Section 2.1

Fields

To define a linear structure, we first need to choose a “base” for developing the notion of linearity. In Euclidean geometry, we work over the real numbers $\mathbb{R}$, roughly meaning that the notion of linearity works over $\mathbb{R}$. For instance, we can take a vector $v \in \mathbb{R}^2$ in the Euclidean plane and multiply it by any real number. A field is supposed to be the scalar number system by which we can multiply a vector.

Definition 2.1.1

A field $k$ is a set with the choice of two elements1 $0, 1 \in k$ and two maps $+, \cdot : k \times k \to k$ satisfying the following conditions: (Here, we denote $+(a, b) = a + b$ and $\cdot (a, b) = a \cdot b$ because they act like addition and multiplication.)

  1. (F0)

    $0 \neq 1$.

  2. (F1)

    For all $a \in k$ we have $a + 0 = 0 + a = a$.

  3. (F2)

    For all $a, b \in k$ we have $a + b = b + a$.

  4. (F3)

    For all $a, b, c \in k$ we have $a + (b + c) = (a + b) + c$.

  5. (F4)

    For all $a \in k$ there exists an $(-a) \in k$ such that $a + (-a) = (-a) + a = 0$.

  6. (F5)

    For all $a \in k$ we have $a \cdot 1 = 1 \cdot a = a$

  7. (F6)

    For all $a, b \in k$ we have $a \cdot b = b \cdot a$.

  8. (F7)

    For all $a, b, c \in k$ we have $a \cdot (b \cdot c) = (a \cdot b) \cdot c$.

  9. (F8)

    For all $a \in k$ with $a \neq 0$ there exists an $a^{-1} \in k$ such that $a \cdot a^{-1} = a^{-1} \cdot a = 1$.

  10. (F9)

    For all $a, b, c \in k$ we have $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$.

The first axiom (F0) is stating that the field is not degenerate, i.e., has at least two elements. The next axioms (F1) to (F4) tell us about how addition behaves. Addition should have an identity element called $0$, be commutative, be associative, and have inverses. The axioms (F5) to (F8) state analogous properties for multiplication. One difference is that $0$ need not have a multiplicative inverse. The last axiom (F9) states that multiplication distributes over addition.

Example 2.1.2

The set of rational numbers $\mathbb{Q}$ is a field under usual addition and multiplication. The set of real numbers $\mathbb{R}$ and the set of complex numbers $\mathbb{C}$ are also fields under normal addition and multiplication.

Example 2.1.3

Let $p$ be a prime number. We consider the set

$$ \mathbb{F}_p = \lbrace 0, 1, 2, \ldots, p-1 \rbrace , $$

and define addition and multiplication modulo $p$. Then $\mathbb{F}_p$ is a field.

Exercise 2.1.A

Let $k$ be a field, and let $a, b, c \in k$. Show that $a + c = b + c$ implies $a = b$. Show that $a \cdot c = b \cdot c$ and $c \neq 0$ imply $a = b$. Show that $a \cdot b = 0$ implies $a = 0$ or $b = 0$.

Exercise 2.1.B

Let $k$ be a field. Show that for all $x \in k$, we have $x \cdot 0 = 0$. (Hint: multiply both sides of $0 + 0 = 0$ by $x$.) So $0$ cannot have a multiplicative inverse, because of (F0).

Exercise 2.1.C

Consider the subset

$$ \mathbb{Q}(i) = \lbrace a + b i : a, b \in \mathbb{Q} \rbrace \subseteq \mathbb{C}. $$

Check that $\mathbb{Q}(i)$ is a field under normal addition and multiplication. (Here, you also need to verify that $x, y \in \mathbb{Q}(i)$ implies $x + y, xy \in \mathbb{Q}(i)$, i.e., that addition and multiplication are well-defined.)

Sometimes, like in $\mathbb{F}_p$, adding $1$ many times can get you back to $0$.

Definition 2.1.4

Let $k$ be a field. If the sequence $1, 1 + 1, 1 + 1 + 1, \ldots$ never reaches $0$, we say that $k$ has characteristic zero, and write $\operatorname{char} k = 0$. If one of $1 + \cdots + 1$ is equal to $0$, we define its characteristic $\operatorname{char} k$ as the least positive integer $n$ such that

$$ \overbrace{1 + \cdots + 1}^{n} = 0. $$

For instance, $\operatorname{char} \mathbb{Q} = 0$ and $\operatorname{char} \mathbb{C} = 0$ but $\operatorname{char} \mathbb{F}_p = p$.

Exercise 2.1.D

Show that the characteristic of a field is either $0$ or a prime number.

Exercise 2.1.E

Show that if $\operatorname{char} k = 0$ for a field $k$, then there exists a map $i : \mathbb{Q} \to k$ satisfying $i(0) = 0$, $i(1) = 1$, $i(a + b) = i(a) + i(b)$, and $i(ab) = i(a) i(b)$ for all $a, b \in \mathbb{Q}$.

  1. Here, $0$ and $1$ are not actually the integers $0$ and $1$. I could have used the symbols $s$ and $t$ to denote them if I wanted, but they play the role of $0$ and $1$. So it is helpful to indicate this fact by using the notations $0$ and $1$. ↩