# Questions tagged [ra.rings-and-algebras]

Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.

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### Is $\mathbb{Q}_p \otimes_{\mathbb{Q}}\mathbb{Q}_p$ coherent?

Let $\mathbb{Q}_p$ denote the field of fractions of $\mathbb{Z}_p$. By the answers to this quesition the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ cannot be a Noetherian ring (...
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### Behavior of invariants under reduction mod p

Let $R$ be a finitely generated $\mathbb{Z}$-algebra with an [edit: linear algebraic] action of $G(\mathbb{Z})$ where $G$ is a split simply-connected semisimple group. Then for any prime $p$ we have a ...
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### Category of modules over an Azumaya algebra and the Brauer group

Let $k$ be a field, and let $\alpha \in \mathrm{Br}(k)$. Let $A$ be an Azumaya algebra representing $\alpha$. Then the category $A$–$\mathrm{mod}$ depends only on $\alpha$. I would like to know ...
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### When the sum of two ideals is indecomposable

I am looking for a commutative ring $R$ and two ideals $I$ and $J$ of $R$ and two different maximal ideals $m_1$ and $m_2$ of $R$ such that $ann(I)=m_1$ and $ann(J)=m_2$ and $I+J$ is an ...
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### Global splitting field for algebras

Let $A$ be a finite dimensional algebra. A field $K$ is a splitting field for an indecomposable $A$-module $M$ in case the local algebra $End_A(M)/(rad(End_A(M))$ is 1-dimensional. $K$ is called a ...
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### Decreasing sequences in a finitely generated closure algebra

I am interested in finitely generated closure algebras (as a special case of Heyting algebras), and in decreasing sequences of elements within such an algebra that have no lower bound. Call two ...
I am copying my question from here: https://math.stackexchange.com/q/3233462/427611. Is it correct that $\mathbb Z/3\mathbb Z$ and $\mathbb Z/4\mathbb Z$ are the only rings with three or more ...