نتایج جستجو برای: strongly z ideal

تعداد نتایج: 445621  

2008
David Donoho Jiashun Jin

We consider two-class linear classification in a high-dimensional, low-sample size setting. Only a small fraction of the features are useful, the useful features are unknown to us, and each useful feature contributes weakly to the classification decision – this setting was called the rare/weak model (RW Model) in [11]. We select features by thresholding feature z-scores. The threshold is set by...

2011
Tariq Shah

We investigate antihomomorphic images and pre images of semiprime, strongly primary, irreducible and strongly irreducible fuzzy ideals of a ring. We also prove that: For a surjective anti-homomorphism f : R → R, if every fuzzy ideal of R is f -invariant and has a fuzzy primary (respectively, strongly primary) decomposition in R, then every fuzzy ideal of R has a fuzzy primary (respectively, str...

Journal: :JSEA 2009
Nazir Ahmad Zafar Ajmal Hussain Amir Ali

Automata theory has played an important role in computer science and engineering particularly modeling behavior of systems since last couple of decades. The algebraic automaton has emerged with several modern applications, for example, optimization of programs, design of model checkers, development of theorem provers because of having properties and structures from algebraic theory of mathemati...

1992
Bruno Buchberger

The method (theory plus algorithms) of Gröbner bases provides a uniform approach to solving a wide range of problems expressed in terms of sets of multivariate polynomials. x y 2y z z 0 y 2 x 2 z x z 0 z 2 y 2 x x 0 Example: kinematic equations of a robot. algebraic geometry, commutative algebra, polynomial ideal theory invariant theory automated geometrical theorem proving Groebner-Bases.nb 1

2007
ELENA GUARDO ADAM VAN TUYL

Let Z be a finite set of double points in P 1 × P 1 and suppose further that X, the support of Z, is arithmetically Cohen-Macaulay (ACM). We present an algorithm, which depends only upon a combinatorial description of X, for the bigraded Betti numbers of I Z , the defining ideal of Z. We then relate the total Betti numbers of I Z to the shifts in the graded resolution, thus answering a special ...

2006
HOLGER BRENNER MORDECHAI KATZMAN

This paper deals with a question regarding tight closure in characteristic zero which we now review. Let R be a commutative ring of prime characteristic p and let I ⊆ R be an ideal. Recall that for e ≥ 0, the e-th Frobenius power of I, denoted I [p e], is the ideal of R generated by all p-th powers of elements in I. We say that f ∈ I∗, the tight closure of I, if there exists a c not in any mini...

Journal: :J. Symb. Comput. 2017
Xiao-Shan Gao Zhang Huang Chun-Ming Yuan

In this paper, binomial difference ideals are studied. Three canonical representations for Laurent binomial difference ideals are given in terms of the reduced Gröbner basis of Z[x]-lattices, regular and coherent difference ascending chains, and partial characters over Z[x]-lattices, respectively. Criteria for a Laurent binomial difference ideal to be reflexive, prime, well-mixed, and perfect a...

2003
Alberto Alzati Alfonso Tortora

We define the monomial invariants of a projective variety Z; they are invariants coming from the generic initial ideal of Z. Using this notion, we generalize a result of Cook [C]: If Z is an integral variety of codimension two, satisfying the additional hypothesis sZ = sΓ, then its monomial invariants are connected.

2008
David A. Cox

Examples • If S = ∅, then Z(S) = An. • If S = {1}, then Z(S) = ∅. • If S = {x1 − a1, . . . , xn − an}, then Z(S) = {(a1, . . . , an)}. Remark. All rings in this course are commutative and have 1. Remark. If A is a ring, then any subset S ⊆ A generates a minimal ideal 〈S〉 ⊆ A. In fact, we have 〈S〉 = {∑ aj xj : aj ∈ A, xj ∈ S}. Lemma. Z(S) = Z(〈S〉) for all S ⊆ k[x1 , . . . , xn]. Proof. Since 〈S〉...

2004
ROBERT OSBURN

Let p > 3 be a prime, ζp a pth primitive root of 1, and ∆ the Galois group of Q(ζp) over Q. Let q 6= p be a prime and n the order of q modulo p. Assume q 6≡ 1 mod p and so n ≥ 2, p(q− 1)|qn − 1, and n|p− 1. Set f = (q − 1)/p and e = (p− 1)/n. Let Q be a prime ideal of Z[ζp] above q and let F = Z[ζp]/Q. Thus F ∼= Fqn , the finite field with q elements. Let α ∈ Z[ζp] be a generator of F such that...

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