نتایج جستجو برای: k extension

تعداد نتایج: 518953  

Journal: :journal of linear and topological algebra (jlta) 0
g soleimani rad young researchers and elite club, central tehran branch, islamic azad university, tehran, iran. department of mathematics, faculty of science, islamic azad university, central tehran branch, po. code 13185-768, tehran, iran.

recently, rahimi et al. [comp. appl. math. 2013, in press] de ned the conceptof quadrupled xed point in k-metric spaces and proved several quadrupled xed pointtheorems for solid cones on k-metric spaces. in this paper some quadrupled xed point resultsfor t-contraction on k-metric spaces without normality condition are proved. obtainedresults extend and generalize well-known comparable result...

1999
MATTHEW A. PAPANIKOLAS

Let k = Fq(t) be the rational function field with finite constant field and characteristic p ≥ 3, and let K/k be a finite separable extension. For a fixed place v of k and an elliptic curveE/K which has ordinary reduction at all places ofK extending v, we consider a canonical height pairing 〈 , 〉v : E(K ) × E(K) → C v which is symmetric, bilinear and Galois equivariant. The pairing 〈 , 〉∞ for t...

2000
Gregor Kemper

Let g(X) ∈ K(t1, . . . , tm)[X] be a generic polynomial for a group G in the sense that every Galois extension N/L of infinite fields with group G and K ≤ L is given by a specialization of g(X). We prove that then also every Galois extension whose group is a subgroup of G is given in this way. Let K be a field and G a finite group. Let us call a monic, separable polynomial g(t1, . . . , tm, X) ...

Journal: :Czechoslovak Mathematical Journal 2021

Let A be a finite-dimensional k-algebra and K/k finite separable field extension. We prove that is derived equivalent to hereditary algebra if only so ⊗kK.

2012
MAREK PYCIA

PROOF: One implication follows from Lemmas 3 and 4. To prove the remaining implication, we show that the domain of all preference profiles that do not admit n-cycles, n= 2 3 satisfies R1 and R2. Fix a profile A that does not admit n-cycles, n= 2 3 To prove that R1 is satisfied, we construct an extension of the partial ordering constructed in the proof of Lemma 5 (the construction did not rely o...

Journal: :Finite Fields and Their Applications 2007
Robert W. Fitzgerald

Let K be a finite extension of F2. We consider quadratic forms written as the trace of xR(x), where R(x) is a linearized polynomial. We determine the K and R(x) where the form has a radical of codimension 2. This is applied to constructing maximal Artin-Schreier curves. Set F = F2 and let K = F2k be an extension of degree k. Let

2010
JONATHAN W. SANDS J. W. SANDS

If p is an odd prime that does not divide the class number of the imaginary quadratic field k , and the cyclotomic Z -extension of k has A-invariant less than or equal to two, we prove that every totally ramified Z extension of k has //-invariant equal to zero and A-invariant less than or equal to two. Combined with a result of Bloom and Gerth, this has the consequence that ß = 0 for every Z -e...

2009
CRISTIANA BERTOLIN

Let S be a site. We introduce the notion of extension of strictly commutative Picard S-stacks. Applying this notion to 1-motives, we get the notion of extension of 1-motives and we prove the following conjecture of Deligne: if MRZ(k) denotes the integral version of the neutral Tannakian category of mixed realizations over an algebraically closed field k, then the subcategory of MRZ(k) generated...

2002
I. Moerdijk

We observe that any regular Lie groupoid G over an manifold M fits into an extension K → G → E of a foliation groupoid E by a bundle of connected Lie groups K. If F is the foliation on M given by the orbits of E and T is a complete transversal to F , this extension restricts to T , as an extension KT → GT → ET of an étale groupoid ET by a bundle of connected groups KT . We break up the classifi...

2012
KEITH CONRAD

Let L/K be a finite extension of fields, with n = [L : K]. We will associate to this extension two important functions L→ K, called the trace and the norm. For each α ∈ L, let mα : L → L be multiplication by α: mα(x) = αx for x ∈ L. Each mα is a K-linear map from L to L, so choosing a K-basis of L lets us write mα as an n×n matrix, which will be denoted [mα], or [mα]L/K if we need to emphasize ...

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