Morphisms of Algebraic Stacks

... What about “separated algebraic stacks”? We have seen in Morphisms of Spaces, Lemma 39.9 that an algebraic space is separated if and only if the diagonal is proper. This is the condition that is usually used to define separated algebraic stacks too. In the example [S/G] → S above this means that G → ...

... What about “separated algebraic stacks”? We have seen in Morphisms of Spaces, Lemma 39.9 that an algebraic space is separated if and only if the diagonal is proper. This is the condition that is usually used to define separated algebraic stacks too. In the example [S/G] → S above this means that G → ...

Incidence structures I. Constructions of some famous combinatorial

... Theorem [Haemers]. Let A be a complete hermitian n × n matrix, partitioned into m2 block matrices, such that all diagonal matrices are square. Let B be the m × m matrix, whose i, j-th entry equals the average row sum of the i, j-th block matrix of A for i, j = 1, . . . , m. Then the eigenvalues α1 ≥ ...

... Theorem [Haemers]. Let A be a complete hermitian n × n matrix, partitioned into m2 block matrices, such that all diagonal matrices are square. Let B be the m × m matrix, whose i, j-th entry equals the average row sum of the i, j-th block matrix of A for i, j = 1, . . . , m. Then the eigenvalues α1 ≥ ...

Algebraic Shift Register Sequences

... Families of recurring sequences over a finite field . . . . 6.7.b Families of Linearly Recurring Sequences over a Ring . . Examples . . . . . . . . . . . . . . . . . . . . . . . 6.8.a Shift registers over a field . . . . . . . . . . . . . . 6.8.b Fibonacci numbers . . . . . . . . . . . . . . . . . E ...

... Families of recurring sequences over a finite field . . . . 6.7.b Families of Linearly Recurring Sequences over a Ring . . Examples . . . . . . . . . . . . . . . . . . . . . . . 6.8.a Shift registers over a field . . . . . . . . . . . . . . 6.8.b Fibonacci numbers . . . . . . . . . . . . . . . . . E ...

An Introduction to Algebraic Number Theory, and the Class Number

... Proposition 1.3. Let A be a commutative unital ring, and let S be a multiplicatively closed subset of A. Then p 7→ S −1 p is a bijection from the set of prime ideals in A which do not meet S to the set of prime ideals in S −1 A. Proof. (Well defined:) Let p be a prime ideal in A which does not meet ...

... Proposition 1.3. Let A be a commutative unital ring, and let S be a multiplicatively closed subset of A. Then p 7→ S −1 p is a bijection from the set of prime ideals in A which do not meet S to the set of prime ideals in S −1 A. Proof. (Well defined:) Let p be a prime ideal in A which does not meet ...

4 Number Theory 1 4.1 Divisors

... A finite field is a field that contains a finite number of elements. There is exactly one finite field of size (order) pn where p is a prime (called the characteristic of the field) and n is a positive integer. If p is a prime Z p is the finite field GF(p) (note here that n = 1 and so is omitted). F ...

... A finite field is a field that contains a finite number of elements. There is exactly one finite field of size (order) pn where p is a prime (called the characteristic of the field) and n is a positive integer. If p is a prime Z p is the finite field GF(p) (note here that n = 1 and so is omitted). F ...

Commutative Algebra

... We have also seen already that this assignment of algebraic to geometric objects is injective in the sense of Remark 0.10 and Construction 0.11 (b). However, not all rings, ideals, and ring homomorphisms arise from this correspondence with geometry, as we will see in Remark 1.10, Example 1.25 (b), a ...

... We have also seen already that this assignment of algebraic to geometric objects is injective in the sense of Remark 0.10 and Construction 0.11 (b). However, not all rings, ideals, and ring homomorphisms arise from this correspondence with geometry, as we will see in Remark 1.10, Example 1.25 (b), a ...

... We have also seen already that this assignment of algebraic to geometric objects is injective in the sense of Remark 0.10 and Construction 0.11 (b). However, not all rings, ideals, and ring homomorphisms arise from this correspondence with geometry, as we will see in Remark 1.10, Example 1.25 (b), a ...

Geometric Constructions from an Algebraic Perspective

... Every construction begins with given points, lines and circles and will be completed with a sequence of the above steps. Now we begin to relate the construction of real numbers to algebra so we begin by constructing our unit measurement OX which has length 1. Although we are restricted to those simp ...

... Every construction begins with given points, lines and circles and will be completed with a sequence of the above steps. Now we begin to relate the construction of real numbers to algebra so we begin by constructing our unit measurement OX which has length 1. Although we are restricted to those simp ...

higher algebra

... 8 Elementary properties of Modules 8.1 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Modules over rings. . . . . . . . . . . . . . . . . . . . 8.1.2 Submodules . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 Factor modules. . . . . . . . . . . . . . . . . . . . . . 8.1 ...

... 8 Elementary properties of Modules 8.1 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Modules over rings. . . . . . . . . . . . . . . . . . . . 8.1.2 Submodules . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 Factor modules. . . . . . . . . . . . . . . . . . . . . . 8.1 ...

, Elementary Number Theory

... Inverse exists only if (a,m)=1 Else we can not get a Bezout representation linking 1, a and m (like 1 = xa+ym) If (a,m)=1 we can use extended Euclidean algorithm to find the inverse of a in Z m ...

... Inverse exists only if (a,m)=1 Else we can not get a Bezout representation linking 1, a and m (like 1 = xa+ym) If (a,m)=1 we can use extended Euclidean algorithm to find the inverse of a in Z m ...

IDEAL FACTORIZATION 1. Introduction

... course n is not prime, so n = ab with a, b > 1. Then a, b < n, so a and b are products of primes. Hence n = ab is a product of primes, which is a contradiction. Uniqueness of the prime factorization requires more work. We will use the same idea (contradiction from a minimal counterexample) to prove ...

... course n is not prime, so n = ab with a, b > 1. Then a, b < n, so a and b are products of primes. Hence n = ab is a product of primes, which is a contradiction. Uniqueness of the prime factorization requires more work. We will use the same idea (contradiction from a minimal counterexample) to prove ...