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We introduce a similarity relation between submodules of a module $M$ over a ring $R$, extending the classical notion of similarity for right ideals. Focusing on (faithfully) projective modules, we establish a sharp lower bound for the…

Rings and Algebras · Mathematics 2026-04-07 Alborz Azarang

We study endomorphisms of a free group of finite rank by means of their action on specific sets of elements. In particular, we prove that every endomorphism of the free group of rank 2 which preserves an automorphic orbit (i.e., acts ``like…

Group Theory · Mathematics 2008-02-03 Vladimir Shpilrain

Let K be the kernel of an epimorphism G -> Z, where G is a finitely presented group. If K has infinitely many subgroups of index 2, 3, or 4, then it has uncountably many. Moreover, if K is the commutator subgroup of a classical knot group…

Geometric Topology · Mathematics 2007-05-23 Daniel S. Silver , Susan G. Williams

We establish new results concerning endomorphisms of a finite chain if the cardinality of the image of such endomorphism is no more than some fixed number k. The semiring of all such endomorphisms can be seen as a k - simplex whose vertices…

Rings and Algebras · Mathematics 2013-05-01 Ivan Dimitrov Trendafilov

Let $R$ be a local ring and let $M$ be a finitely generated $R$-module. Appealing to the natural left module structure of $M$ over its endomorphism ring and corresponding center $Z(\operatorname{End}_R(M))$, we study when various…

Commutative Algebra · Mathematics 2025-10-06 Souvik Dey , Justin Lyle

We describe the automorphism group of the endomorphism semigroup $\End(K[x_1,...,x_n])$ of ring $K[x_1,...,x_n]$ of polynomials over an {\it arbitrary} field $K$. A similar result is obtained for automorphism group of the category of…

Rings and Algebras · Mathematics 2017-12-05 A. Belov-Kanel , R. Lipyanski

We study the problem of generating the endomorphism ring of a supersingular elliptic curve by two cycles in $\ell$-isogeny graphs. We prove a necessary and sufficient condition for the two endomorphisms corresponding to two cycles to be…

Given a ring homomorphism $B \to A$, consider its centralizer $R = A^B$, bimodule endomorphism ring $S = \End {}_BA_B$ and sub-tensor-square ring $T = (A \o_B A)^B$. Nonassociative tensoring by the cyclic modules $R_T$ or ${}_SR$ leads to…

Rings and Algebras · Mathematics 2007-05-23 Lars Kadison

Let $A=A(x_{1},...,x_{n})$ be a free associative algebra in $\mathcal{A}$ freely generated over $K$ by a set $X=\{x_{1},...,x_{n}\}$, $End A$ be the semigroup of endomorphisms of $A$, and $Aut End A$ be the group of automorphisms of the…

Rings and Algebras · Mathematics 2017-12-05 A. Kanel-Belov , A. Berzins , R. Lipyanski

Let $V$ be a vector space with countable dimension over a field, and let $u$ be an endomorphism of it which is locally finite, i.e. $(u^k(x))_{k \geq 0}$ is linearly dependent for all $x$ in $V$. We give several necessary and sufficient…

Rings and Algebras · Mathematics 2021-07-12 Clément de Seguins Pazzis

The endomorphism ring End(A) of an abelian variety A is an order in a semi-simple algebra over Q. The co-index of End(A) is the index to a maximal order containing it. We show that for abelian varieties of fixed dimension over any…

Number Theory · Mathematics 2014-07-03 Chia-Fu Yu

This paper studies the set of $n\times n$ matrices for which all row and column sums equal zero. By representing these matrices in a lower dimensional space, it is shown that this set is closed under addition and multiplication, and…

Rings and Algebras · Mathematics 2008-10-02 Samuel N. Cohen , Robert J. Elliott , Charles E. M. Pearce

Let S be a subring of the ring R. We investigate the question of whether S intersected by U(R) is equal to U(S) holds for the units. In many situations our answer is positive. There is a special emphasis on the case when R is a full matrix…

Rings and Algebras · Mathematics 2007-07-04 Jeno Szigeti , Leon van Wyk

The authors [3] proved that the endomorphism semiring of a nontrivial semilattice is always subdirectly irreducible and described its monolith. Here we prove that the endomorphism semiring of a commutative inverse semigroup with at least…

Rings and Algebras · Mathematics 2020-09-18 M. K. Sen , S. K. Maity , Sumanta Das

For every semilattice $\mathcal{A}=(A,+)$, the set $\mathrm{End}(\mathcal{A})$ of its endomorphisms forms a semiring under pointwise addition and composition. We prove that that if $\mathcal{A}$ is finite, then the endomorphism semiring…

Rings and Algebras · Mathematics 2026-03-10 Igor Dolinka , Sergey V. Gusev , Mikhail V. Volkov

In [2] Su Gao proves that the following are equivalent for a countable $M$ (cf. theorem 1.2 too): (I)There is an uncountable model of the Scott sentence of $M$. (II) There exists some $j\in \overline{Aut(M)}\setminus Aut(M)$, where…

Logic · Mathematics 2015-06-09 Ioannis Souldatos

We calculate the rank and idempotent rank of the semigroup $E(X,P)$ generated by the idempotents of the semigroup $T(X,P)$, which consists of all transformations of the finite set $X$ preserving a non-uniform partition $P$. We also classify…

Group Theory · Mathematics 2017-12-14 Igor Dolinka , James East , James D. Mitchell

We use pullbacks of rings to realize the submonoids $M$ of $(\N_0\cup\{\infty\})^k$ which are the set of solutions of a finite system of linear diophantine inequalities as the monoid of isomorphism classes of countably generated projective…

Rings and Algebras · Mathematics 2011-05-19 Dolors Herbera , Pavel Prihoda

Let $\widehat{\mathbb{F}\mathbb{S}et}$ be the groupoid of finite sets and bijections between them equipped with the canonical symmetric rig category structure given by the disjoint union and the cartesian product of finite sets. We prove…

Category Theory · Mathematics 2020-04-21 Josep Elgueta

For every semilattice $\mathcal{S}=(S,+)$, the set $\mathrm{End}(\mathcal{S})$ of its endomorphisms forms a semiring under point-wise addition and composition. We prove that the semiring of all endomorphisms of the 3-element chain has no…

Group Theory · Mathematics 2026-05-05 Sergey V. Gusev , Mikhail V. Volkov
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