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Results tagged with ra.rings-and-algebras
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user 15934
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.
1
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Accepted
Examples of reduced associative algebras
Let $Q$ be the quaternion group of order $8$ and let $P(Q)$ be the power set of $Q$ viewed as a monoid by $AB=\{ab\mid a\in A,b\in B\}$. Let $\mathbb QP(Q)$ be the contracted monoid algebra of $P(Q)$ …
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6
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rings in which every element is a sum of two commuting idempotents
The trivial/elementary proof (in particular, it does not use the axiom choice).
A ring $R$ satisfies your condition iff it satisfies the identity $x^3=x$.
Pf.
Will shows above with a short computatio …
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3
votes
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Standard method and name for extending a semiring to a ring
I think it is often called forming the Grothendieck ring. The construction is really defined for commutative monoids, where one calls it the Grothendieck group and then you carry the multiplication al …
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6
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Finite Dimensional Simple nonunital associative Algebras
Here is a direct argument avoiding adjoining a unit and explicit use of the Wedderburn theorem. In what follows left ideals and ideals are always subspaces. If $A^2=0$, then any subspace is an ideal. …
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3
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Are corner rings of (semi)perfect rings (semi)perfect?
Rowen shows in Lemma 2.7.34 of his book Ring Theory that $R$ is right perfect iff for each idempotent $e$ one has both $eRe$ and $(1-e)R(1-e)$ are right perfect. Hence $R$ right perfect implies $eRe$ …
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The tensor product of two path algebras
What you want is
here.
Basically you take the 1-skeleton of the products of the quivers and you say that certain paths e.g. (edge,vertex)(vertex,edge)=(vertex,edge)(edge,vertex).
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14
votes
Accepted
Non-isomorphic algebras becoming isomorphic after adding identities: mistake in an exercise
The counterexample is correct. I had put the same counterexample on mathoverflow in a comment last year, unaware of the above paper, on a question asking about rings not isomorphic to their opposite. …
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7
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Infinite dimensional simple algebras
There are lots of examples among Leavitt path algebras. The easiest is the Leavitt algebra generated by x,x',y,y' and relations x'x=1=y'y and x'y=0=y'x and xx'+yy'=1. See https://arxiv.org/abs/math/0 …
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9
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Consequences of not requiring ring homomorphisms to be unital?
There are natural situations where non-unit preserving homomorphisms are needed. For instance if $R$ is a unital ring and $e$ an idempotent, then $eRe$ is a unital ring with identity $e$ and the incl …
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5
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In modules over finite rings $Rv = Rw \iff R^\times v = R^\times w$
Let $r,s\in R$ with $rv=w$ and $sw=v$. Since $R$ is finite, there exists $n>0$ such that $f=(rs)^n$ and $e=(sr)^n$ are idempotent. Note that $ev=v$ and $fw=w$. Let $r'=fre$ and $s'=esf$. Notice tha …
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An isomorphic invariant in ring theory
If you have a finite ring $R$ that is von Neumann regular, then $[0]=N(R)$ or more generally if you have a von Neumann regular ring $R$ such that each element has a power that belongs to a subsemigrou …
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Can a semigroup with zero be globally isomorphic to a semigroup without zero?
This is a little too long for a comment.
Here is an idea to prove the minimal ideal can be recovered. Let $G$ be the minimal ideal of $S$ and let $e$ be its identity; it is a central idempotent. Th …
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7
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1
answer
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Looking for citable reference for a well-known fact about tensor product of finite dimension...
Let $K$ be an algebraically closed field and let $A$ and $B$ be finite dimensional algebras over $K$. Let $e_1,\ldots, e_n$ be orthogonal primitive idempotents of $A$ summing to $1$ and $f_1,\ldots, …
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Cancellable elements of a power semigroup
Here is an example of a cancellable set with a non cancellable element. Take the semigroup with presentation $S=\langle a,b,c\mid ab=ac, ba=ca\rangle$
One checks that $ac\to ab$ and $ca\to ba$ is a …
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3
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A BF-monoid $H$ s.t. $H^\times$ is not divisor-closed
Your monoid cannot be atomic if $H^\times$ is not divisor-closed.
Suppose $xy$ is a unit. Assume $x$ is not a unit. The other case is similar. Then $xyz=1$ for some $z$. Let $a$ be an atom. Then $a= …
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