Proof of Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field

Each result cited below is universally quantified over the data in its own statement and is applied to the data indicated where it is cited.

Throughout we use the axioms of a commutative ring freely. By commutativity, $0+x=x$, $1\cdot x=x$ and $(y+z)x=yx+zx$.

Zero. Let $a=x\cdot0$. Then $a=x\cdot(0+0)=a+a$. Add $-a$ to both sides (Negatives, Differences, Reciprocals and Quotients §negative): $0=a+(-a)=(a+a)+(-a)=a+(a+(-a))=a+0=a$. Hence $x\cdot0=0$, and $0\cdot x=x\cdot0=0$ by commutativity.

Signs. By Additive and Multiplicative Inverses Are Unique §negative, for each $u\in R$ the negative $-u$ is the only $w$ with $u+w=0$, so to prove $-u=v$ it suffices to show $u+v=0$. First, $(-x)+x=x+(-x)=0$, so $-(-x)=x$. Second, $(x+y)+((-x)+(-y))=(x+(-x))+(y+(-y))=0+0=0$ by associativity and commutativity, so $-(x+y)=(-x)+(-y)$. Third, $xy+(-x)y=(x+(-x))y=0\cdot y=0$ by Zero, so $(-x)y=-(xy)$. Fourth, by the third rule and commutativity, $x(-y)=(-y)x=-(yx)=-(xy)$, so $(-x)(-y)=-(x(-y))=-(-(xy))=xy$. Fifth, $-(x-y)=-(x+(-y))=(-x)+(-(-y))=(-x)+y=y-x$. Sixth, $(x-y)z=xz+(-y)z=xz+(-(yz))=xz-yz$.

Squares. By Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, the element $2$ of $R$ in the statement is the image $2_{R}$ of the natural number $2$, $2=1+1$ in $\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §digits, so $2_{R}=1+1$ by Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals, since the image of a sum is the sum of the images and $1_{R}$ is the unit of $R$. Also, by Powers with Exponents in the Natural Numbers with Zero §power and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, $u^{2}=\prod_{k=1}^{1+1}u=\big(\prod_{k=1}^{1}u\big)\cdot u=u\cdot u$ for every $u\in R$. Hence $u+u=1\cdot u+1\cdot u=(1+1)u=2_{R}\,u$. By distributivity, associativity and commutativity, $(x+y)^{2}=(x+y)x+(x+y)y=xx+(xy+xy)+yy=x^{2}+2_{R}\,xy+y^{2}$. Apply this to $x$ and $-y$. By Signs, $x(-y)=-(xy)$, so $2_{R}\,x(-y)=-(2_{R}\,xy)$, and $(-y)(-y)=yy$. This gives $(x-y)^{2}=x^{2}-2_{R}\,xy+y^{2}$. Lastly, by Signs, $(x-y)(x+y)=x(x+y)+(-y)(x+y)=xx+xy+(-(yx))+(-(yy))=x^{2}+(xy+(-(xy)))-y^{2}=x^{2}-y^{2}$.

Field. Let $R$ be a field, let $xy=0$, and suppose $x\neq0$. By Fields §field there is $w\in R$ with $xw=1$. Then $y=1\cdot y=(wx)y=w(xy)=w\cdot0=0$ by Zero.