[permalink] [id link]
Let A be an integral domain.
from
Wikipedia
Some Related Sentences
Let and be
Let every policeman and park guard keep his eye on John and Jane Doe, lest one piece of bread be placed undetected and one bird survive.
Let us assume that it would be possible for an enemy to create an aerosol of the causative agent of epidemic typhus ( Rickettsia prowazwki ) over City A and that a large number of cases of typhus fever resulted therefrom.
Let p be the minimal polynomial for T, Af, where the Af, are distinct irreducible monic polynomials over F and the Af are positive integers.
Let V be a finite-dimensional vector space over an algebraically closed field F, e.g., the field of complex numbers.
Let N be a positive integer and let V be the space of all N times continuously differentiable functions F on the real line which satisfy the differential equation Af where Af are some fixed constants.
Let Q be a nonsingular quadric surface bearing reguli Af and Af, and let **zg be a Af curve of order K on Q.
Let us take a set of circumstances in which I happen to be interested on the legislative side and in which I think every one of us might naturally make such a statement.
Let the state of the stream leaving stage R be denoted by a vector Af and the operating variables of stage R by Af.
Let it be granted then that the theological differences in this area between Protestants and Roman Catholics appear to be irreconcilable.
Let us therefore put first things first, and make sure of preserving the human race at whatever the temporary price may be ''.
Let and integral
Let φ ( ξ, η, ζ ) be an arbitrary function of three independent variables, and let the spherical wave form F be a delta-function: that is, let F be a weak limit of continuous functions whose integral is unity, but whose support ( the region where the function is non-zero ) shrinks to the origin.
Let p be the nth decimal of the nth number of the set E ; we form a number N having zero for the integral part and p + 1 for the nth decimal, if p is not equal either to 8 or 9, and unity in the contrary case.
Theorem: Let R be a Dedekind domain with fraction field K. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.
Let R be an integral domain with fraction field K. A fractional ideal is a nonzero R-submodule I of K for which there exists a nonzero x in K such that
Let X be a g-dimensional torus given as X = V / L where V is a complex vector space of dimension g and L is a lattice in V. Then X is an abelian variety if and only if there exists a positive definite hermitian form on V whose imaginary part takes integral values on L × L.
Let R be the ring of integers of an algebraic number field K and P a prime ideal of R. For each extension field L of K we can consider the integral closure S of R in L and the ideal PS of S. This may or may not be prime, but assuming is finite it is a product of prime ideals
Let A be a Dedekind domain with the field of fractions K and B be the integral closure of A in a finite separable extension L of K. ( In particular, B is Dedekind then.
Let M < sub > m </ sub > be the set of 2 × 2 integral matrices with determinant m and Γ = M < sub > 1 </ sub > be the full modular group SL ( 2, Z ).
Let us note that the surface integral of this 2-form is the same as the surface integral of the vector field which has as components, and.
Let us notice that we defined the surface integral by using a parametrization of the surface S. We know that a given surface might have several parametrizations.
Let G be a semisimple Lie group or algebraic group over, and fix a maximal torus T along with a Borel subgroup B which contains T. Let λ be an integral weight of T ; λ defines in a natural way a one-dimensional representation C < sub > λ </ sub > of B, by pulling back the representation on T = B / U, where U is the unipotent radical of B.
Let R be a commutative ring with prime characteristic p ( an integral domain of positive characteristic always has prime characteristic, for example ).
Let L / K be a finite extension of number fields, and let B and A be the corresponding ring of integers of L and K, respectively, which are defined to be the integral closure of the integers Z in the field in question.
Let X be a vector field on M of class C < sup > r − 1 </ sup > and let p ∈ M. An integral curve for X passing through p at time t < sub > 0 </ sub > is a curve α: J → M of class C < sup > r − 1 </ sup >, defined on an open interval J of the real line R containing t < sub > 0 </ sub >, such that
* Let R ⊂ S be an integral extension of commutative rings, and P a prime ideal of R. Then there is a prime ideal Q in S such that Q ∩ R = P. Moreover, Q can be chosen to contain any prime Q < sub > 1 </ sub > of S such that Q < sub > 1 </ sub > ∩ R ⊂ P.
1.552 seconds.