# Fields

We have already defined Groups and Rings, and now we will define our next algebraic structure of interest - fields!

Definition: If $+$ and $*$ are binary operations on the set $F$, then $F$ is called a Field under $+$ and $*$ denoted $(F, +, *)$ if $F$ under $+$ and $*$ satisfies the following properties:1. For all $a, b \in F$ we have that $(a + b \in F)$ (Closure under $+$).2. For all $a, b, c \in F$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $F$ under $+$).3. There exists an $0 \in F$ such that for all $a \in F$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $F$ under $+$).4. For all $a \in F$ there exists a $-a \in F$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $F$ under $+$).5. For all $a, b \in F$ we have that $a + b = b + a$ (Commutativity of elements in $F$ under $+$).6. For all $a, b \in F$ we have that $a * b = b * a$ (Closure under $*$).7. For all $a, b, c \in F$, $a * (b * c) = (a * b) * c$ (Associativity of elements in $F$ under $*$).8. There exists a $1 \in F$ such that for all $a \in F$ we have that $a * 1 = a$ and $1* a = a$ (The existence of an identity element $1$ of $F$ under $*$).9. For all $a \in F \setminus \{ 0 \}$ there exists an $a^{-1} \in F$ such that $a * a^{-1} = 1$ and $a^{-1} * a = 1$. (The existence of inverses for each element in $F$ under $*$).10. For all $a, b \in F$ we have that $a * b = b * a$ (Commutativity of elements in $F$ undr $*$).11. For all $a, b, c \in F$ we have that $a * (b + c) = (a * b) + (b * c)$ and $(a + b) * c = (a * c) + (b * c)$ (Distributivity of $*$ over $+$). |

Notice that the first four field axioms are identical to the group axioms. Therefore, the field $F$ with respect to the operation $+$, i.e., $(F, +)$ is a group. Similarly, we note that the first eight axioms and the eleventh axiom together are all identical to the ring axioms. Therefore, the field $F$ with respect to the operations $+$ and $*$, i.e., $(F, +, *)$ is a ring.

The only new axioms that are added to make a ring a field is the necessity that $*$ is a commutative operation and if $0$ is the identity of $+$ then every element $a \in F \setminus \{ 0 \}$ must be invertible, that is there exists an element $b = a^{-1} \in F$ such that $a * a^{-1} = 1$ and $a^{-1} * a = 1$. In essence, a field $F$ is a commutative ring in which there exists a inverse under $*$ for each element in $F$ apart from the identity $0$ of $+$.

The simplest field is yet again the field of real numbers $(\mathbb{R}, +, *)$ which we are already familiar with as a group under $+$ and as a ring under $+$ and $*$. We know that standard multiplication is commutative and we noted on the Multiplicatively Invertible Elements in Rings page that every nonzero element in $\mathbb{R}$ has a multiplicative inverse.

Now recall that in a ring $(R, +, *)$ where $0$ is the identity of $+$ that a zero divisor is an element $a \in R \setminus \{ 0 \}$ such that there exists an element $b \in R \setminus \{ 0 \}$ such that $a * b = 0$. In the following theorem we will show that if all fields contain no zero divisors.

Theorem 1: If $(F, +, *)$ is a field then $F$ contains no zero divisors. |

**Proof:**Let $0$ be the identity of $+$ and let $a, b \in F \setminus \{ 0 \}$. Suppose that:

- Since $a \neq 0$ we have that since $(F, +, *)$ is a field then there exists an $a^{-1} \in F$ such that $a * a^{-1} = 1$. Therefore:

- Therefore $b = 0$. But $b \in F \setminus \{ 0 \}$. Therefore our assumption that $a, b \in F \setminus \{ 0 \}$ was false and so one of $a$ or $b$ is equal to $0$ and so $F$ contains no zero divisors. $\blacksquare$