TheoremBase

Existence comes from the ring axiom on additive inverses and from the definition of a field; uniqueness follows by inserting one inverse into the other and reassociating.

Proof

Negative. Let x∈rx\in r. By Commutative Rings Β§ring there is w∈rw\in r with x+w=0x+w=0. Let w,wβ€²βˆˆrw,w'\in r with x+w=0x+w=0 and x+wβ€²=0x+w'=0. Then, by the axioms of Commutative Rings Β§ring,

w=w+0=w+(x+wβ€²)=(w+x)+wβ€²=(x+w)+wβ€²=0+wβ€²=wβ€²+0=wβ€².w=w+0=w+(x+w')=(w+x)+w'=(x+w)+w'=0+w'=w'+0=w'.

Hence there is exactly one w∈rw\in r with x+w=0x+w=0.

Reciprocal. Let rr be a field and let x∈rx\in r with xβ‰ 0x\neq0. By Fields Β§field there is y∈ry\in r with xβ‹…y=1x\cdot y=1, and rr is a commutative ring. Let y,yβ€²βˆˆry,y'\in r with xβ‹…y=1x\cdot y=1 and xβ‹…yβ€²=1x\cdot y'=1. Then, by the axioms of Commutative Rings Β§ring,

y=yβ‹…1=yβ‹…(xβ‹…yβ€²)=(yβ‹…x)β‹…yβ€²=(xβ‹…y)β‹…yβ€²=1β‹…yβ€²=yβ€²β‹…1=yβ€².y=y\cdot1=y\cdot(x\cdot y')=(y\cdot x)\cdot y'=(x\cdot y)\cdot y'=1\cdot y'=y'\cdot1=y'.

Hence there is exactly one y∈ry\in r with xβ‹…y=1x\cdot y=1.

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