Under addition and multiplication. the set of rational numbers and the set of real numbers share some common properties. What is perhaps surprising is that a great number of these properties are also shared by \(GF(2)\) under multiplication and addition with the rule \(1+1 = 0\). and many other structures. Over the years, these properties have been extracted to form the definition of a field. Because there are so many different examples of fields, by studying fields and their properties, one can deduce in one stroke what is true for all the examples without looking at them individually.

It is possible to study linear algebra just over the real numbers. However, it is in fact more convenient to study linear algebra over different fields early on so we do not have to reinvent the wheel.

More formally, a field \(\mathbb{F}\) is a set equipped with two binary operations \(+\) and \(\cdot\) (called addition and multiplication, respectively) such that the elements are closed under such operations; i.e. for all \(a,b\in \mathbb{F}\), \(a+b\) and \(a\cdot b\) are again elements of \(\mathbb{F}\) and satisfying the following properties:

Properties of a field

1. For all \(a, b,\) and \(c \in \mathbb{F}\), the following hold: \(a + (b+c) = (a+b)+ c\) and \(a\cdot (b\cdot c) = (a\cdot b)\cdot c\). (Associativity of addition and multiplication.)

2. For all \(a\) and \(b \in \mathbb{F}\), the following hold: \(a + b = b + a\) and \(a\cdot b = b\cdot a\). (Commutativity of addition and multiplication.)

3. For all \(a, b,\) and \(c \in \mathbb{F}\), the following holds: \(a \cdot (b+c) = a \cdot b+ a\cdot c\). (Left-distributivity.)

4. There exists an element of \(\mathbb{F}\), denoted by 0, such that for all \(a \in \mathbb{F}, a + 0 = a\). (Existence of additive identity.)

5. There exists an element of \(\mathbb{F}\) not equal to 0, denoted by 1, such that for all \(a \in \mathbb{F}, a \cdot 1 = a\). (Existence of multiplicative identity.)

6. For each \(a \in \mathbb{F}\), there exists an element \(b \in \mathbb{F}\), such that \(a+b = 0\). (Existence of additive inverse, often denoted as \(-a\).)

7. For each \(a \in \mathbb{F}\) not equal to 0, there exists an element \(b \in \mathbb{F}\), such that \(a\cdot b = 1\). (Existence of multiplicative inverse, often denoted as \(a^{-1}\).)

For convenience, one often writes \(ab\) instead of \(a\cdot b\).

Note that one can prove from these properties that a field also satisfies right-distributivity: \((a+b)\cdot c = a\cdot c + b \cdot c\) for all \(a,b,c\in \mathbb{F}\). Also, there is a unique additive identity and a unique multiplicative identity.

If we look at the set of integers \(\mathbb{Z}\) with the usual addition and multiplication, we see that properties 1 to 6 are satisfied but property 7 is not. Therefore, \(\mathbb{Z}\) is not a field. But \(\mathbb{Q}\) and \(\mathbb{R}\) do satisfy all seven properties and therefore they are fields. One can also check that \(GF(2)\) satisfies all the above properties and is therefore a field. We will later see that the complex numbers also form a field.

Powers of field elements

Let \(a \in \mathbb{F}\) where \(\mathbb{F}\) denotes a field. \(a^0\) is defined to be \(1\), the multiplicative identity in \(\mathbb{F}\). For every positive integer \(n\), \(a^n\) is defined to be \(a\cdot a^{n-1}\) and if \(a \neq 0\), \(a^{-n}\) is defined to be the multiplicative inverse of \(a^n\).

The set \(\mathbb{F}^n\)

Very often, we consider \(n\)-tuples whose entries are from the same field. We use \(\mathbb{F}^n\) to denote the set of all \(n\)-tuples whose entries are from \(\mathbb{F}\).

Examples

1. \(\mathbb{R}^3\) consists of all tuples of the form \(\begin{bmatrix} a\\b\\c \end{bmatrix}\) where \(a,b,c\in \mathbb{R}\).

2. \(\begin{bmatrix} 1\\0\\0 \\1\end{bmatrix}\) is an element of \(GF(2)^4\).

Quick Quiz

Exercise

1. Let \(a,b \in \mathbb{Q}\).

   1. Give the additive inverse of \(a + b\sqrt{2}\) in the form \(x+y\sqrt{2}\) where \(x, y \in \mathbb{Q}\).  

      \((-a)+(-b)\sqrt{2}\).

   2. Suppose that not both \(a\) and \(b\) are zero. Give the multiplicative inverse of \(a + b\sqrt{2}\) in the form \(x+y\sqrt{2}\) where \(x, y \in \mathbb{Q}\).  

      \(\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}\).

2. Check that \(GF(2)\) is a field. (Recall that “1+1 = 0”.)  

   We show how to check property 1 for addition, proprty 4, and property 7.

   Consider the following table: \[ \begin{array}{|c|c|c|c|c|c|c|} \hline a & b & c & b+c & a+b & a+(b+c) & (a+b) + c \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 & 1 \\ \hline \end{array} \] Since the last two columns are the same, property 1 for addition holds.

   For property 4, note that the element \(0\) in \(GF(2)\) is an additive identity since \(0 + 0 = 0\) and \(1 + 0= 1\).

   For property 7, note that the only element in \(GF(2)\) not equal to \(0\) is \(1\). We need to show that \(1\) has a multiplicative inverse. But \(1 \cdot 1 = 1\). Hence, \(1\) is the multplicative inverse of \(1\).