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Results tagged with ra.rings-and-algebras
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user 84349
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.
7
votes
When does a ring surjection imply a surjection of the group of units?
Addendum I. I am raising this addendum at the top because it is more relevant to the question than what I have written before.
We denote by $R^{\times}$ the unit group of a unital ring $R$.
Claim 1. …
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15
votes
Accepted
Finite non-commutative ring with few invertible (unit) elements
This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity must be a Boolean ring, hence commutative. The proof given bel …
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9
votes
Accepted
Principal ideal ring, does there exist an invertible matrix such that certain matrix is uppe...
The answer to your initial question is yes, and Igor Rivin's comment is right and good. The result follows from the fact that a commutative PIR (Principal Ideal Ring) with identity is a Hermite ring i …
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3
votes
Ring with vanishing $K_0$
This answer only adds more references and remarks, serving as a complement to the accepted answer of Steven Landsburg.
Rings are supposed to be associative and unital, but not necessarily commutative. …
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4
votes
Accepted
For a finite-type $\mathbb{Z}$-algebra $A$, is the intersection of all ideals $I$ such that ...
A proof of the proposed result that is similar (if not identical) to Peter Kropholler's and YCor's proofs can be derived from two well-known results, namely Lemma 2 and Theorem 3 below, together with …
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19
votes
Accepted
When is $A\otimes R$ a free $R$-module?
Here is a 7-line proof of your statement.
Claim. Let $R$ be a commutative ring with identity $1_R$. Let $A \simeq \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z}$ be a …
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4
votes
Accepted
Finite distributive lattices as lattice of ideals of a finite ring
The answer is yes, there is a finite distributive lattice which is not isomorphic to the lattice
of right ideals in a non-commutative ring with identity,
of ideals in a commutative ring with identit …
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2
votes
A property similar to arithmetical property
Let us not leave this question as unanswered:
A ring $R$ has the $X$-property if and only if $R$ is arithmetical.
The proof is straightforward.
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6
votes
Localization and intersection
With a view to understand to which extent OP's identity can fail, I am sharing these easy positive results. The emphasis is on the fact that OP's question tightly relates to (generalizations of) the p …
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8
votes
For which rings $R$ is $\mathrm{SL}_n(R)$ generated by transvections?
The goal of my answer is only to provide recent references.
I warmly recommend these two bits of T. Y Lam's book [2]:
§I.8, for examples where transvections fail to generate $SL_n(R)$
the second to l …
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6
votes
Accepted
Modules over infinite rings which can not be a finite union of their proper submodules
The two questions can be answered in the positive.
Claim 1. Let $R$ be an infinite division ring. Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.
…
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3
votes
Every non-zero submodule of $R_R$ has an indecomposable direct summand: True when $R$ is von...
No, a free Boolean algebra $R$ on an infinite cardinal $\kappa$ (e.g., if $\kappa = \aleph_0$, $R$ is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as a …
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5
votes
Accepted
Non-associative commutative "group"
The answer to the first question is yes:
Claim. Let $a, b \in \{0, 1 \}^{\ast}$ and let $n \ge 0$ be the least integer such that $a(i) = b(i) = 0$ for every $i > n$. Then the equation $$a +_2 x = b$$ …
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10
votes
Bass' stable range of $\mathbf Z[X]$
The goal of my answer is to elaborate on steps (1) and (2) of the accepted answer of Steven Landsburg. In particular, I would like to make clear how a "small enough ideal" looks like and how to produc …
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3
votes
Any ideal as an intersection of ideals primary to maximal ones
The answer is yes for any excellent Hilbert ring, hence for $\mathbb{C}[x_1, \dots, x_n]$.
This follows from:
Eisenbud and Hochster's Theorem [Corollary 2, 1]. Let $A$ be a ring finitely generated ov …
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