Solutions
... of K containing A. Then either x ∈ B or x−1 ∈ B. If x ∈ B, we are done. If x−1 ∈ B, then by the integrality of x we have xn + a1 xn−1 + · · · + an−1 x + an = 0 with ai ∈ A, and thus x = −(a1 + a2 x−1 + · · · + an (x−1 )n−1 ) ∈ B as desired. Conversely, suppose x ∈ K is not integral over A, and let A ...
... of K containing A. Then either x ∈ B or x−1 ∈ B. If x ∈ B, we are done. If x−1 ∈ B, then by the integrality of x we have xn + a1 xn−1 + · · · + an−1 x + an = 0 with ai ∈ A, and thus x = −(a1 + a2 x−1 + · · · + an (x−1 )n−1 ) ∈ B as desired. Conversely, suppose x ∈ K is not integral over A, and let A ...
Permutations and groups
... b) Find a formula for the inverse of τ = (a1 , a2 , · · · , ak ). c) Show that (στ )−1 = τ −1 σ −1 . Students found Question (a) confusing so we first did (b) and (c). 3.1.1. inverse of a k-cycle. The inverse of a cycle is given by writing the cycle backwards: τ −1 = (ak , ak−1 , · · · , a2 , a1 ) T ...
... b) Find a formula for the inverse of τ = (a1 , a2 , · · · , ak ). c) Show that (στ )−1 = τ −1 σ −1 . Students found Question (a) confusing so we first did (b) and (c). 3.1.1. inverse of a k-cycle. The inverse of a cycle is given by writing the cycle backwards: τ −1 = (ak , ak−1 , · · · , a2 , a1 ) T ...
Lecture 1
... an algorithm that, given a FO formula Φ(x1,…,xn), builds an automaton that recognizes the set {(a1,…,an) | A satisfies Φ(a1,…,an)}. Proof. By induction on the length of the formula Φ. The disjunction corresponds to the union, negation to the complementation, and to projection operations. ...
... an algorithm that, given a FO formula Φ(x1,…,xn), builds an automaton that recognizes the set {(a1,…,an) | A satisfies Φ(a1,…,an)}. Proof. By induction on the length of the formula Φ. The disjunction corresponds to the union, negation to the complementation, and to projection operations. ...
INTRODUCTION TO MODEL THEORY FOR REAL ANALYTIC
... Model theory uses mathematical logic to formalise the underlying language of mathematics. The natural objects of study are the definable sets, and the key is to choose an appropriate language so that the definable sets are both tractable and the natural objects of study in another branch of mathemat ...
... Model theory uses mathematical logic to formalise the underlying language of mathematics. The natural objects of study are the definable sets, and the key is to choose an appropriate language so that the definable sets are both tractable and the natural objects of study in another branch of mathemat ...
Actions of Groups on Sets
... Definitions. For a given x ∈ X, xG = {xg : g ∈ G} is called the orbit of x. The orbits partition X, i.e., either xG = yG or xG ∩ yG = ∅. Equivalently, x ∼ y if x = yg for some g ∈ G is an equivalence relation on X. The orbit set X 0 of X is X/ ∼, the set of the equivalence classes under ∼. It may b ...
... Definitions. For a given x ∈ X, xG = {xg : g ∈ G} is called the orbit of x. The orbits partition X, i.e., either xG = yG or xG ∩ yG = ∅. Equivalently, x ∼ y if x = yg for some g ∈ G is an equivalence relation on X. The orbit set X 0 of X is X/ ∼, the set of the equivalence classes under ∼. It may b ...
An algebraic topological proof of the fundamental theorem of al
... We outline the idea of the proof. Choose the base point at (1,0). Let f : [0, 1] → S 1 be a loop. When n = 1, f completes a turn and returns to (1,0). Therefore the total number of times it wounds around the circle is an integer. This number (called the winding number) is unchanged by the deformatio ...
... We outline the idea of the proof. Choose the base point at (1,0). Let f : [0, 1] → S 1 be a loop. When n = 1, f completes a turn and returns to (1,0). Therefore the total number of times it wounds around the circle is an integer. This number (called the winding number) is unchanged by the deformatio ...
Field Theory
... in every row and column, that means that that for every a, there is another b (also known as a−1 ) such that a ∗ b = e. Finally, we now have a subset that is closed under the group’s operator, an identity element, and inverses for every element. Therefore a finite subset of a group that is closed un ...
... in every row and column, that means that that for every a, there is another b (also known as a−1 ) such that a ∗ b = e. Finally, we now have a subset that is closed under the group’s operator, an identity element, and inverses for every element. Therefore a finite subset of a group that is closed un ...
Topology/Geometry Aug 2014
... 1. Answer each of the six questions on a separate page. Turn in a page for each problem even if you cannot do the problem. 2. Label each answer sheet with the problem number. 3. Put your number, not your name, in the upper right hand corner of each page. If you have not received a number, please cho ...
... 1. Answer each of the six questions on a separate page. Turn in a page for each problem even if you cannot do the problem. 2. Label each answer sheet with the problem number. 3. Put your number, not your name, in the upper right hand corner of each page. If you have not received a number, please cho ...
Topology, MM8002/SF2721, Spring 2017. Exercise set 4 Exercise 1
... Exercise 1. Let X be topological space, Y be a set and f : X → Y be a surjective map. Recall that a subset U ⊆ Y is open in the quotient topology, if and only if f −1 (U ) is open in X. • Show that the quotient topology is in fact a topology. • Show that the quotient topology is the finest topology ...
... Exercise 1. Let X be topological space, Y be a set and f : X → Y be a surjective map. Recall that a subset U ⊆ Y is open in the quotient topology, if and only if f −1 (U ) is open in X. • Show that the quotient topology is in fact a topology. • Show that the quotient topology is the finest topology ...
Why is addition of fractions defined the way it is? Two reasons
... Division Algorithm for the Gaussian integers Let a and b be Gaussian integers, with b 6= 0. Then there exist q and r such that a = bq + r where N(r ) < N(b). Algorithm: Compute a/b in the complexes, say a/b = u + iv . Now let u 0 be the closest integer to u, and v 0 be the closest integer to v . Se ...
... Division Algorithm for the Gaussian integers Let a and b be Gaussian integers, with b 6= 0. Then there exist q and r such that a = bq + r where N(r ) < N(b). Algorithm: Compute a/b in the complexes, say a/b = u + iv . Now let u 0 be the closest integer to u, and v 0 be the closest integer to v . Se ...