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Proof of Zero Products and Elementary Identities in a Field

lemmalem:field-zero-product-2026a
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Β· 1,864 chars Β· 1 dep Β· depth 1 Reason: Initial publication: uniqueness of additive inverses and right distributivity are recorded first, then each of the five claims is derived from the numbered axioms of def:field-c54-2026b.

Proof

All axiom numbers below refer to Field.

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

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

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

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

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

Claim 1. By axiom 2, 0+0=00+0=0, so axiom 9 gives

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

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

0=x0+(βˆ’(x0))=(x0+x0)+(βˆ’(x0))=x0+(x0+(βˆ’(x0)))=x0+0=x0.0=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=00x=x0=0.

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

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

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

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

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

y=1y=(xβˆ’1x)y=xβˆ’1(xy)=xβˆ’10=0.y=1y=\bigl(x^{-1}x\bigr)y=x^{-1}(xy)=x^{-1}0=0 .

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

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

(xβˆ’y)(x+y)=(x+(βˆ’y))(x+y)=x(x+y)+(βˆ’y)(x+y)=(xx+xy)+((βˆ’(yx))+(βˆ’(yy))).(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=xyyx=xy by axiom 8, the middle terms satisfy xy+(βˆ’(xy))=0xy+\bigl(-(xy)\bigr)=0, so axioms 1, 4, 3 and 2 give

(xβˆ’y)(x+y)=xx+(βˆ’(yy))=x2βˆ’y2.(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)=(xx+xy)+(yx+yy)=x2+(xy+xy)+y2,(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=xyyx=xy together with axioms 1 and 4.

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lemmaZero Products and Elementary Identities in a Fieldlem:field-zero-product-2026a

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