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1 vote

A result of Schofield in the case of quivers with relations

The question doesn't seem to be a generalization of Schofield's result (which doesn't involve fixing a brick $B$), and seems to be false already for quivers without relations. Take $Q$ to be the ...
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1 vote
Accepted

Automorphisms of special egg-box diagrams

It turns out (surprisingly) that the answer to my question is yes, and there are even finite examples. I came up with the following diagram $$ \begin{array}{|c|c|c|c|c|c|} \hline \circ & \circ &...
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3 votes

Subrings, submodules, and flatness

No it need not be an isomorphism. Something more than flatness is required. For example, taking $M$ and $N$ to be copies of the ring $R$ you are asking for the multiplication $R\otimes_SR\to R$ to be ...
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11 votes

Is any finite-dimensional algebra a sub-algebra of a finite-group algebra?

Assuming that by "sub-algebra" you mean "unital sub-algebra": Every group algebra has a one-dimensional module (the trivial module), so any subalgebra has a one-dimensional module. ...
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6 votes
Accepted

Distributive lattice of subspaces

Just to not leave this open, a proof can be found in Proposition 7.1 of Chapter 1 of Quadratic Algebras by Alexander Polishchuk and Leonid Positselski as alluded to by Mariano on MSE https://math....
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5 votes

An example of a local integral domain with special spectrum

Put $D_0 = \mathbb{Q}[T,X_1,X_2,\cdots]/I_0$, where $I_0$ is generated by the elements $X_i^2-TX_{i+1}$, for every $i \in \mathbb{N}$. This is a domain. Let $\mathfrak{m}_0$ $=$ $(T,X_1,X_2,\cdots)$, ...
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7 votes
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An example of a local integral domain with special spectrum

Let $k$ be a field. Put $R=k[[x,y]]$ and let $D\subset R$ be the subring $k+xR$. It consists of power series without any term $ay^n$ ($a\in k^\times$, $n>0$), or (equivalently) series $f(x,y)$ such ...
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10 votes
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What is the centralizer of a Young subgroup of $S_n$?

This answer is largely inspired by the wonderful paper of Samuel Creedon, The Farahat-Higman Algebra of Centralizers of Symmetric Group Algebras, which studies in detail the case of $\mathbb{C}S_n^{S_{...
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0 votes

Is there any non-commutative ring such that every element other than the identity is a zero divisor?

Time for a second attempt, here's hoping for the best. Algebras will be strictly non-unital, i.e. they have no global identity element. Claim 1: Let $R$ be a $\mathbb{Z}/2$-algebra with the EV ...
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