Proof of The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products

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

Throughout, $x,y,z$ range over $R$, and we use the identities of Commutative Rings §ring: $(x+y)+z=x+(y+z)$, $x+y=y+x$, $x+0=x$, $xy=yx$, $x\cdot1=x$, $x(y+z)=xy+xz$, and for every $x$ some $w\in R$ with $x+w=0$. Computations with elements of $\mathbb{N}_{0}$ use the laws of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §laws. For $n\in\mathbb{N}$ the image $n_{R}$ is the finite sum, for the addition of $R$, of the constant map $k\mapsto1$ on $[n]$, by The Image of a Natural Number in a Commutative Ring §image; the constant maps on the various $[n]$ all have the value $1$ of $R$.

Constants. By The Image of a Natural Number in a Commutative Ring §image, $0_{R}=0$. Since $1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets, $1_{R}=\sum_{k=1}^{1}1$, and the first part of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, applied with $n=1$ to the constant map on $[1]$, gives $1_{R}=1$.

Successor step. We show $(n+1)_{R}=n_{R}+1$ for every $n\in\mathbb{N}_{0}$. If $n=0$, then $n+1=0+1=1$ in $\mathbb{N}_{0}$, and $0+1=1+0=1$ in $R$; so, by the constants clause, $(0+1)_{R}=1_{R}=1=0+1=0_{R}+1$, where the sum $0+1$ in the first term is formed in $\mathbb{N}_{0}$ and the sum $0+1$ in the fourth term in $R$. If $n\neq0$, then $n\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets, and $n+1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §operations. Let $c$ be the constant map $k\mapsto1$ on $[n+1]$. The second part of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, applied to $c$, gives

$$(n+1)_{R}=\sum_{k=1}^{n+1}c_{k}=\Big(\sum_{k=1}^{n}c_{k}\Big)+c_{n+1}=n_{R}+1,$$

because $\sum_{k=1}^{n}c_{k}$ is the finite sum of $c|_{[n]}$ by Iterated Operations: Finite Sums and Finite Products §restriction, which applies as $[n]\subseteq[n+1]$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §successor, and $c|_{[n]}$ is the constant map $k\mapsto1$ on $[n]$, whose finite sum is $n_{R}$; and $c_{n+1}=1$.

Sum. Fix $m\in\mathbb{N}_{0}$; we prove $(m+n)_{R}=m_{R}+n_{R}$ for every $n\in\mathbb{N}_{0}$ by induction from $0$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction. For $n=0$, $(m+0)_{R}=m_{R}=m_{R}+0=m_{R}+0_{R}$. If the claim holds for $n$, then, using $m+(n+1)=(m+n)+1$, the successor step twice, the induction hypothesis and associativity in $R$,

$$(m+(n+1))_{R}=((m+n)+1)_{R}=(m+n)_{R}+1=(m_{R}+n_{R})+1=m_{R}+(n_{R}+1)=m_{R}+(n+1)_{R}.$$

Zero factor in $R$. For every $x\in R$, $x\cdot0=0$. Indeed $0+0=0$, so $x\cdot0=x(0+0)=x\cdot0+x\cdot0$. Let $w\in R$ with $x\cdot0+w=0$. Then

$$0=x\cdot0+w=(x\cdot0+x\cdot0)+w=x\cdot0+(x\cdot0+w)=x\cdot0+0=x\cdot0.$$

Product. Fix $m\in\mathbb{N}_{0}$; we prove $(mn)_{R}=m_{R}\,n_{R}$ for every $n\in\mathbb{N}_{0}$ by induction from $0$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction. For $n=0$, $m\cdot0=0$ in $\mathbb{N}_{0}$, so $(m\cdot0)_{R}=0_{R}=0=m_{R}\cdot0=m_{R}\,0_{R}$ by the zero factor in $R$. If the claim holds for $n$, then, using $m(n+1)=mn+m\cdot1=mn+m$ in $\mathbb{N}_{0}$, the sum clause, the induction hypothesis, $x\cdot1=x$, distributivity in $R$ and the successor step,

$$(m(n+1))_{R}=(mn+m)_{R}=(mn)_{R}+m_{R}=m_{R}\,n_{R}+m_{R}\cdot1=m_{R}(n_{R}+1)=m_{R}\,(n+1)_{R}.$$

Difference. Let $m\le n$ and $d=n-m$, so that $m+d=n$ in $\mathbb{N}_{0}$ by The Difference of Two Natural Numbers with Zero §difference. By the sum clause, $n_{R}=(m+d)_{R}=m_{R}+d_{R}$. By Negatives, Differences, Reciprocals and Quotients §negative, $-m_{R}$ is the element of $R$ with $m_{R}+(-m_{R})=0$, and $n_{R}-m_{R}=n_{R}+(-m_{R})$. Using commutativity and associativity of $+$ in $R$ and $x+0=x$,

$$n_{R}-m_{R}=(m_{R}+d_{R})+(-m_{R})=(d_{R}+m_{R})+(-m_{R})=d_{R}+(m_{R}+(-m_{R}))=d_{R}+0=d_{R}=(n-m)_{R}.$$

Power step. Let $\cdot$ be a binary operation on a set $X$ with a neutral element $e$, let $u\in X$, and let powers of $u$ be as in Powers with Exponents in the Natural Numbers with Zero §power and Powers with Exponents in the Natural Numbers with Zero §zero, so that $u^{0}=e$. We show $u^{k+1}=u^{k}\cdot u$ for every $k\in\mathbb{N}_{0}$. If $k=0$, then $k+1=1\in\mathbb{N}$, and the first part of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, applied with $n=1$ to the constant map $j\mapsto u$ on $[1]$, gives $u^{1}=\prod_{j=1}^{1}u=u=e\cdot u=u^{0}\cdot u$. If $k\neq0$, then $k\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets, and $k+1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §operations. Let $c$ be the constant map $j\mapsto u$ on $[k+1]$. The second part of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion, applied to $c$, gives

$$u^{k+1}=\prod_{j=1}^{k+1}c_{j}=\Big(\prod_{j=1}^{k}c_{j}\Big)\cdot c_{k+1}=u^{k}\cdot u,$$

because $\prod_{j=1}^{k}c_{j}$ is the finite product of $c|_{[k]}$ by Iterated Operations: Finite Sums and Finite Products §restriction, which applies as $[k]\subseteq[k+1]$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §successor, and $c|_{[k]}$ is the constant map $j\mapsto u$ on $[k]$, whose finite product is $u^{k}$ by Powers with Exponents in the Natural Numbers with Zero §power; and $c_{k+1}=u$.

The power step applies to the multiplication of $\mathbb{N}_{0}$, whose neutral element is $1$ by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §one, as $1=S(0)$ by The Set of Natural Numbers and the Number One §one; and to the multiplication of $R$, whose neutral element is the unit $1$ of $R$, since $x\cdot1=x$ and $1\cdot x=x\cdot1$ for every $x\in R$. In both cases the $1$ of Powers with Exponents in the Natural Numbers with Zero §zero is this neutral element, which is unique by A Binary Operation Has at Most One Neutral Element §unique.

Power. Fix $m\in\mathbb{N}_{0}$; we prove $(m^{k})_{R}=(m_{R})^{k}$ for every $k\in\mathbb{N}_{0}$ by induction from $0$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction. For $k=0$, $m^{0}=1$ in $\mathbb{N}_{0}$ and $(m_{R})^{0}=1$ in $R$ by Powers with Exponents in the Natural Numbers with Zero §zero, so $(m^{0})_{R}=1_{R}=1=(m_{R})^{0}$ by the constants clause. If the claim holds for $k$, then, by the power step in $\mathbb{N}_{0}$, the product clause, the induction hypothesis and the power step in $R$,

$$(m^{k+1})_{R}=(m^{k}\,m)_{R}=(m^{k})_{R}\,m_{R}=(m_{R})^{k}\,m_{R}=(m_{R})^{k+1}.$$

Finite sums and products. Let $\varphi:\mathbb{N}_{0}\to R$ be the map $n\mapsto n_{R}$ of Sets and Maps: Ordinary Notation §maps. The addition of $\mathbb{N}_{0}$ is associative and commutative with neutral element $0$, by Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §associative, Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §commutative and Arithmetic of Addition on Omega: Recursion Rules, Associativity, Commutativity, Cancellation and Compatibility with the Order §zero; its multiplication is associative and commutative with neutral element $1$, by Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §associative, Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §commutative and Arithmetic of Multiplication on Omega: Recursion Rules, Distributivity, Associativity, Commutativity, No Zero Divisors, Cancellation and Compatibility with the Order §one. The addition and the multiplication of $R$ are associative and commutative with neutral elements $0$ and $1$, by the identities of Commutative Rings §ring listed above, neutrality holding on both sides by commutativity. By the constants clause $\varphi(0)=0$ and $\varphi(1)=1$, and by the sum and product clauses $\varphi(m+n)=\varphi(m)+\varphi(n)$ and $\varphi(mn)=\varphi(m)\,\varphi(n)$ for all $m,n\in\mathbb{N}_{0}$. Hence Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §homomorphism, applied to $\varphi$ once with the additions of $\mathbb{N}_{0}$ and $R$ and once with their multiplications, gives $\varphi\big(\sum_{a\in A}f(a)\big)=\sum_{a\in A}\varphi(f(a))$ and $\varphi\big(\prod_{a\in A}f(a)\big)=\prod_{a\in A}\varphi(f(a))$ for every finite set $A$ and every $f:A\to\mathbb{N}_{0}$, which is the claim.