Fields

Computations involving only integers or only rational numbers are much easier than those with real numbers. Could other algebraic structures, such as the integers or the rationals, work in the place of $ℝ$ in the definition of a vector space?

If we take “work” to mean that the results of this chapter remain true then there is a natural list of conditions that a structure (that is, number system) must have in order to work in the place of $ℝ$ . A field is a set $ℱ$ with operations ‘ $+$ ’ and ‘ $\cdot$ ’ such that

for any $a, b \in ℱ$ the result of $a + b$ is in $ℱ$ , and $a + b = b + a$ , and if $c \in ℱ$ then $a + (b + c) = (a + b) + c$
for any $a, b \in ℱ$ the result of $a \cdot b$ is in $ℱ$ , and $a \cdot b = b \cdot a$ , and if $c \in ℱ$ then $a \cdot (b \cdot c) = (a \cdot b) \cdot c$
if $a, b, c \in ℱ$ then $a \cdot (b + c) = a \cdot b + a \cdot c$
there is an element $0 \in ℱ$ such that if $a \in ℱ$ then $a + 0 = a$ , and for each $a \in ℱ$ there is an element $- a \in ℱ$ such that $(- a) + a = 0$
there is an element $1 \in ℱ$ such that if $a \in ℱ$ then $a \cdot 1 = a$ , and for each element $a \neq 0$ of $ℱ$ there is an element $a^{- 1} \in ℱ$ such that $a^{- 1} \cdot a = 1$ .

For example, the algebraic structure consisting of the set of real numbers along with its usual addition and multiplication operation is a field. Another field is the set of rational numbers with its usual addition and multiplication operations. An example of an algebraic structure that is not a field is the integers, because it fails the final condition.

Some examples are more surprising. The set $𝔹 = {0, 1}$ under these operations:

$+$	$0$	$1$
$0$	$0$	$1$
$1$	$1$	$0$

$\cdot$	$0$	$1$
$0$	$0$	$0$
$1$	$0$	$1$

is a field; see Exercise 4.

We could in this book develop Linear Algebra as the theory of vector spaces with scalars from an arbitrary field. In that case, almost all of the statements here would carry over by replacing ‘ $ℝ$ ’ with ‘ $ℱ$ ’, that is, by taking coefficients, vector entries, and matrix entries to be elements of $ℱ$ (the exceptions are statements involving distances or angles, which would need additional development). Here are some examples; each applies to a vector space $V$ over a field $ℱ$ .

$*$ For any $\vec{v} \in V$ and $a \in ℱ$ , (i) $0 \cdot \vec{v} = \vec{0}$ , (ii) $- 1 \cdot \vec{v} + \vec{v} = \vec{0}$ , and (iii) $a \cdot \vec{0} = \vec{0}$ .
$*$ The span, the set of linear combinations, of a subset of $V$ is a subspace of $V$ .
$*$ Any subset of a linearly independent set is also linearly independent.
$*$ In a finite-dimensional vector space, any two bases have the same number of elements.

(Even statements that don’t explicitly mention $ℱ$ use field properties in their proof.)

We will not develop vector spaces in this more general setting because the additional abstraction can be a distraction. The ideas we want to bring out already appear when we stick to the reals.

The exception is Chapter Five. There we must factor polynomials, so we will switch to considering vector spaces over the field of complex numbers.

Exercises

Exercise 1 Worked answer

Check that the real numbers form a field.

Back to Exercise 1

Answer. Going through the five conditions shows that they are all familiar from elementary mathematics.
Exercise 2 Worked answer

Prove that these are fields.
1. The rational numbers $ℚ$
2. The complex numbers $ℂ$
Back to Exercise 2

Answer. As with the prior question, going through the five conditions shows that for both of these structures, the properties are familiar.
Exercise 3 Worked answer

Give an example that shows that the integer number system is not a field.

Back to Exercise 3

Answer. The integers fail condition (5). For instance, there is no multiplicative inverse for $2$ —while $2$ is an integer, $1 / 2$ is not.
Exercise 4 Worked answer

Check that the set $𝔹 = {0, 1}$ is a field under the operations listed above,

Back to Exercise 4

Answer. We can do these checks by listing all of the possibilities. For instance, to verify the first half of condition (2) we must check that the structure is closed under addition and that addition is commutative $a + b = b + a$ , we can check both of these for all possible pairs $a$ and $b$ because there are only four such pairs. Similarly, for associativity, there are only eight triples $a$ , $b$ , $c$ , and so the check is not too long. (There are other ways to do the checks; in particular, you may recognize these operations as arithmetic modulo $2$ . But an exhaustive check is not onerous)
Exercise 5 Worked answer

Give suitable operations to make the set ${0, 1, 2}$ a field.

Back to Exercise 5

Answer. These will do.

$+$ $0$ $1$ $2$

$0$ $0$ $1$ $2$

$1$ $1$ $2$ $0$

$2$ $2$ $0$ $1$

$\cdot$ $0$ $1$ $2$

$0$ $0$ $0$ $0$

$1$ $0$ $1$ $2$

$2$ $0$ $2$ $1$

As in the prior item, we could verify that they satisfy the conditions by listing all of the cases.