Proof of Zero Products and Elementary Identities in a Field

All axiom numbers below refer to Field.

Uniqueness of additive inverses. If $u,v,w\in F$ satisfy $u+v=0$ and $u+w=0$, then, using axioms 2, 1 and 4,

$$v=v+0=v+(u+w)=(v+u)+w=(u+v)+w=0+w=w .$$

In particular $-u$ is the only element $v$ with $u+v=0$, and since $(-u)+u=0$ by axioms 3 and 4, we get $-(-u)=u$.

Right distributivity. For all $p,q,s\in F$, axioms 8 and 9 give

$$(p+q)s=s(p+q)=sp+sq=ps+qs .$$

Claim 1. By axiom 2, $0+0=0$, so axiom 9 gives

$$x0=x(0+0)=x0+x0 .$$

Adding $-(x0)$ to both sides and using axioms 1, 3 and 2 gives

$$0=x0+\bigl(-(x0)\bigr)=\bigl(x0+x0\bigr)+\bigl(-(x0)\bigr)=x0+\Bigl(x0+\bigl(-(x0)\bigr)\Bigr)=x0+0=x0 .$$

By axiom 8, $0x=x0=0$.

Claim 2. By right distributivity, axiom 3 and claim 1,

$$xy+(-x)y=\bigl(x+(-x)\bigr)y=0y=0 ,$$

so $(-x)y$ is an additive inverse of $xy$, and by the uniqueness recorded above $(-x)y=-(xy)$. Applying this twice, together with axiom 8,

$$(-x)(-y)=-\bigl(x(-y)\bigr)=-\bigl((-y)x\bigr)=-\bigl(-(yx)\bigr)=yx=xy .$$

Claim 3. Suppose $xy=0$ and $x\ne0$. By axiom 7 there is $x^{-1}\in F$ with $xx^{-1}=1$, and $x^{-1}x=1$ by axiom 8. Using axioms 6, 8, 5 and claim 1,

$$y=1y=\bigl(x^{-1}x\bigr)y=x^{-1}(xy)=x^{-1}0=0 .$$

Hence $x=0$ or $y=0$.

Claim 4. By right distributivity, axiom 9, claim 2 and axiom 8,

$$(x-y)(x+y)=\bigl(x+(-y)\bigr)(x+y)=x(x+y)+(-y)(x+y)=\bigl(xx+xy\bigr)+\Bigl(\bigl(-(yx)\bigr)+\bigl(-(yy)\bigr)\Bigr).$$

Since $yx=xy$ by axiom 8, the middle terms satisfy $xy+\bigl(-(xy)\bigr)=0$, so axioms 1, 4, 3 and 2 give

$$(x-y)(x+y)=xx+\bigl(-(yy)\bigr)=x^{2}-y^{2}.$$

Claim 5. By right distributivity, axiom 9 and axiom 8,

$$(x+y)^{2}=(x+y)(x+y)=x(x+y)+y(x+y)=\bigl(xx+xy\bigr)+\bigl(yx+yy\bigr)=x^{2}+(xy+xy)+y^{2},$$

the last step using $yx=xy$ together with axioms 1 and 4.

Claim 6. The identity $-(-x)=x$ was recorded with the uniqueness of additive inverses above. Using axioms 1, 2, 3 and 4,

$$(x-y)+(y-x)=x+\bigl((-y)+y\bigr)+(-x)=x+0+(-x)=0 ,$$

so $y-x$ is an additive inverse of $x-y$, and by uniqueness $-(x-y)=y-x$.