An Introduction to Real Analysis
Section 1.3.2 notes
2025-12-31
This purpose of these notes is to derive algebraic properties of a general field \(F\).1 Since the preceding Section 1.3.1 notes for Katznelson and Katznelson (2024) were primarily concerned with developing arithmetic in \(\symbb{Q}\), these notes will confirm that \(\symbb{Q}\) is a field and then catalog algebraic properties of a general field \(F\).
Because \(\symbb{Q}\) is a field all of these properties apply to \(\symbb{Q}\) in particular. Moreover, the two number systems developed after \(\symbb{Q}\)—the real numbers \(\symbb{R}\) and the complex numbers \(\symbb{C}\)—are themselves fields so all of the results cataloged here apply to those number systems as well.2
After cataloging algebraic properties of fields these notes then discuss field isomorphisms, which are special bijections that show that two fields \(F\) and \(\tilde{F}\) are interchangeable. The notes then conclude with brief discussions of fields commonly encountered in analysis and algebra.
The Section 1.1 notes and exercises for Schröder (2008) are also concerned with algebra in a general field, so they are complementary to these notes.
Informally, Katznelson and Katznelson (2024) define a field as a non-empty set \(F\) together with two binary operations, addition \(+ \colon F \times F \to F\) and multiplication \(\cdot \colon F \times F \to F\) with the following properties:
It is interesting to note that even when formalized, this definition does not preclude the possibility that \(F\) is a singleton set. Because of this Katznelson and Katznelson (2024) stipulate that in a field \(F\), \(0 \neq 1\). This is a curiosity as it pertains to fields in general but it can be shown that the same holds for the rational numbers as well. Together with the algebraic properties of \(\symbb{Q}\) that were proven in the Section 1.3.1 notes for Katznelson and Katznelson (2024) this will prove that \(\symbb{Q}\) is a field.
To see that the rational additive- and multiplicative-identity elements are distinct, first note that in Exercise 1.2.7 of Katznelson and Katznelson (2024) it was proven that for the integers \(0, 1 \in \symbb{Z}\), \(1 > 0\). It follows from the trichotomy law for the integers that \(0 \neq 1\). There is a bijection (in fact an isomorphism) \(\iota \colon \symbb{Z}\to \tilde{\symbb{Z}} \subseteq \symbb{Q}\) as described in the Section 1.3.1 notes for Katznelson and Katznelson (2024). Since a bijection is an injection it follows that \(0 \neq 1\) implies that \(\iota(0) \neq \iota(1)\) as well, where \(\iota(0), \iota(1) \in \tilde{\symbb{Z}} \subseteq \symbb{Q}\). It only remains to show that \(\iota(0)\) and \(\iota(1)\) are additive and multiplicative identities in the rational numbers. By definition of \(\iota\), for \(m \in \symbb{Z}\), \(\iota(m) = [(m, 1)]\). Therefore \(\iota(0) = [(0, 1)]\) and \(\iota(1) = [(1, 1)]\). Since it was shown in the Section 1.3.1 notes for Katznelson and Katznelson (2024) that \([(0, 1)]\) and \([(1, 1)]\) are additive and multiplicative identities in the rational numbers, respectively, the conclusion follows.
Because the preceding three properties of a general field \(F\) listed above were all proven for the rational numbers in the Section 1.3.1 notes for Katznelson and Katznelson (2024), and it has now been shown that the rational additive and multiplicative identities are distinct, it follows that \(\symbb{Q}\) is a field.
Proposition 1.3.1 in Katznelson and Katznelson (2024, 10–11) proves five algebraic properties of a general field \(F\). These properties are restated and proven below as Proposition 1, Proposition 2, Proposition 3, Proposition 4 and Proposition 5. This is done to give a different presentation of and commentary on these properties than that given by Katznelson and Katznelson (2024).
Proposition 1 Let \(x, 0 \in F\). Then \(x \cdot 0 = 0\).
Proof. Using the fact that \(0\) is an additive identity element and that multiplication distributes over addition, \[\begin{align*} x \cdot 0 &= x \cdot (0 + 0)\\[0.75em] &= x \cdot 0 + x \cdot 0. \end{align*}\] Adding \(-(x \cdot 0)\) to both sides of the equation, and using associativity of addition and properties of additive inverses and zero, shows that \[\begin{align*} \underbrace{-(x \cdot 0) + (x \cdot 0)}_{= 0} &= \underbrace{-(x \cdot 0) + (x \cdot 0 + x \cdot 0)}_{= (-(x \cdot 0) + x \cdot 0) + x \cdot 0}\\[0.75em] \implies 0 &= \underbrace{(-(x \cdot 0) + x \cdot 0)}_{= 0} + x \cdot 0\\[0.75em] \implies 0 &= \underbrace{0 + x \cdot 0}_{= x \cdot 0}\\[0.75em] \implies 0 &= x \cdot 0, \end{align*}\] so \(x \cdot 0 = 0\). \(\blacksquare\)
The proof of Proposition 1 is essentially the same as that given in Proposition 1.2.1 of Katznelson and Katznelson (2024, 6) for the integers, although the former is more detailed. The proof in fact shows that \(0 \cdot x = 0\) in any ring because the proof only uses properties of rings, so this result is not unique to fields.
Proposition 2 The additive-identity element \(0 \in F\) is unique.
Proof. Let \(x, y \in F\) be such that \(x + y = x\). Adding \(-x\) to both sides of the equation, and using associativity of addition and properties of additive inverses and zero, shows that \[\begin{align*} \underbrace{(-x) + (x + y)}_{= ((-x) + x) + y} &= \underbrace{(-x) + x}_{= 0}\\[0.75em] \implies \underbrace{((-x) + x)}_{= 0} + y &= 0\\[0.75em] \implies \underbrace{0 + y}_{= y} &= 0\\[0.75em] \implies y &= 0. \end{align*}\] Since any element \(y \in F\) acting as an additive identity must be equal to \(0\), it follows that \(0\) is unique. \(\blacksquare\)
The proof of Proposition 2 is essentially the same as that given in Exercise 1.2.2 for Katznelson and Katznelson (2024) for the corresponding result for the integers. As mentioned in that exercise the proof shows that the identity element in any group is unique because the proof uses only properties of groups, so this result is not unique to fields.
Proposition 3 Let \(x \in F\). Then \(-x \in F\) is unique.
Proof. Let \(y \in F\) be such that \(x + y = 0\). Then as in the proof of Proposition 2, \[\begin{align*} \underbrace{(-x) + (x + y)}_{= ((-x) + x) + y} &= \underbrace{(-x) + 0}_{= -x}\\[0.75em] \implies \underbrace{((-x) + x)}_{= 0} + y &= -x\\[0.75em] \implies \underbrace{0 + y}_{= y} &= -x\\[0.75em] \implies y &= -x. \end{align*}\] Since any element \(y \in F\) acting as an additive inverse for \(x \in F\) must be equal to \(-x\), it follows that \(-x\) is unique. \(\blacksquare\)
The proof of Proposition 3 is essentially the same as that given in Exercise 1.2.3 for Katznelson and Katznelson (2024) for the corresponding result for the integers. As mentioned in that exercise the proof shows that additive-inverse elements in any group are unique because the proof uses only properties of groups, so this result is not unique to fields.
Proposition 4 The multiplicative-identity element \(1 \in F\) is unique.
Proof. Let \(x \in F \setminus \{ 0 \}\) and \(y \in F\) be such that \(x \cdot y = x\). Multiplying both sides of the equation by \(x^{-1}\), and using associativity of multiplication and properties of multiplicative inverses and one, shows that \[\begin{align*} \underbrace{x^{-1} \cdot (x \cdot y)}_{= (x^{-1} \cdot x) \cdot y} &= \underbrace{x^{-1} \cdot x}_{= 1}\\[0.75em] \implies \underbrace{(x^{-1} \cdot x)}_{= 1} \cdot y &= 1\\[0.75em] \implies \underbrace{1 \cdot y}_{= y} &= 1\\[0.75em] \implies y &= 1. \end{align*}\] Since any element \(y \in F\) acting as a multiplicative identity must be equal to \(1\), it follows that \(1\) is unique. \(\blacksquare\)
The proof of Proposition 4 is unique to fields because rings do not necessarily have multiplicative-inverse elements for every non-zero element.
Proposition 5 Let \(x \in F \setminus \{ 0 \}\). Then \(x^{-1} \in F\) is unique.
Proof. Let \(y \in F \setminus \{ 0 \}\) be such that such that \(x \cdot y = 1\). Then as in the proof of Proposition 4, \[\begin{align*} \underbrace{x^{-1} \cdot (x \cdot y)}_{= (x^{-1} \cdot x) \cdot y} &= \underbrace{x^{-1} \cdot 1}_{= x^{-1}}\\[0.75em] \implies \underbrace{(x^{-1} \cdot x)}_{= 1} \cdot y &= x^{-1}\\[0.75em] \implies \underbrace{1 \cdot y}_{= y} &= x^{-1}\\[0.75em] \implies y &= x^{-1}. \end{align*}\] Since any element \(y \in F \setminus \{ 0 \}\) acting as a multiplicative inverse for \(x \in F \setminus \{ 0 \}\) must be equal to \(x^{-1}\), it follows that \(x^{-1}\) is unique. \(\blacksquare\)
The proof of Proposition 5 is unique to fields because rings do not necessarily have multiplicative-inverse elements for every non-zero element.
Proposition 6 shows that the additive inverse of the additive inverse of an element \(x \in F\) is \(x\) itself.
Proposition 6 Let \(x \in F\). Then \(-(-x) = x\).
Proof. By definition \(x + (-x) = 0\), so \(x\) is an additive inverse of \(-x\). By Proposition 3 additive-inverse elements are unique, whence \(x = -(-x)\). \(\blacksquare\)
The proof of Proposition 6 shows that the inverse of the inverse of an element in a group is the element itself because the proof only uses properties of groups, so this result is not unique to fields.
Theorem 1.3.2 in Katznelson and Katznelson (2024, 11) proves five further algebraic properties of a general field \(F\). These properties are restated and proven below as Proposition 7, Proposition 8, Proposition 9, Proposition 10 and Proposition 11. As before this is done to give a different presentation of and commentary on these properties than that given by Katznelson and Katznelson (2024).
Proposition 7 (Cancellation law of addition) Let \(x, y, z \in F\). If \(x + y = x + z\) then \(y = z\).
Proof. Adding \(-x\) to both sides of the equation, and using associativity of addition and properties of additive inverses and zero, shows that
\[\begin{align*} \underbrace{(-x) + (x + y)}_{= ((-x) + x) + y} &= \underbrace{(-x) + (x + z)}_{= ((-x) + x) + z}\\[0.75em] \implies \underbrace{((-x) + x)}_{= 0} + y &= \underbrace{((-x) + x)}_{= 0} + z\\[0.75em] \implies \underbrace{0 + y}_{= y} &= \underbrace{0 + z}_{= z}\\[0.75em] \implies y &= z, \end{align*}\] so \(x\) can be cancelled from both sides of an equation. \(\blacksquare\)
Similar cancellation laws of addition were proven for the natural numbers in the Section 1.1.2 notes for Katznelson and Katznelson (2024), for the integers in the Section 1.2.1 notes, and for the rational numbers in the Section 1.3.1 notes. Proposition 7 is in fact true for any group because the proof only uses properties of groups, so this result is not unique to fields. From this perspective it is not surprising that the integers and rational numbers each have cancellation laws of addition and moreover that their proofs are essentially identical to the proof of Proposition 7, since \((\symbb{Z}, +)\) and \((\symbb{Q}, +)\) are groups.3 However, the natural numbers are different because \((\symbb{N}, +)\) is not a group. The Section 1.1.2 notes show that there is no identity element of addition for the natural numbers so there can be no additive inverses in the natural numbers; \((\symbb{N}, +)\) is only a semigroup.
Proposition 8, the cancellation law of multiplication for fields, can be proven in two different ways. The first proof is more natural but the second proof is more general.
Proposition 8 (Cancellation law of multiplication) Let \(x \in F \setminus \{ 0 \}\) and \(y, z \in F\). If \(x \cdot y = x \cdot z\) then \(y = z\).
Proof (First). Multiplying both sides of the equation by \(x^{-1}\), and using associativity of multiplication and properties of multiplicative inverses and one, shows that \[\begin{align*} \underbrace{x^{-1} \cdot (x \cdot y)}_{= (x^{-1} \cdot x) \cdot y} &= \underbrace{x^{-1} \cdot (x \cdot z)}_{= (x^{-1} \cdot x) \cdot z}\\[0.75em] \implies \underbrace{(x^{-1} \cdot x)}_{= 1} \cdot y &= \underbrace{(x^{-1} \cdot x)}_{= 1} \cdot z\\[0.75em] \implies \underbrace{1 \cdot y}_{= y} &= \underbrace{1 \cdot z}_{= z}\\[0.75em] \implies y &= z, \end{align*}\] so non-zero \(x\) can be cancelled from both sides of an equation. \(\blacksquare\)
The second proof of Proposition 8 relies on Proposition 9 and Proposition 10.
Proof (Second). Adding \(-(x \cdot z)\) to both sides of the equation, and using properties of additive inverses and zero, shows that \[\begin{align} x \cdot y + (-(x \cdot z)) &= \underbrace{x \cdot z + (-(x \cdot z))}_{= 0} \notag \\[0.75em] \implies x \cdot y + (-(x \cdot z)) &= 0 \label{eq:multcancel-1} . \end{align}\] By Proposition 10 \(-(x \cdot z) = (-1) \cdot (x \cdot z)\), and because multiplication is associative and commutative, it follows that \((-1) \cdot (x \cdot z) = x \cdot ((-1) \cdot z) = x \cdot (-z)\). Connecting this to \(\eqref{eq:multcancel-1}\) and applying the distributive property, \[\begin{align*} 0 &= x \cdot y + (-(x \cdot z)) \\[0.75em] &= x \cdot y + x \cdot (-z)\\[0.75em] &= x \cdot (y + (-z)). \end{align*}\] By Proposition 9 \(x \cdot (y + (-z)) = 0\) implies \((x = 0) \lor (y + (-z) = 0)\). Since by assumption \(x \neq 0\) it follows that \(y + (-z) = 0\). Adding \(z\) to both sides of this equation, and using associativity of addition and properties of additive inverses and zero, shows that \[\begin{align*} (y + (-z)) + z &= 0 + z\\[0.75em] \implies y + ((-z) + z) &= z\\[0.75em] \implies y + 0 &= z\\[0.75em] \implies y = z, \end{align*}\] so unlike the first proof, canceling non-zero \(x\) from both sides of an equation can be justified without invoking \(x^{-1}\) at all. \(\blacksquare\)
Similar cancellation laws of multiplication were proven for the natural numbers in the Section 1.1.2 notes for Katznelson and Katznelson (2024), for the integers in the Section 1.2.1 notes, and for the rational numbers in the Section 1.3.1 notes. The techniques used in the proofs have varied. The first proof of Proposition 8 is unique to fields (such as the rational numbers) because of the use of the multiplicative inverse element \(x^{-1}\), which is not guaranteed to exist for rings (such as the integers). However, the second proof of Proposition 8 is essentially the same as the second proof given for the integers. This implies that the cancellation law of multiplication is true in any commutative ring with no zero divisors. (Such a ring is called an integral domain.)
Proposition 9 (No zero divisors) Let \(x, y \in F\). Then \((x \cdot y = 0) \iff ((x = 0) \lor (y = 0))\).
Proof. Suppose \((x = 0) \lor (y = 0)\) is true. Since at least one of \(x\) and \(y\) is equal to zero, by Proposition 1 it follows that \(x \cdot y = 0\). Conversely, suppose \(x \cdot y = 0\). Either \(x = 0\) or \(x \neq 0\). If \(x = 0\) then the conclusion follows. If \(x \neq 0\) then \(x^{-1}\) exists, so multiplying both sides of the equation by \(x^{-1}\), and using associativity of multiplication and properties of multiplicative inverses and one, and Proposition 1, shows that \[\begin{align*} \underbrace{x^{-1} \cdot (x \cdot y)}_{= (x^{-1} \cdot x) \cdot y} &= \underbrace{x^{-1} \cdot 0}_{= 0}\\[0.75em] \implies \underbrace{(x^{-1} \cdot x)}_{= 1} \cdot y &= 0\\[0.75em] \implies \underbrace{1 \cdot y}_{= y} &= 0\\[0.75em] \implies y &= 0. \end{align*}\] Therefore it must be that at least one of \(x\) and \(y\) is equal to zero. \(\blacksquare\)
Proposition 9 is not unique to fields. In Exercise 1.2.9 for Katznelson and Katznelson (2024) it was proven that for the integers, if \(m, n \in \symbb{Z}\) then \((m \cdot n = 0) \implies (m = 0) \lor (n = 0)\). The converse is implied by Proposition 1.2.1 from Katznelson and Katznelson (2024, 6), so the ring of integers \(\symbb{Z}\) has exactly the same property as Proposition 9 states for fields. However, the proof of Proposition 9 is unique to fields because rings do not necessarily have multiplicative-inverse elements for every non-zero ring element. The proof of Exercise 1.2.9 is in fact completely different from the proof of Proposition 9, relying instead on the order on \(\symbb{Z}\) established in the Section 1.2.2 notes for Katznelson and Katznelson (2024); the proof establishes the contrapositive \((m \neq 0) \land (n \neq 0) \implies (m \cdot n \neq 0)\) and proceeds by case analysis given that \(m\) and \(n\) can each be greater than or less than zero according to the trichotomy law for the integers.
Proposition 10 Let \(x \in F\). Then \((-1) \cdot x = -x\).
Proof. Using that multiplication distributes over addition, properties of additive inverses and 1, and Proposition 1, shows that \[\begin{align*} x + (-1) \cdot x &= 1 \cdot x + (-1) \cdot x\\[0.75em] &= (1 + (-1)) \cdot x\\[0.75em] &= 0 \cdot x\\[0.75em] &= 0, \end{align*}\] so \((-1) \cdot x\) is an additive inverse of \(x\). By Proposition 3 it follows that \((-1) \cdot x\) is the additive inverse of \(x\), whence \((-1) \cdot x = -x\). \(\blacksquare\)
The proof of Proposition 10 is essentially the same as that given in Exercise 1.2.4 for Katznelson and Katznelson (2024). As mentioned in that exercise the proof shows that multiplication by \(-1\) gives the additive inverse in any ring, because the proof uses only properties of rings, so this result is not unique to fields.
Proposition 11 can be proven in two different ways. The first proof is more natural but the second proof is more general.
Proposition 11 Let \(x, y \in F\). Then \((-x) \cdot (-y) = x \cdot y\).
Proof (First). The first part of the proof is to show that \((-1) \cdot (-1) = 1\). Using Proposition 10 it follows that \((-1) \cdot (-1) = -(-1)\), so \((-1) \cdot (-1)\) is an additive inverse of \(-1\). On the other hand, since \(1 + (-1) = 0\), \(1\) is also an additive inverse of \(-1\). By Proposition 3 it follows that \((-1) \cdot (-1) = 1\).
Next, using this result, Proposition 10 and that multiplication is associative and commutative shows that \[\begin{align*} (-x) \cdot (-y) &= ((-1) \cdot x) \cdot ((-1) \cdot y)\\[0.75em] &= ((-1) \cdot (-1)) \cdot (x \cdot y)\\[0.75em] &= 1 \cdot (x \cdot y)\\[0.75em] &= x \cdot y, \end{align*}\] so in product of additive inverses the minus signs cancel. \(\blacksquare\)
Proof (Second). The first part of the proof is to show that \(x \cdot (-y) = (-x) \cdot y = -(x \cdot y)\). Using distributivity of multiplication over addition and Proposition 1 it follows that \(x \cdot y + x \cdot (-y) = x \cdot (y + (-y)) = x \cdot 0 = 0\), so \(x \cdot (-y)\) is an additive inverse of \(x \cdot y\). By Proposition 3 it follows that \(x \cdot (-y) = -(x \cdot y)\). Similarly it follows that \((-x) \cdot y + x \cdot y = ((-x) + x) \cdot y = 0 \cdot y = 0\), so \((-x) \cdot y\) is also an additive inverse of \(x \cdot y\), whence \((-x) \cdot y = -(x \cdot y)\). It follows that \(x \cdot (-y) = (-x) \cdot y = -(x \cdot y)\).
Using this result twice and Proposition 6 shows that \[\begin{align*} (-x) \cdot (-y) &= -(x \cdot (-y))\\[0.75em] &= -(-(x \cdot y))\\[0.75em] &= x \cdot y, \end{align*}\] so in a product of additive inverses the minus signs can be cancelled without appealing to commutativity of multiplication. \(\blacksquare\)
The first proof of Proposition 11 only uses properties of commutative rings so this result holds for commutative rings more generally, not just fields.4 The second proof never uses commutativity of multiplication, that \((-1) \cdot (-1) = 1\) or even the notion of a multiplicative identity at all, so Proposition 11 is in fact true in general (not necessarily commutative) rings, possibly even without unity.
Incidentally, the second proof of Proposition 11 also shows that in any ring, and hence in any field, \(x \cdot (-y) = (-x) \cdot y = -(x \cdot y)\), so it’s unambiguous to simply write \(-x \cdot y\) or \(-xy\).
Because all non-zero elements in a field \(F\) have multiplicative inverses, it is possible to form “fractions” in \(F\). A fraction is special notation for the multiplication of one element by the inverse of another element. That is, if \(x \in F\) and \(y \in F \setminus \{ 0 \}\), then \[\begin{equation}\label{eq:fracdef} \frac{x}{y} \coloneq x/y \coloneq x \cdot y^{-1}. \end{equation}\] It follows from \(\eqref{eq:fracdef}\) that \(y^{-1} = 1 \cdot y^{-1} = 1/y\). As with the rational numbers in the Section 1.3.1 notes for Katznelson and Katznelson (2024), \(x\) is called the numerator of the fraction and \(y\) is called the denominator.
The following propositions demonstrate the algebraic properties of fractions. Because fractions are in general unique to fields most of the following propositions do not require comparison with previous results or with more rudimentary structures such as rings, groups or semigroups.
Proposition 12 is similar to Proposition 6 because the two propositions show that for both addition and multiplication, the inverse of the inverse is the original element.
Proposition 12 Let \(x \in F \setminus \{ 0 \}\). Then \((x^{-1})^{-1} = x\). Equivalently, \[ 1 \bigg/ \frac{1}{x} = x. \]
Proof. By definition \(x \cdot x^{-1} = 1\), so \(x\) is a multiplicative inverse of \(x^{-1}\). By Proposition 5 multiplicative-inverse elements are unique, whence \(x = (x^{-1})^{-1}\). \(\blacksquare\)
Proposition 13 Let \(x, y \in F \setminus \{ 0 \}\). Then \((x \cdot y^{-1})^{-1} = y \cdot x^{-1}\). Equivalently, \[ \left( \frac{x}{y} \right)^{-1} = 1 \bigg/ \frac{x}{y} = \frac{y}{x}. \]
Proof. Because multiplication in fields is associative and commutative, \[\begin{align*} (x \cdot y^{-1}) \cdot (y \cdot x^{-1}) &= \underbrace{(x \cdot x^{-1})}_{= 1} \cdot \underbrace{(y^{-1} \cdot y)}_{= 1}\\[0.75em] &= 1, \end{align*}\] so \(y \cdot x^{-1}\) is a multiplicative inverse of \(x \cdot y^{-1}\). By Proposition 5 multiplicative-inverse elements are unique, whence \((x \cdot y^{-1})^{-1} = y \cdot x^{-1}\). \(\blacksquare\)
Proposition 14 Let \(x, y \in F \setminus \{ 0 \}\). Then \((x \cdot y)^{-1} = x^{-1} \cdot y^{-1}\). Equivalently, \[ \frac{1}{x \cdot y} = \frac{1}{x} \cdot \frac{1}{y}. \]
Proof. Because multiplication in fields is associative and commutative, \[\begin{align*} (x \cdot y) \cdot (x^{-1} \cdot y^{-1}) &= \underbrace{(x \cdot x^{-1})}_{= 1} \cdot \underbrace{(y \cdot y^{-1})}_{= 1}\\[0.75em] &= 1, \end{align*}\] so \(x^{-1} \cdot y^{-1}\) is a multiplicative inverse of \(x \cdot y\). By Proposition 5 multiplicative-inverse elements are unique, whence \((x \cdot y)^{-1} = x^{-1} \cdot y^{-1}\). \(\blacksquare\)
Proposition 15 Let \(x, z \in F\) and \(y \in F \setminus \{ 0 \}\). Then \(z \cdot y = x \iff z = x / y\).
Proof. If \(z \cdot y = x\), then multiplying both sides by \(y^{-1}\), and using that multiplication is associative and properties of one, shows that \[\begin{align*} \underbrace{(z \cdot y) \cdot y^{-1}}_{= z \cdot (y \cdot y^{-1})} &= \underbrace{x \cdot y^{-1}}_{= x / y}\\[0.75em] \implies z \cdot \underbrace{(y \cdot y^{-1})}_{= 1} &= x / y\\[0.75em] \implies \underbrace{z \cdot 1}_{= z} &= x / y\\[0.75em] \implies z &= x / y, \end{align*}\] so a non-zero factor can be moved “down” to be a denominator on the other side of the equation. Conversely, if \(z = x / y = x \cdot y^{-1}\), then multiplying both sides by \(y\) and using the same properties as as those mentioned above shows that \[\begin{align*} z \cdot y &= \underbrace{(x \cdot y^{-1}) \cdot y}_{= x \cdot (y^{-1} \cdot y)}\\[0.75em] &= x \cdot \underbrace{(y^{-1} \cdot y)}_{= 1}\\[0.75em] &= \underbrace{x \cdot 1}_{= x}\\[0.75em] &= x, \end{align*}\] so a denominator can be moved “up” to the other side of the equation. \(\blacksquare\)
Proposition 16 shows that equality of two fractions in a field is similar to the equivalence relation used to construct the rational numbers in the Section 1.3 notes for Katznelson and Katznelson (2024).
Proposition 16 Let \(x, z \in F\) and \(y, \alpha \in F \setminus \{ 0 \}\). Then \[ \frac{x}{y} = \frac{z}{\alpha} \iff x \cdot \alpha = y \cdot z. \]
Proof. If \(x / y = z / \alpha\) then applying Proposition 15 shows that \((x / y) \cdot \alpha = z\). Using associativity and commutativity of multiplication to rearrange the factors shows that \((x / y) \cdot \alpha = (x \cdot \alpha) / y = z\), so applying Proposition 15 again shows that \(x \cdot \alpha = y \cdot z\).
Conversely, if \(x \cdot \alpha = y \cdot z\) then applying Proposition 15 and rearranging factors shows that \(x = (y \cdot z) / \alpha = (z / \alpha) \cdot y\). Applying Proposition 15 again shows that \(x / y = z / \alpha\). \(\blacksquare\)
Proposition 17 and Proposition 18 show that given the field axioms, arithmetic with fractions works just like the arithmetic of rational numbers as defined in the Section 1.3.1 notes for Katznelson and Katznelson (2024).
Proposition 17 Let \(x, z \in F\) and \(y, \alpha \in F \setminus \{ 0 \}\). Then \[ \frac{x}{y} \pm \frac{z}{\alpha} = \frac{x \cdot \alpha \pm y \cdot z}{y \cdot \alpha}. \]
Proof. Using properties of multiplicative inverses, one, associativity and commutativity of multiplication, Proposition 14, that multiplication distributes over addition shows that for addition of fractions, \[\begin{align*} \frac{x}{y} + \frac{z}{\alpha} &= (x \cdot y^{-1}) \cdot (\alpha \cdot \alpha^{-1}) + (z \cdot \alpha^{-1}) \cdot (y \cdot y^{-1})\\[0.75em] &= (x \cdot \alpha) \cdot \underbrace{(y^{-1} \cdot \alpha^{-1})}_{= (y \cdot \alpha)^{-1}} + (y \cdot z) \cdot \underbrace{(y^{-1} \cdot \alpha^{-1})}_{= (y \cdot \alpha)^{-1}}\\[0.75em] &= (x \cdot \alpha) \cdot (y \cdot \alpha)^{-1} + (y \cdot z) \cdot (y \cdot \alpha)^{-1}\\[0.75em] &= (x \cdot \alpha + y \cdot z) \cdot (y \cdot \alpha)^{-1}\\[0.75em] &= \frac{x \cdot \alpha + y \cdot z}{y \cdot \alpha}. \end{align*}\] For subtraction, using the same properties as those mentioned above plus Proposition 10 shows that \[\begin{align*} \frac{x}{y} - \frac{z}{\alpha} &= (x \cdot y^{-1}) \cdot (\alpha \cdot \alpha^{-1}) + ((-1) \cdot (z \cdot \alpha^{-1})) \cdot (y \cdot y^{-1})\\[0.75em] &= (x \cdot \alpha) \cdot \underbrace{(y^{-1} \cdot \alpha^{-1})}_{= (y \cdot \alpha)^{-1}} + ((-1) \cdot (y \cdot z)) \cdot \underbrace{(y^{-1} \cdot \alpha^{-1})}_{= (y \cdot \alpha)^{-1}}\\[0.75em] &= (x \cdot \alpha + (-1) \cdot (y \cdot z)) \cdot (y \cdot \alpha)^{-1}\\[0.75em] &= \frac{x \cdot \alpha - y \cdot z}{y \cdot \alpha}, \end{align*}\] so subtraction of fractions works just like addition. \(\blacksquare\)
Proposition 18 Let \(x, z \in F\) and \(y, \alpha \in F \setminus \{ 0 \}\). Then \[ \frac{x}{y} \cdot \frac{z}{\alpha} = \frac{x \cdot z}{y \cdot \alpha}. \]
Proof. Using associativity and commutativity of multiplication and Proposition 14 shows that \[\begin{align*} \frac{x}{y} \cdot \frac{z}{\alpha} &= (x \cdot y^{-1}) \cdot (z \cdot \alpha^{-1})\\[0.75em] &= (x \cdot z) \cdot \underbrace{(y^{-1} \cdot \alpha^{-1})}_{= (y \cdot \alpha)^{-1}}\\[0.75em] &= (x \cdot z) \cdot (y \cdot \alpha)^{-1}\\[0.75em] &= \frac{x \cdot z}{y \cdot \alpha}, \end{align*}\] so the numerators and denominators of two fractions can be multiplied together. \(\blacksquare\)
Proposition 19 Let \(x \in F\) and \(y, z \in F \setminus \{ 0 \}\). Then \[ \frac{x}{y} = \frac{x \cdot z}{y \cdot z}. \]
Proof. Using properties of multiplicative inverses, one, and Proposition 18 shows that \[\begin{align*} \frac{x}{y} &= \frac{x}{y} \cdot (z \cdot z^{-1})\\[0.75em] &= \frac{x}{y} \cdot \frac{z}{z}\\[0.75em] &= \frac{x \cdot z}{y \cdot z}, \end{align*}\] so both numerator and denominator can be multiplied by a common non-zero factor. \(\blacksquare\)
Proposition 20 shows that in a fraction of fractions, the denominator can be flipped and multiplied with the numerator.
Proposition 20 Let \(x \in F\) and \(y, z, \alpha \in F \setminus \{ 0 \}\). Then \[ \frac{x}{y} \bigg/ \frac{z}{\alpha} = \frac{x \cdot \alpha}{y \cdot z}. \]
Proof. Using Proposition 13 and Proposition 18 shows that \[\begin{align*} \frac{x}{y} \bigg/ \frac{z}{\alpha} &= \frac{x}{y} \cdot \underbrace{\left( \frac{z}{\alpha} \right)^{-1}}_{= \frac{\alpha}{z}}\\[0.75em] &= \frac{x}{y} \cdot \frac{\alpha}{z}\\[0.75em] &= \frac{x \cdot \alpha}{y \cdot z}, \end{align*}\] so when dividing one fraction by another, the fraction in the denominator can be flipped and multiplied with the fraction in the numerator. \(\blacksquare\)
The Section 1.3.1 notes for Katznelson and Katznelson (2024) showed that linear equations in one unknown are solvable in the rational numbers. Proposition 21 shows that this is true in general fields as well.
Proposition 21 Let \(x \in F \setminus \{ 0 \}\) and \(y, z \in F\). Then there exists unique \(\alpha \in F\) such that \(x \cdot \alpha + y = z\).
Proof. Let \(\alpha \coloneq x^{-1} \cdot (z + (-y))\). Then using many of the algebraic properties of fields shows that \[\begin{align*} x \cdot \alpha + y &= x \cdot (x^{-1} \cdot (z + (-y))) + y\\[0.75em] &= \underbrace{(x \cdot x^{-1})}_{= 1} \cdot (z + (-y)) + y\\[0.75em] &= \underbrace{1 \cdot (z + (-y))}_{= z + (-y)} + y\\[0.75em] &= (z + (-y)) + y. \end{align*}\] Continuing, \[\begin{align*} (z + (-y)) + y &= z + \underbrace{((-y) + y)}_{= 0}\\[0.75em] &= z + 0\\[0.75em] &= z, \end{align*}\] so at least one solution to the equation exists.
Suppose there were another \(\beta \in F\) such that \(\beta \neq \alpha\) and \(x \cdot \beta + y = z\). Then using the same arithmetic properties shows that \[\begin{align*} x \cdot \beta + y &= z\\[0.75em] \implies (x \cdot \beta + y) + (-y) &= z + (-y)\\[0.75em] \implies x \cdot \beta + (y + (-y)) &= z + (-y)\\[0.75em] \implies x \cdot \beta &= z + (-y)\\[0.75em] \implies x^{-1} \cdot (x \cdot \beta) &= x^{-1} \cdot (z + (-y)). \end{align*}\] Continuing, \[\begin{align*} (x^{-1} \cdot x) \cdot \beta &= x^{-1} \cdot (z + (-y))\\[0.75em] \implies \beta &= \underbrace{x^{-1} \cdot (z + (-y))}_{= \alpha}\\[0.75em] \implies \beta &= \alpha, \end{align*}\] which is a contradiction. Therefore \(\beta = \alpha\) so the solution to the equation is unique. \(\blacksquare\)
To this point in the Chapter 1 notes for Katznelson and Katznelson (2024), isomorphisms have found use in justifying the conventions of writing \(\symbb{N}\subseteq \symbb{Z}\) and \(\symbb{Z}\subseteq \symbb{Q}\) in the Section 1.2 and Section 1.3.1 notes for Katznelson and Katznelson (2024), respectively. Specifically, in the former case there was a special subset \(\tilde{\symbb{N}} \subseteq \symbb{Z}\) and an isomorphism between \(\symbb{N}\) and \(\tilde{\symbb{N}}\) was used to establish that \(\symbb{N}\) and \(\tilde{\symbb{N}}\) are not meaningfully different. Similarly, in the latter case there was a special subset \(\tilde{\symbb{Z}} \subseteq \symbb{Q}\) and an isomorphism between \(\symbb{Z}\) and \(\tilde{\symbb{Z}}\) was used to establish that \(\symbb{Z}\) and \(\tilde{\symbb{Z}}\) are not meaningfully different.
Considering two objects or structures \(X\) and \(\tilde{X}\) of the same type—whether they be sets, Peano systems, groups, rings, fields, topological spaces, metric spaces, manifolds or something else besides—and trying to show that they are essentially the same is a recurring theme in mathematics.5 After using isomorphisms to prove that two objects are not meaningfully different in the specific cases of the natural numbers and integers, isomorphisms can now be discussed in general. If there is an isomorphism between structures \(X\) and \(\tilde{X}\) of the same type, instead of saying that \(X\) and \(\tilde{X}\) are equal one says that they are isomorphic. Thus instead of writing \(X = \tilde{X}\) one writes \(X \cong \tilde{X}\).
The specific definition of an isomorphism will depend on the type of objects under consideration. For fields \((F, +, \cdot)\) and \((\tilde{F}, \mathbin{\tilde{+}}, \mathbin{\tilde{\cdot}})\) an isomorphism \(\iota \colon F \to \tilde{F}\) is a bijection such that for all \(x, y \in F\), \(\iota(x + y) = \iota(x) \mathbin{\tilde{+}}\iota(y)\) and \(\iota(x \cdot y) = \iota(x) \mathbin{\tilde{\cdot}}\iota(y)\). This is consistent with the isomorphisms previously encountered for natural numbers and integers. In the case of fields an isomorphism formalizes the idea that it makes no difference whether field arithmetic is done on elements of \(F\) within \(F\) and the result is considered as an element of \(\tilde{F}\), or elements of \(F\) are considered as elements of \(\tilde{F}\) and the corresponding arithmetic operations are done in \(\tilde{F}\). Stated another way, an isomorphism formalizes the idea that if two fields \(F\) and \(\tilde{F}\) are essentially the same, then there should be no difference between arithmetic done on arbitrary \(x, y \in F\) and the same arithmetic done on the corresponding \(\iota(x), \iota(y) \in \tilde{F}\).
The three most important fields in analysis are the rational numbers \(\symbb{Q}\), the real numbers \(\symbb{R}\) and the complex numbers \(\symbb{C}\). The rational numbers are introduced in the Section 1.3 notes for Katznelson and Katznelson (2024); the real numbers are introduced in the Section 1.5, and the complex numbers are introduced in the Section 1.9 notes.
However, merely describing \(\symbb{Q}\), \(\symbb{R}\) and \(\symbb{C}\) as fields is to sell short their most salient qualities. Both \(\symbb{Q}\) and \(\symbb{R}\) are ordered fields, although \(\symbb{C}\) is not ordered. In addition \(\symbb{R}\) and \(\symbb{C}\) are complete, whereas \(\symbb{Q}\) is not complete. Pithily it may be said that analysis, broadly interpreted, is undergirded by the additional qualities that \(\symbb{R}\) and \(\symbb{C}\) have relative to \(\symbb{Q}\). The elucidation of these differences proceeds over many sections but it begins with the discussion of ordered fields in the Section 1.3.3 notes for Katznelson and Katznelson (2024).
A variety of fields are encountered in algebra; one example is \(\symbb{Z}/ p\symbb{Z}\) where \(p \in \symbb{N}\) is prime. For proof that \(\symbb{Z}/ p\symbb{Z}\) is a field see Exercise 1.1.4 from Katznelson and Katznelson (2008). This leads to the theory of finite fields. Fields in general are discussed briefly in Section 1.1 of Katznelson and Katznelson (2008) and Section 3.2 of Artin (1991). Finite fields are discussed in greater depth in Section 13.6 of Artin (1991) and Appendix A.4 of Taylor (2020).
Fields are also prominent in linear algebra. Linear algebra revolves around the notion of a vector space, which is informally an abelian group \(V\) and a field \(F\) that interact well together. The elements of the group are called vectors and the elements of the field are called scalars. As an example of the interaction between vectors and scalars, if \(v \in V\) and \(s, t \in F\) then it is required axiomatically for a vector space that \((s + t) \cdot v = s \cdot v + t \cdot v\).6
Commonly the study of linear algebra proceeds assuming the field of scalars is \(\symbb{R}\) or \(\symbb{C}\), but it is not always necessary to assume the field of scalars is any field in particular. Linear algebra using a general scalar field \(F\) is studied in Katznelson and Katznelson (2008) and the accompanying notes, and in Chapter 6 of Taylor (2020).
Katznelson and Katznelson (2024) use the symbol \(\symbb{F}\) for a general field. However, keeping with the convention of using italic type for generic mathematical objects and roman type for special mathematical objects, these notes use the symbol \(F\) for a general field instead.↩︎
The real numbers are introduced in the Section 1.5 notes for Katznelson and Katznelson (2024), and the complex numbers are introduced in the Section 1.9 notes for Katznelson and Katznelson (2024).↩︎
In particular it is the second, more sophisticated proof of the cancellation law of addition for the integers given in the Section 1.2.1 notes for Katznelson and Katznelson (2024) that is like the proofs of the corresponding result for rational numbers and general fields. The first proof given for the integers uses integers in terms of their construction as equivalence classes of ordered pairs of natural numbers and relies on the cancellation law of addition for natural numbers.↩︎
Incidentally, a lemma in Exercise 1.2.5 of Katznelson and Katznelson (2024) proves that in the integers \((-1) \cdot (-1) = 1\), although the lemma was used towards a different end. Because the integers are a commutative ring Proposition 11 is true for \(\symbb{Z}\) as well.↩︎
Here \(X\) and \(\tilde{X}\) are sets, and writing \(X\) and \(\tilde{X}\) is shorthand for the full structures of \(X\) and \(\tilde{X}\) according to their common type. For example, if they were rings then \(X\) and \(\tilde{X}\) would be shorthand for \((X, +, \cdot)\) and \((\tilde{X}, \mathbin{\tilde{+}}, \mathbin{\tilde{\cdot}})\). If they were topological spaces then \(X\) and \(\tilde{X}\) would be shorthand for \((X, O)\) and \((\tilde{X}, \tilde{O})\).↩︎
The notation in linear algebra is overloaded. The group \((V, +)\) is usually written with an additive binary operation \(+ \colon V \times V \to V\), which clashes with the notation for the additive binary operation of the field \((F, +, \cdot)\) which is also written as \(+ \colon F \times F \to F\). Furthermore the multiplication of scalars with vectors as in the expressions \((s + t) \cdot v\), \(s \cdot v\) and \(t \cdot v\) means that there is binary operation \(\cdot \colon F \times V \to V\) which clashes with the notation for the field multiplication \(\cdot \colon F \times F \to F\). The scalar multiplication and scalar-vector multiplication are often left implicit.↩︎