ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...

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ationals from the integers. This make it possible to take the stand that you can define the rationals if only you take the integers for granted. This fits into a program where you want to have all concepts defined. This includes a way of constructing the integers from the natural numbers. And so on till you reach the foundations of mathematic based on set theory. But this is quite another story to be told in another afsnit. It is them important to say that the construction can be used in a lot of other cases, where we don’t have the result of the construction already. First we have to clarify the notions with the following 145. Definition: Field of fractions Let R be a ring and let L be et field. Suppose there exists an injective homomorphism i : R → L where L is the underlying ring with the property that any x ∈ L may be written in the form i(p)/i(q) where p and q are in R. Now we use i to identify R with i(R), that is we consider p and i(p) to be the same. Then R is a subring of L and any element in L can be written in the form p/q, where p and q belongs to R. We shall say that L via i is a field of fractions for R. We shall call i the associated embedding of R into L Now follows the construction, which has the form of a 146. Theorem: Existence of field of fractions There exists a field of fractions for the ring R if and only if R is an integral domain Proof : The necessity of the condition follows from the fact that L as a field is also an integral domain and so the i(R) as a substructure of an integral domain is itself an integral domain, and then of course this also holds for R by isomorphism. To prove the sufficiency assume that R is an integral domain. We define the set S = {(p, q) ∈ R × R : q ̸= 0}. On S we define an addition + by the assignment (a, b) + (c, d) = (ad + bc, bd) and a multiplication by the assignment (a, b)(c, d) = (ac, bd). That this in fact defines operations rely heavily on R being an integral domain, since the non existence of zero divisors ensures that bd ̸= 0. We also define the opposite operation −(a, b) = (−a, b) and a reciprocal operation (a, b) −1 = (b, a). Finally we define a zero element (0, 1) and et unit-element (1, 1). 70
Then it is routine to check that S equipped with these operations is a ring. Now the set I consisting of all elements of the form (0, b) can be checked to be an ideal. We let L denote the quotient ring of S wrt this ideal. Let ∼ denote the associated equivalence relation, then the following calculation (x, y) ∼ (u, v) ⇔ (x, y) − (u, v) ∈ I ⇔ (xv − yu, yv) ∈ I ⇔ xv − yu = 0 shows that (x, y) ∼ (u, v) ⇔ xv = yu In what follows we shall use [p/q] to denote the equivalence class containing (p, q) ∈ S. We notice that [1/1] is the class which contains the unit element (1, 1) in the ring S, and so is the unit element, 1, in the quotient L. Furthermore this class also contains all the elements (a, a) because (a, a) ∼ (1, 1) since a · 1 = 1 · a. Analogously you see that [0/1] is the zero element,0. If [a/b] ̸= 0 the a ̸= 0, and then (b, a) ∈ S, and [b/a][a/b] = [ab/ab] = 1, and this shows that [a/b] is invertible with [a/b] −1 = [b/a] . And the L is a field. The mapping defined by h(a) = (a, 1) is a homomorphism of R into S, since h(a) + h(b) = (a, 1) + (b, 1) = (a · 1 + b · 1, 1 · 1) = (a + b, 1) = h(a + b) h(a)h(b) = (a, 1)(b, 1) = (ab, 1) = h(ab) The mapping i : R → L defined by i(a) = [h(a)] = [a/1] is a composition of two homomorphisms and therefore itself a homomorphism. To show that it is injective lets assume that i(a) = i(b). Using the definitions we see that i(a) = i(b) ⇔ [a/1] = [b/1] ⇔ (a, 1) ∼ (b, 1) ⇔ a · 1 = 1 · b ⇔ a = b. which proves injectivity As a result of the identification of a with i(a) we have justified the the following 147. Definition: Canonical field of fractions for an integral domain The field constructed in the previous theorem is said to be the canonical field of fractions for R. The mapping i is said to be the canonical embedding of R. 71
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Anders Madsen OPERATIONS AND STRUCT
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Table of contents 1 Introduction 2
Page 5 and 6:
1: Introduction The word algebra is
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This is connected to the fact that
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Let f : B → A1 × . . . × An be
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Let ♢ and ♡ be operations on A
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to yield g ◦ ♡ = ♠ ◦ g, usi
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Let ♢ be an operation on A with n
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2. 8: Induced operation on product
Page 19 and 20: Suppose that ♢ is an operation on
Page 21 and 22: 22. Definition: Compatibility Suppo
Page 23 and 24: 26. Theorem: Criterion for homomorp
Page 25 and 26: Proof : Assume f surjective and ♢
Page 27 and 28: Let (A, ♢ 1 , . . . , ♢ n ) and
Page 29 and 30: Proof : It follows from the definit
Page 31 and 32: Proof : The factors have per defini
Page 33 and 34: 23 SU(n) special unitary matrices 2
Page 35 and 36: 4. 3: Tables with homomorphisms In
Page 37 and 38: We consider the unary operation −
Page 39 and 40: Show that the determinant considere
Page 41 and 42: Lets take a look on the two example
Page 43 and 44: Anders Madsen CATEGORIES OF STRUCTU
Page 45 and 46: Use the table of contents in the fi
Page 47 and 48: is always possible to extend (doubl
Page 49 and 50: 67. Definition: Neutral element for
Page 51 and 52: For product structures we must insp
Page 53 and 54: Proof : The theorem is already prov
Page 55 and 56: When dealing with abstract groups o
Page 57 and 58: A homomorphism f : A ↦→ B is in
Page 59 and 60: If H is a subgroup of the finite gr
Page 61 and 62: Proof : This is just T91 111. Defin
Page 63 and 64: 10. 2: Subring 117. Definition: Del
Page 65 and 66: A subset I is said to be et ideal o
Page 67 and 68: 133. Definition: Integral domain A
Page 69: 143. Theorem: Any substructure of a
Page 73 and 74: The prototype example of field exte
Page 75 and 76: Consider the possibility of the map
Page 77 and 78: Now it is easily checked that Fx is
Page 79 and 80: We shall use this to illustrate an
Page 81 and 82: What is said above may be generaliz
Page 83 and 84: The field of fractions for the ring
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