# Field Axioms

We will now look at a very important algebraic structure known as a **Field**.

Definition: A field is a nonempty set $F$ containing at least 2 elements alongside the two binary operations of addition, $f_+ : F \times F \to F$ such that $f_+(x, y) = x + y$ and multiplication $f_{\cdot}(x, y) = x \cdot y$ that satisfy all of the axioms below. |

**Field Axiom for Addition 1.**The operation of addition is*closed*, that is $\forall x, \forall y$ $x + y \in F$.

**Field Axiom for Addition 2.**The operation of addition is*commutative*, that is $\forall x, \forall y$, $x + y = y + x$ ().*Commutativity of addition*

**Field Axiom for Addition 3**The operation of addition is*associative*, that is $\forall x, \forall y, \forall z$, $x + (y + z) = (x + y) + z$ ().*Associativity of addition*

**Field Axiom for Addition 4**The operation of addition has the additive identity element of 0 such that $\forall x$, $x + 0 = x$ ().*Existence of an additive identity*

**Field Axiom for Addition 5**The operation of addition has the additive inverse element of -x such that $\forall x$, $x + (-x) = 0$ ().*Existence of an additive inverse*

**Field Axiom for Multiplication 1.**The operation of multiplication is*closed*, that is $\forall x, \forall y$ $xy \in F$.

**Field Axiom for Multiplication 2.**The operation of multiplication is*commutative*, that is $\forall x, \forall y$, $xy = yx$ ().*Commutativity of multiplication*

**Field Axiom for Multiplication 3**The operation of multiplication is*associative*, that is $\forall x, \forall y, \forall z$, $x(yz) = (xy)z$ ().*Associativity of multiplication*

**Field Axiom for Multiplication 4**The operation of multiplication has the multiplicative identity element of 1 such that $\forall x$, $1x = x$ ().*Existence of an multiplicative identity*

**Field Axiom for Multiplication 5**The operation of multiplication has the multiplicative inverse element of $\frac{1}{x}$ such that $\forall x$, $x \cdot \frac{1}{x} = 1$ ().*Existence of a multiplicative inverse*

**Field Axiom for Distributivity**The operation of multiplication is*distributive*over addition, that is $\forall x, \forall y, \forall z$, $x(y + z) = xy + xz$ ().*Distributive law*

So we have established 11 field axioms. These axioms are statements that aren't intended to be proved but are to be taken as given. We will consequentially build theorems based on these axioms, and create more complex theorems by referring to these field axioms and other theorems we develop. Let's first look at one of the simplest fields, the field of real numbers $\mathbb{R}$ whose operations are standard addition and standard multiplication.

It is not difficult to verify that axioms 1-11 hold for the field of real numbers. We already know that addition of real numbers is commutative, that is $\forall a, b \in \mathbb{R}$, $a + b = b + a$, for example $2 + 5 = 5 + 2 = 7$. The same goes for the commutativity of real number multiplication, that is $a \cdot b = b \cdot a$, for example $3 \cdot 6 = 6 \cdot 3 = 18$.

Sometimes it may not be extremely obvious as to where a set with defined operations of addition and multiplication is in fact a field though, so it may be necessary to verify all 11 axioms.

# Example 1

**Verify that the field of rational numbers $\mathbb{Q}$ under the operations of standard addition and standard multiplication form a field.**

Recall that $\mathbb{Q} \subset \mathbb{R}$ and the set of rational numbers is defined as $\mathbb{Q} := \{ \frac{a}{b} \: a, b, \in \mathbb{Z} , \: b \neq 0 \}$. Let's verify a few of the axioms, and the rest will be left for the reader to verify.

First let $a, b \in \mathbb{Q}$ where $a = \frac{m_1}{n_1}$ and $b = \frac{m_2}{n_2}$.

**A1:**$a + b = \frac{m_1}{n_1} + \frac{m_2}{n_2} = \frac{m_1n_2 + m_2n_1}{n_1n_2}$, and $\frac{m_1n_2 + m_2n_1}{n_1n_2} \in \mathbb{Q}$, so the operation of addition is closed.

**A2:**$a + b = \frac{m_1}{n_1} + \frac{m_2}{n_2} = \frac{m_1n_2 + m_2n_1}{n_1n_2} = \frac{n_1m_2 + n_2m_1}{n_2n_1} = \frac{m_2}{n_2} + \frac{m_1}{n_1} = b + a$.

**A4:**The additive identity is $\frac{0}{1}$. That is $a + 0 = \frac{m_1}{n_1} + \frac{0}{1} = \frac{m_1 \cdot 1 + n_1 \cdot 0}{1 \cdot n_1} = \frac{m_1}{n_1} = a$.

**A5:**The inverse element for any element $a = \frac{m_1}{n_1}$ is $-a = -\frac{m_1}{n_1}$, since $a + (-a) = \frac{m_1}{n_1} + \left ( - \frac{m_1}{n_1} \right ) = \frac{0}{1}$.

**M1:**$a\cdot b = \frac{m_1}{n_1} \cdot \frac{m_2}{n_2} = \frac{m_1 \cdot m_2}{n_1 \cdot n_2}$ and $\frac{m_1 \cdot m_2}{n_1 \cdot n_2} \in \mathbb{Q}$.

**M2:**$a \cdot b = \frac{m_1 \cdot m_2}{n_1 \cdot n_2} = \frac{m_2 \cdot m_1}{n_2 \cdot n_1} = b \cdot a$.