Field Axioms

The following axioms must be proven for a Field to be defined. Most definitions of a field use addition ($+$) and multiplication ($\times$) as their binary operators, because it’s easy to understand how they work.

Assume $F$ is the set that which the field to be defined ($\mathbb{F}$) uses.

Addition Axioms (A) §

A1. Closure under Addition §

A2. Commutativity of Addition §

A3. Associativity of Addition §

$$\forall x,y,z\in F,x+(y+z)=(x+y)+z$$

A4. Existence of Additive Identity §

$$\exists 0_{\mathbb{F}}\in F:\forall x\in F:x+0=0+x=x$$

$0$ is also called the field identity

Different zeroes!

The $0$ element here does not have to be the number zero. Because mathematicians like to be confusing, the additive identity is 0, which is the same symbol as the number zero. However, we could define (for example), $4$ to be our additive identity.

A5. Existence of Additive Inverse §

$$\forall x\in F:\exists x^{\prime }\in F:x+x^{\prime }=0_F=x^{\prime }+x$$

A lot of textbooks and internet sites denote the additive inverse as $-x$. I feel like it causes you to associate it with normal subtraction, which isn’t always the case.

Multiplication Axioms (M) §

M1. Closure under Multiplication §

$$\forall x,y\in F,x\times y\in F$$

M2. Commutativity of Multiplication §

$$\forall x,y\in F,x\times y=y\times x$$

M3. Associativity of Multiplication §

$$\forall x,y,z\in F,x\times (y\times z)=(x\times y)\times z$$

M4. Existence of Multiplicative Identity §

$$\exists 1\in F:\forall x\in F:x\times 1=x$$

$1$ is also called the field unity of $F$

M5. Existence of Multiplicative Inverse §

$$\forall x\in F\backslash \{0\},\exists x^{-1}\in F,x\times x^{-1}=1$$

Other Axioms (D) §

D1. Distributivity of Multiplication and Addition §

$$\forall x,y,z\in F:x\times (y+z)=(x\times y)+(x\times z)$$

D2. Distinct Additive and Multiplicative Identities §

$$0_F\ne 1_F$$

Now, because we have the additive inverse ($x^{\prime }$) and multiplicative inverse ($x^{-1}$), we can define new binary operators: subtraction and division:

$$a-b=a+(b^{\prime })$$

$$a/b=a\times b^{-1}$$

And that basically stems down to how a field is defined. If you have a set of anything, and you can show that the 10 field axioms are proved, that set can be called a field.

Lemmas §

L1: Additive Cancellation ^l1

$$x+y=x+z\to y=z$$

Proof

$$\begin{array}{rl}x+y& =x+z\\ (x+y)+(-x)& =(x+z)+(-x)\\ (x+(-x))+y& =(x+(-x))+z\\ 0+y& =0+z\\ y& =z\end{array}\text{\hspace{1em}(A5. Existence of additive inverse)(A3. Associativity of addition )(A4. Existence of additive identity)}$$

L2: Uniqueness of additive identity ^l2

$$x+y=x\to y=0$$

Proof

$$\begin{array}{rl}x+y& =x\\ (x+y)+(-x)& =x+(-x)\\ (x+(-x))+y& =x+(-x)\\ 0+y& =0\\ y& =0\end{array}\text{\hspace{1em}(A5. Existence of additive inverse)(A3. Associativity of addition )(A4. Existence of additive identity)}$$

L3: Uniqueness of additive inverse ^l3

$$x+y=0\to y=-x$$

Proof

$$\begin{array}{rl}x+y& =0\\ (x+y)+(-x)& =0+(-x)\\ (x+(-x))+y& =0+(-x)\\ 0+y& =-x\\ y& =-x\end{array}\text{\hspace{1em}(A5. Existence of additive inverse)(A3. Associativity of addition )(A4. Definition of additive identity)}$$

L4: Cancellation of additive identities ^l4

$$-(-x)=x$$

$$\begin{array}{r}(-(-x))=x\\ (-(-x))+(-x)=x+(-x)\\ 0=0\\ \\ \text{Use L1 }\end{array}$$

#todo Prove the bottom the same as L1-L4

L5: Multiplicative Cancellation ^l5

$$x\ne 0\wedge xy=xz\to y=z$$

L6: Uniqueness of multiplicative identity ^l6

$$x\ne 0\wedge xy=x\to y=1$$

L7: Uniqueness of multiplicative inverse ^l7

$$x\ne 0\wedge xy=1\to y=x^{-1}$$

L8: Cancellation of multiplicative identities ^l8

$$x\ne 0\wedge (x^{-1}{)}^{-1}=x$$

L9: The product involving the additive identity is zero ^l9

$$0\times x=0$$

Proof

$$\begin{array}{rl}& \\ 0x& =0\\ 0x+0x& =0+0x\\ 0x+0x& =0x\\ 0x& =0\end{array}\text{\hspace{1em}(By L2)}$$

See Lemma 2

L10: The product of non-zero elements is non-zero ^l10

$$x\ne 0\wedge y\ne 0\to x\times y\ne 0$$

L11: Associativity of additive inverses ^l11

$$(-x)\times y=-(x\times y)=x\times (-y)$$

L12: The product of elements is the product of additive inverses ^l12

$$(-x)\times (-y)=x\times y$$