ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...

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The canonical field of fractions L for an integral domain R has the property that it is a field which is an extension of R. Furthermore it is minimal wrt to this property in the sense that it does not contain more than absolutely necessary, namely the fractions made from elements in R. In some sense it is also the only one possible. This is stated in the following 148. Theorem: ”Uniqueness” of field of fractions Assume that L, M are fields of fractions for the ring R via i, j. Then there exists an isomorphism F between the fields L and M such that i = F ◦ j, which in particular means that the fields are isomorphic. Proof : It is enough to show the theorem when L is the canonical fields of fractions for R. We use the notation from the proof of the existence part. So let M be any field of fractions for R and let j be the embedding of R into M. We define the map f of S into M by f(p, q) = j(p)/j(q) The first step is to show that f is a homomorphism (between rings). This is straight forward using that j is a homomorphism and L is a field: f ((p, q) + (r, s)) = j(ps + qr) j(qs) = f(p, q) + f(r, s) = j(p)j(s) + j(q)j(r) j(q)j(s) = j(p) j(r) + j(q) j(s) Next we note that the equivalence relation ∼ is generated by f in the sense that (p, q) ∼ (r, s) ⇔ f(p, q) = f(r, s) This means that we can define a function F on S/I by F ([p/q]) = f(p, q) since the value does not depend on the chosen representative of the class. Then f = F ◦k and therefore F is a homomorphism by a general theorem, T392, since f is surjective as a result of the definition D145 and since k is a homomorphism ( T56) Next we show that F is injective. So assume that F (a) = F (b) with a = [p/q], b = [r/s]. Then f(p, q) = F (a) = F (b) = f(r, s) and so (p, q) ∼ (r, s) which means that a = b. Since f is surjective, so is F . Now we have shown that F is an isomorphism between fields 72
The prototype example of field extension is the extension from Z to Q (see E61) A very important field of fractions is the one resulting from the ring of polynomials : E62 12: Examples and exercises Example 22: Words of equal length (see nr 74) Th relation x and y have the same length is an equivalence relation on the monoid of words and it is compatible with concatenation En equivalence class consists of all words of a certain length. This classification is induced by the homomorphism which is defined to be the length of the word. Example 23: The semigroup of words (see nr 74) The set S of words over a given alphabet with concatenation as operation is a semigroup. Example 24: The semigroup of words is not a monoid, but ... (see nr 74) If the semigroup of words is extended with the empty word you will get a monoid with the empty word as the neutral element. Example 25: The set of naturals with addition is a semigroup (see nr 74) The set of naturals with addition is a semigroup Example 26: The word length is a homomorphism (see nr 74) The mapping x ↦→ |x| which to a word assigns its length (number of letters) is a semigroup homomorphism. It can in an obvious way be extended to a monoid homomorphism Exercise 27: Greatest common divisor gives a semigroup (see nr 74) On the integers you have the binary operation x ⊓ y which to x and y assigns the greatest common divisor of x and y. Show that Z with this operation is a semigroup. Exercise 28: Greatest common divisor does not yield a monoid (see nr 74) Show that the semigroup i X27 is not (the underlying semigroup of) a monoid Exercise 29: Minimum gives a semigroup (see nr 74) We define x ∧ y as the minimum of x and y. 73
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Anders Madsen OPERATIONS AND STRUCT
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Table of contents 1 Introduction 2
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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
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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 and 70: 143. Theorem: Any substructure of a
Page 71: Then it is routine to check that S
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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