# Questions tagged [ring-theory]

This tag is for questions about rings, which are a type of algebraic structure studied in abstract algebra and algebraic number theory.

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### Does every ring of integers sit inside a ring of integers that has a power basis?

Given a finite extension of the rationals, $K$, we know that $K=\mathbb{Q}[\alpha]$ by the primitive element theorem, so every $x \in K$ has the form $$x = a_0 + a_1 \alpha + \cdots + a_n \alpha^n,$$ ...
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### Classification of local Artin (commutative) rings which are finite over an algebraically closed field.

A result in deformation theory states that if every morphism $Y=\operatorname{Spec}(A)\rightarrow X$ where $A$ is a local Artin ring finite over $k$ can be extended to every $Y'\supset Y$ where $Y'$ ...
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### Semirings induced by symmetric monoidal categories with finite coproducts

A symmetric monoidal category with finite coproducts is by definition a symmetric monoidal category $(\mathcal{C},\otimes,1,\dotsc)$ such that the underlying category $\mathcal{C}$ has finite ...
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### What properties do the rings of infinite, upper-triangular matrices have?

I'm very familiar with the ring of $n\times n$ matrices over a field, the ring of $n\times n$ upper triangular matrices over a field, and the ring of infinite column-finite matrices over a field. But ...
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### What is the most general algebraic structure that a finite set has?

For an object $X$ in a category with finite products, define its endomorphism Lawvere theory to be the Lawvere theory generated by $X$: its $n$-ary operations are given by $\text{Hom}(X^n, X)$, and so ...
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### Is there an elementary way to prove that the algebraic integers are a Bézout domain?

Well, the title of my question says all, but let me give some context. Right now I'm writing some lecture notes on ring theory with a little of commutative algebra. I wrote a few results about ...
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### Can we characterize all infinite PID s whose group of units is singleton?

I am looking for a way to characterize all infinite PID s having exactly one unit i.e. invertible element ( finite PID s are not interesting , they are all fields ) . The only such example I know of ...
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### How to effectively compute fundamental units in rings?

Consider the ring of integers of $K=\mathbb{Q}(\sqrt 2,\sqrt 3)$. By Dirichlet's unit theorem the units of $\mathcal O_K$ have rank 3, so they are expressible as $\pm u_1^au_2^bu_3^c$ for suitable ...
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### Does there exist a non-field Noetherian domain whose field of fraction is the field of real numbers?

Does there exist a Noetherian domain (but not a field) whose field of fractions is the field of real numbers $\mathbb{R}$ ? Any help will be appreciated. Thanks
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### Applications of the Dedekind-Hasse criterion

It is a fact that an integral domain $R$ is a principal ideal domain if and only if there is a Dedekind-Hasse function $|R|\setminus\{0\}\xrightarrow{\ \ \delta\ \ }\mathbb{N}$ on $R$, i.e. a function ...
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### Relative chinese remainder theorem and the lattice of ideals

Let $R$ be a ring with two-sided ideals $I,J$. The proof of the "absolute" chinese remainder theorem revolves around the fact that if $I,J$ cover $R$ in the lattice of ideals, i.e $I+J=R$, then the ...
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### Classification of all subrings

Let $R$ be an integral domain whose underlying additive group is finitely generated free and whose field of fractions $K$ is a finite Galois extension of $\mathbb{Q}$. Is there a method of ...
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I was wondering if "rings" with noncommutative addition are studied at all? Of course, if a ring $R$ has a $1$, then for all $a, b\in R$, $a+a+b+b=(1+1)a+(1+1)b=(1+1)(a+b)=(a+b)+(a+b)=a+b+a+b$, from ...
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### Regular Noetherian local rings are integral domains - questions about the proof

I am reading a proof that if $(A,\mathfrak m)$ is a regular local ring, then $A$ is an integral domain. I put the major questions I'm worried about in bold, but there are a lot of little things I'm ...
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### An example of a compact multiplicatively unbounded ring

My teacher asked me to build an associative topological Hausdorff compact ring $R$ with $1$, which is multiplicatively unbounded. That means there is a neighborhood $U\ni 1$ such that $FU\not=R$ for ...
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