Fields

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The sets \(\mathbb{Q}\), \(\mathbb{R}\), and \(\mathbb{C}\) behave in a similar manner. They all have addition, subtraction, multiplication, and division that satisfy certain properties. This common behavior is summarised in the notion of fields. Thus, \(\mathbb{Q}\), \(\mathbb{R}\), and \(\mathbb{C}\) are examples of fields.




Definition (Field)

A set \(F\) is called a field if it is endowed with binary operations \(+\) and \(*\) and satisfies the following axioms.
  1. \(F\) is closed under \(+\) (closure): \(\forall a, b \in F\), \(a + b \in F\).
  2. \(+\) is commutative: \(\forall a, b\in F\), \(a + b = b + a\).
  3. \(+\) is associative: \(\forall a, b, c\in F\), \(a + (b + c) = (a + b) + c\).
  4. There is an additive identity element \(0\) with respect to \(+\): \(\exists 0 \in F\) such that \(\forall a\in F\), \(a + 0 = a\).
  5. There is an inverse of each element with respect to \(+\): \(\forall a\in F,\)\(\exists -a \in F\) such that \(a + (-a) = 0\).
  6. \(F\) is closed under \(*\) (closure): \(\forall a,b \in F\), \(a*b \in F\).
  7. \(*\) is commutative: \(\forall a,b\in F\), \(a * b = b*a\).
  8. \(*\) is associative: \(\forall a,b, c\in F\), \(a * (b*c) = (a*b)*c\).
  9. There is a multiplicative identity \(1\) w.r.t. \(*\): \(\exists 1 \in F\) such that \(\forall a\in F\), \(a*1 = a\).
  10. There is an inverse of each element except \(0\) w.r.t. \(*\): \(\forall a \in F\setminus \{0\},\)\(\exists a^{-1}\) such that \(a * a^{-1} = 1\).
  11. Distributive law: \(\forall a, b, c\in F\), \(a*(b + c) = a*b + a*c\).
We usually call \(+\) and \(*\) ``addition'' and ``multiplication'', respectively.

Remark. Multiplication \(*\) may be denoted by \(\cdot\) or \(\times\); or these symbols may be omitted (as in \(ab\) for \(a*b\)).  □

Example. 
  1. \(\mathbb{Q}\) is a field.
  2. \(\mathbb{R}\) is a field.
  3. \(\mathbb{C}\) is a field.
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Example. 
  1. \(\mathbb{N}\) is not a field. Because \(0\not\in\mathbb{N}\). Even if we adopt the convention that \(0\in \mathbb{N}\), the additive inverse of \(1\) is \(-1\not\in\mathbb{N}\).
  2. \(\mathbb{Z}\) is not a field. Because the multiplicative inverse of \(2\) is \(1/2\not\in\mathbb{Z}\).
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Example. Let \(A = \{1, 2, 3\}\). \(A\) is not closed under the usual addition and multiplication of numbers. For example, \(2 + 3 = 5 \notin A\) and \(2\times 3 = 6 \notin A\). □

What's nice about defining the field is that we can prove some properties common to all fields so that we don't have to repeat the same proof for each of, say, \(\mathbb{Q}\), \(\mathbb{R}\) and \(\mathbb{C}\).

Proposition 

Let \(F\) be a field. We have the following:
  1. For all \(f \in F\), \(f\cdot 0 = 0\).
  2. For all \(h, k \in F\), if \(hk = 0\), then \( h= 0\) or \(k = 0\).
Proof. 
  1. By the definition of \(0\) (additive identity), for any \(f\) we have \[f\cdot 0 = f\cdot (0 + 0).\] By the distributive law, \[f\cdot 0 = f\cdot 0 + f \cdot 0.\] Add the inverse of \(f\cdot 0\) on both sides, \[-(f\cdot 0) + f\cdot 0 = -(f\cdot 0) + (f\cdot 0 + f \cdot 0).\] By the associativity of addition, \[-(f\cdot 0) + f\cdot 0 = (-(f\cdot 0) + f\cdot 0) + f \cdot 0 = 0 + f\cdot 0.\] By the definitions of an additive inverse and \(0\), \[0 = f\cdot 0.\] And we are done.
  2. Suppose \(hk = 0\). If \(h = 0\), we are done. Suppose \(h\neq 0\). Then its multiplicative inverse \(h^{-1}\) exists. Then we have, by using Part 1, \[h^{-1}(hk) = h^{-1}0 = 0.\] But we have \[\begin{align} h^{-1}(hk) &= (h^{-1}h)k &\text{(associativity of multiplication)}\\ &= 1\cdot k &\text{(definition of multiplicative inverse)}\\ &= k &\text{(definition of multiplicative identity)}. \end{align}\] Thus, \(k = 0\).
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