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For any compact set $K \subseteq \mathbb{R}^3$ we define a number $r(K)$ that is either a nonnegative integer or $\infty$. Intuitively, $r(K)$ provides some information on how wildly $K$ sits in $\mathbb{R}^3$. We show that attractors for…

Dynamical Systems · Mathematics 2014-06-23 J. J. Sánchez-Gabites

If, for a subset S of Z^k, we compare the conditions of being parametrizable (a) by a single k-tuple of polynomials with integer coefficients, (b) by a single k-tuple of integer-valued polynomials and, (c) by finitely many k-tuples of…

Number Theory · Mathematics 2011-06-29 Sophie Frisch

Let X, Y be nonsingular real algebraic sets. A map fi:X-->Y is said to be k-regulous, where k is a nonnegative integer, if it is of class C^k and the restriction of fi to some Zariski open dense subset of X is a regular map. Assuming that Y…

Algebraic Geometry · Mathematics 2023-02-03 Wojciech Kucharz

A domain $R$ is said to have the finite factorization property if every nonzero non-unit element of $R$ has at least one and at most finitely many distinct factorizations up to multiplication of irreducible factors by central units. Let $k$…

Rings and Algebras · Mathematics 2019-03-06 Jason P. Bell , Albert Heinle , Viktor Levandovskyy

Let $K$ be a complete non-trivially valued non-Archimedean field. Given an algebraic group over $K$ on which every regular function is constant, any rigid analytic function is shown to be constant too. It follows that an algebraic group…

Algebraic Geometry · Mathematics 2022-12-13 Marco Maculan

A mapping $f:X\to Y$ between metric spaces is called \emph{little Lipschitz} if the quantity $$ \operatorname{lip}(f(x)=\liminf_{r\to0}\frac{\operatorname{diam} f(B(x,r))}{r} $$ is finite for every $x\in X$. We prove that if a compact (or,…

Classical Analysis and ODEs · Mathematics 2018-02-23 Jan Malý , Ondřej Zindulka

Let $f:\mathbb{K}^n\rightarrow\mathbb{K}^m$ be a generically finite polynomial map of degree $d$ between affine spaces. In arXiv:1411.5011 we proved that if $\mathbb{K}$ is the field of complex or real numbers, then the set $S_f$ of points…

Algebraic Geometry · Mathematics 2021-04-06 Zbigniew Jelonek , Michał Lasoń

A continuous map from R^m to R^N or from C^m to C^N is called k-regular if the images of any $k$ points are linearly independent. Given integers m and k a problem going back to Chebyshev and Borsuk is to determine the minimal value of N for…

Differential Geometry · Mathematics 2016-11-08 Jarosław Buczyński , Tadeusz Januszkiewicz , Joachim Jelisiejew , Mateusz Michałek

A graph is called $(k,t)$-regular if it is $k$-regular and the induced subgraph on the neighbourhood of every vertex is $t$-regular. We find new conditions on $(k,t)$ for the existence of such graphs and provide a wide range of examples.

Combinatorics · Mathematics 2021-12-02 Marston Conder , Jeroen Schillewaert , Gabriel Verret

It is shown that if A is an AF algebra then a crossed product of A by the integers can be embedded into an AF algebra if and only if the crossed product is stably finite. This equivalence follows from a simple K-theoretic characterization…

funct-an · Mathematics 2007-05-23 Nathanial P. Brown

Let K be a field and \tilde{K} denote the set of all r \in K for which there exists a finite set A(r) with {r} \subseteq A(r) \subseteq K such that each mapping f:A(r) \to K that satisfies: if 1 \in A(r) then f(1)=1, if a,b \in A(r) and a+b…

Logic · Mathematics 2007-05-23 Apoloniusz Tyszka

We prove that whenever $G$ is a graph from a nowhere dense graph class $\mathcal{C}$, and $A$ is a subset of vertices of $G$, then the number of subsets of $A$ that are realized as intersections of $A$ with $r$-neighborhoods of vertices of…

A stable smooth map $f:N\to M$ is called "$k$-realizable" if its composition with the inclusion $M\subset M\times\Bbb R^k$ is $C^0$-approximable by smooth embeddings; and a "$k$-prem" if the same composition is $C^\infty$-approximable by…

Geometric Topology · Mathematics 2021-05-13 Peter M. Akhmetiev , Sergey A. Melikhov

We prove that Z in definable in Q by a formula with 2 universal quantifiers followed by 7 existential quantifiers. It follows that there is no algorithm for deciding, given an algebraic family of Q-morphisms, whether there exists one that…

Number Theory · Mathematics 2017-04-03 Bjorn Poonen

Let $K$ be a global field and $n > 1$ an integer. We show $n$ is composite if and only if there is an irreducible polynomial $f(x) \in K[x]$ of degree $n$ which is reducible $q$-adically for all the primes $q$ of $K$.

Number Theory · Mathematics 2007-05-23 R. Guralnick , M. Schacher , J. Sonn

Let $n > k > 1$ be integers, $[n] = \{1, \ldots, n\}$. Let $\mathcal F$ be a family of $k$-subsets of~$[n]$. The family $\mathcal F$ is called intersecting if $F \cap F' \neq \emptyset$ for all $F, F' \in \mathcal F$. It is called almost…

Combinatorics · Mathematics 2021-03-22 Peter Frankl , Andrey Kupavskii

For positive integers $n>k>t$ let $\binom{[n]}{k}$ denote the collection of all $k$-subsets of the standard $n$-element set $[n]=\{1,\ldots,n\}$. Subsets of $\binom{[n]}{k}$ are called $k$-graphs. A $k$-graph $\mathcal{F}$ is called…

Combinatorics · Mathematics 2022-10-21 Peter Frankl , Jian Wang

A ring $R$ is said to be clean if each element of $R$ can be written as the sum of a unit and an idempotent. In a recent article (J. Algebra, 405 (2014), 168-178), Immormino and McGoven characterized when the group ring $\mathbb…

Rings and Algebras · Mathematics 2019-11-13 Yuanlin Li , Qinghai Zhong

Let $R$ be a regular ring containing a field $k$. Let $\mathbf{x} = x_1, \ldots, x_r$ be a regular sequence in $R$ such that $R/(\mathbf{x})$ is a regular ring. Fix $m \geq 1$. Set $A_m = R/(\mathbf{x})^m$. We show that for any ideal $Q$ of…

Commutative Algebra · Mathematics 2025-03-27 Tony J. Puthenpurakal

A ring is *unit-additive* if a sum of units is always either a unit or nilpotent. For example, $k[X]$ and $k[X]/(X^2)$ are unit-additive, but $\mathbb Z$ is not. We prove a wide-ranging theorem about unit-additivity in semigroup rings,…

Commutative Algebra · Mathematics 2025-04-22 Neil Epstein , Jay Shapiro