tensors by $$n otimes r mapsto sigma(n)r$$. We can take them all together to produce

$$mathcal{O}_K otimes R to prod_{sigma in Sigma} R, qquad n otimes r mapsto (sigma(n)r)_sigma.$$

Katz then states that this is an isomorphism if the discriminant of $$K$$ is invertible in $$R$$.

I’m trying to figure out what the image of this map is when the discriminant is not invertible. If $$R$$ is flat over $$mathbb{Z}$$, it should be injective, just because each $$sigma$$ is injective. I thought I had found an obstruction to surjectivity using the homomorphism of Abelian groups
$$prod_sigma R to R, qquad T((r_sigma)_sigma) = sum_sigma r_sigma$$
where the composition $$mathcal{O}_K otimes R to prod_sigma R to R$$ is $$n otimes r mapsto operatorname{Tr}_{mathcal{O}_K/mathbb{Z}}(n)r$$. The image is then the subgroup of $$R$$ generated by traces of elements of $$mathcal{O}_K$$. However, the subgroup of $$mathbb{Z}$$ generated by traces of elements of $$mathcal{O}_K$$ is not the different ideal as I had assumed (e.g. $$Tr(1) = d$$ is there regardless of whether any primes dividing $$d$$ ramify), and so it doesn’t have as tight a connection to ramification in $$K$$ as I hoped.

Is there a nice description of the image of this map in general or in certain cases? I am most interested in the case when $$R$$ is the ring of integers in a finite extension of $$mathbb{Q}_p$$.

## matrices – Can all finite dimensional non commutative algebras be embedded into matrix rings?

Suppose I have a finite (non-)commutative ring $$R/k$$ (over a field $$k$$ of char $$0$$) with a linear “trace” function $$t: R to k$$. Can I find square matrices $$A_1,dots,A_n$$ (of some dimension $$r$$) so that I have an embedding $$f: R to M_r(k)$$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $$R$$ is that it should be invariant under cyclic permutations : $$t(a_1a_2dots a_n) = t(a_2dots a_na_1)$$. Is this the only restriction?

## gr.group theory – How to classify rings by combinatorial structures?

There are many ways to encode information about algebraic structures such as groups, rings, etc… in combinatorial form. For example the Cayley graph of a group with a subset of generators, or the various graphs associated to rings, as can be found in, e.g., the answers to
Why do we associate a graph to a ring?. So I was wondering about the converse questions, which for groups and rings take the form:

First question(s): Given $$X$$ a graph is there a way to discover, intrinsically, constructively and algorithmically, whether it is the Cayley graph of a group? How to recover the group structure from the graph? Is there a unique group $$G$$ such that $$X = CG(G)$$, the Cayley graph of $$G$$? How to find such a group $$G$$? Which graphs are the Cayley Graphs of some Group?

Second Question: Is there a combinatorial structure (such as a system of graphs) associated to rings (or algebra or module of an algebra) from which you can recover the full ring (or module) in a similar manner as in the first question? Preferably in an intrinsic, constructive and algorithmic way. Assuming one could find such a combinatorial category, how to find out which objects in it are the objects associated to rings (or modules)?

I would also be interested in considering similar types of questions for general well-known algebraic structures (some kind of combinatorial informational encoding for these algebraic structures) in the sense that you can define precisely combinatorial structures out of algebraic structures, intrinsically constructively and algorithmically, but from which you can recover the original structure, also intrinsically constructively and algorithmically and intrinsically.

For groups, there is positive answer given by Sabidussi’s theorem, as mentioned in https://en.wikipedia.org/wiki/Cayley_graph#Characterization, which characterizes graphs which are Cayley Graphs of groups. This theorem would suffice in terms of instrisic, constructive and algorithmic profile of the proof, for question 1.

I would be satisfied with partial answers.

## ag.algebraic geometry – Local rings \$R subsetneq S\$ with \$R\$ regular and \$S\$ Cohen-Macaulay, non-regular

Let $$R subseteq S$$ be local rings with maximal ideals $$m_R$$ and $$m_S$$.
Assume that:

(1) $$R$$ and $$S$$ are (Noetherian) integral domains.

(2) $$dim(R)=dim(S) < infty$$, where $$dim$$ is the Krull dimension.

(3) $$R$$ is regular (hence a UFD).

(4) $$S$$ is Cohen-Macaulay.

(5) $$R subseteq S$$ is simple, namely, $$S=R(w)$$ for some $$w in S$$.

(6) $$R subseteq S$$ is free.

(7) $$R subseteq S$$ is integral, namely, every $$s in S$$ satisfies a monic polynomial over $$R$$.

(8) $$m_RS=m_S$$, namely, the extension of $$m_R$$ to $$S$$ is $$m_S$$.

(9) It is not known whether the fields of fractions of $$R$$ and $$S$$, $$Q(R)$$ and $$Q(S)$$, are equal or not.

(10) It is not known if $$R subseteq S$$ is separable or not.

Remark: It is known that if a (commutative) integral domains ring extension $$A subseteq B$$ is integral+flat, then it is faithfully flat, and if also $$Q(A)=Q(B)$$, then $$A=B$$.
This is why I did not want to assume that $$Q(R)=Q(S)$$, since in this case $$R=S$$ immediately.

Question:
Is it true that, assuming (1)(10) imply that $$S$$ is regular or $$R=S$$?

Example:
$$R=mathbb{C}(x(x-1))_{x(x-1)}$$ and $$S=mathbb{C}(x)_{(x)}$$,
with $$R neq S$$ and $$S$$ is regular.

Non-example: $$R=mathbb{C}(x^2)_{(x^2)}$$ and $$S=mathbb{C}(x^2,x^3)_{(x^2,x^3)}$$, but condition (8) is not satisfied.

Relevant questions, for example: a, b, c, d.

Thank you very much! I have asked the above question here, with no comments (yet).

## When are rings of the form \$K[x_1,…,x_n]/(Q)\$ principal ideal domains when \$Q\$ is a quadratic form?

By a result of Klein-Nagata rings of the form $$A_Q=K(x_1,…,x_n)/(Q)$$ are factorial when $$K$$ is a field, $$n geq 5$$ and $$Q$$ is a non-degenerate quadratic form.

Question: When is $$A_Q$$ a principal ideal domain or even an euclidean ring?

## ac.commutative algebra – Rings with terminating division chains of a given length

Let $$R$$ be an integral domain. Given $$a,bin R$$, then a division chain for $$(a,b)$$ is a sequence where we take $$r_{-1}=a$$, $$r_0=b$$, and for each $$n>0$$ we take $$r_n=r_{n-1}s_n+r_{n-2}$$ for some $$s_nin R$$. We say that the division chain terminates if $$r_n=0$$, and it terminates at length $$n$$ when $$r_ineq 0$$ for $$i.

These concepts, of course, have a lot to do with Euclidean domains, etc…

I ran across the following very interesting fact, proved by Cooke and Weinberger in 1975. Let $$K$$ be a number field, and let $$R=mathscr{O}_{K,S}$$, the ring of $$S$$-integers, where $$S$$ contains all the infinite places of $$K$$. Assume (some appropriate version of) GRH. If the unit group of $$R$$ is infinite, and $$aR+bR=R$$, then there is a terminating division chain for the pair $$(a,b)$$ of length $$5$$. Under some additional restrictions, like $$S$$ has a non-infinite place, or $$K$$ has a real embedding, the number $$5$$ can be lowered to $$4$$ or $$3$$. Moreover, they give examples showing that these numbers are best possible in some cases.

I’m interested if the following appears anywhere in the literature: For each $$n>5$$, there exists an integral domain $$R$$ such that for any $$a,bin R$$ with $$aR+bR=R$$, there is a terminating division chain for the pair $$(a,b)$$ of length $$n$$; and there is some pair of comaximal elements in $$R$$ that doesn’t have a smaller terminating length. The example will have to be somewhat complicated, since it won’t be a ring of integers over a number field.

## ag.algebraic geometry – Local rings \$R subseteq S\$ with \$m_RS=m_S\$

Let $$R subseteq S$$ be two Noetherian local rings, not necessarily regular, which are integral domains,
with $$m_RS=m_S$$, namely, the ideal in $$S$$ generated by $$m_R$$ (= the maximal ideal of $$R$$) is $$m_S$$ (= the maximal ideal of $$S$$).

Further assume that $$R$$ and $$S$$ are $$mathbb{C}$$-algebras, $$R subseteq S$$ is flat and algebraic but not integral,
where algebraic non-integral means: Every element of $$S$$ satisfies a polynomial with coefficients in $$R$$, with non-invertible leading coefficient.

Could one find an example of such rings?

Unfortunately, the examples I find are integral, for example:
$$R=mathbb{C}(x(x-1))_{(x(x-1))}$$, $$S=mathbb{C}(x)_{(x)}$$.

Remarks:

(i) I am interested in both cases where $$R$$ and $$S$$ have the same fields of fractions or different fields of fractions.

(ii) Recall the following results, which are not applicable here, since I assume that $$R subseteq S$$ is non-integral:
If $$A subseteq B$$ is integral and flat, then $$A subseteq B$$ is faithfully flat, and if in addition $$Q(A)=Q(B)$$ (same fields of fractions), then $$A=B$$.

Relevant questions: a, b and c.