نتایج جستجو برای: primary zariski topology

تعداد نتایج: 708371  

2008
Avinash Sathaye

A hypersurface is said to be quasihomogeneous if in suitable coordinates with assigned weights, its equation becomes weighted homogeneous in its variables. For an irreducible quasihomogeneous plane curve, the equation necessarily becomes a two term equation of the form aY n + bX where n,m are necessarily coprime. Zariski, in a short paper, established a criterion for an algebroid curve to be qu...

Journal: :Journal of algebraic hyperstructures and logical algebras 2021

In this paper, we introduce the notions of Belluce lattice associated with a bounded $BCK$-algebra and reticulation $BCK$-algebra. To do this, first, define operations $curlywedge ,$ $curlyvee $ $sqcup on $BCK$-algebras study some algebraic properties them. Also, for $A$ Zariski topology Spec(A)$ induced $tau _{A,Max(A)}$ $Max(A)$. We prove $(Max(A),tau_{A,Max(A)})$ is compact topological space...

2015
Torsten Ekedahl

0965 Contents 1. Introduction 1 2. Some topology 2 3. Local isomorphisms 4 4. Ind-Zariski algebra 6 5. Constructing w-local affine schemes 6 6. Identifying local rings versus ind-Zariski 10 7. Ind-´ etale algebra 14 8. Constructing ind-´ etale algebras 15 9. Weaklyétale versus pro-´ etale 18 10. Constructing w-contractible covers 18 11. The pro-´ etale site 21 12. Points of the pro-´ etale site...

2006
SUSUMU ODA

Zariski Problem (Cancellation of indeterminates) is settled affirmatively, that is, it is proved that : Let k be an algebraically closed field of characteristic zero and let n, m ∈ N. If R[Y1, . . . , Ym] ∼=k k[X1, . . . , Xn+m] as k-algebras, where Y1, . . . , Ym, X1, . . . , Xn+m are indetermoinates, then R ∼=k k[X1, . . . , Xn]. Zariski Problem is the following : Zariski Problem. Let k be an...

2007
MICHAEL TEMKIN

Let k be an algebraically closed field andK be a finitely generated k-field. In the first half of the 20-th century, Zariski defined a Riemann variety RZK(k) associated to K as the projective limit of all projective k-models of K. Zariski showed that this topological space, which is now called a Riemann-Zariski (or Zariski-Riemann) space, possesses the following set-theoretic description: to gi...

2008
ALEX KÜRONYA

Based on a recent work of Thomas Bauer’s [1] reproving the existence of Zariski decompositions for surfaces, we construct a b-divisorial analogue of Zariski decomposition in all dimensions.

Journal: :Israel Journal of Mathematics 2021

A complex affine Poisson algebra is said to satisfy the Dixmier-Moeglin equivalence if cores of maximal ideals are precisely those prime that locally closed in spectrum P.spec and if, moreover, these whose extended centers exactly numbers. In this paper, we provide some topological criteria for terms poset (P.spec A, ⊆) symplectic leaf or core stratification on its spectrum. particular, prove Z...

2008
MANUEL BLICKLE

is an isomorphism. Here F : X −→ X denotes the Frobenius morphism on X and H denotes the a cohomology sheaf of F∗Ω•X . If the variety is not smooth, not much is known about the properties of the Cartier operator and the poor behaviour of the deRham complex in this case makes its study difficult. If one substitutes the deRham complex with the Zariski-deRham complex the situation is better. For e...

2010
MAHAN MJ

We begin by showing that commensurators of Zariski dense subgroups of isometry groups of symmetric spaces of non-compact type are discrete provided that the limit set on the Furstenberg boundary is not invariant under the action of a (virtual) simple factor. In particular for rank one or simple Lie groups, Zariski dense subgroups with non-empty domain of discontinuity have discrete commensurato...

2009
RAYMOND HOOBLER

1.1. Sites. We begin with the underlying idea for a Grothendieck topology. Consider your favorite topological space X. Its topology, don’t use the Zariski topology, is determined by the partially ordered category UX consisting of all the open subsets of X ordered with respect to inclusion. In order to verify that a presheaf is a sheaf on X, we must first construct all possible coverings U = (Ui...

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