Field axioms

Definition: Field

A field ( $\mathbb{F},+,\cdot$ ) consists of the following three pieces of data:

• (D1) A set $\mathbb{F}$, whose contents are “the scalars” or “the elements of the field”
• (D2) An addition map $+:\mathbb{F}\times \mathbb{F}\to \mathbb{F}$, meaning for every pair of scalars $a,b\in \mathbb{F}$, we define a sum $a+b\in \mathbb{F}$
• (D3) A product map $\cdot :\mathbb{F}\times \mathbb{F}\to \mathbb{F}$, meaning for every pair of scalars $a,b\in F$, we define a product $a\cdot b\in \mathbb{F}$

Axioms: Fields

• (F1) Associativity. Addition and multiplication are “associative,” meaning that for all $a,b,c\in \mathbb{F}$, we have $(a+b)+c=a+(b+c)$ and $(a\cdot b)\cdot c=a\cdot (b\cdot c)$.
• (F2) Commutativity. Addition and multiplication are “commutative,” meaning that for all $a,b\in \mathbb{F}$, we have $a+b=b+a$ and $a\cdot b=b\cdot a$.
• (F3) Distributivity. Multiplication distributes over addition, meaning that for all $a,b,c\in \mathbb{F}$, we have $a\cdot (b+c)=a\cdot b+a\cdot c$.
• (F4) Identities. There exist elements $0,1\in \mathbb{F}$ which serve as “additive and multiplicative identities,” meaning that for all $a\in \mathbb{F}$ we have

$$0+a=a1\cdot a=a$$

• (F5) Additive inverses. For all $a\in \mathbb{F}$, there exists an element $b\in \mathbb{F}$ with $a+b=0$. We call $b$ the “additive inverse” of $a$, and denote it as $-a$.
• (F6) Multiplicative inverses. For all $a\in \mathbb{F}$, there exists an element $b\in \mathbb{F}$ with $ab=1$. We call $b$ the “multiplicative inverse” of $a$, and denote it as $1/a$ or $a^{-1}$.
• (F7) Nontriviality. There are at least two elements of $\mathbb{F}$.

Field lemmas §

• (Lemma 17) The additive identity in a field is unique.
• (Lemma 18) Additive inverses in a field are unique.
• (Lemma 19) Let $\mathbb{F}$ be a field, and write $0\in \mathbb{F}$ for the additive identity. Then $0\cdot x=0$ for all $x\in \mathbb{F}$.
• (Lemma 20) If $a\in \mathbb{F}$, then the additive inverse is given by $-a=(-1)\cdot a$.
• (Proposition 21) If $a,b\in \mathbb{F}$, then $a\cdot b=0$ if and only if $a=0$ or $b=0$.