Proof of Constants, Powers, Sums, Scalar Multiples and Products of Polynomial Functions

Throughout, $S$ is the successor map of Natural Numbers, $\le$ is the order on $\mathbb{N}$, and $[n]$ is the initial segment determined by $n$. Sums are the finite sums of $K$ and powers are those of Natural Number Power of an Element of a Field; commutativity, associativity and distributivity are the axioms of Field.

By Polynomial Function on a Field we may fix natural numbers $M,N$, elements $a_{0},b_{0}\in K$ and maps $a:[M]\to K$ and $b:[N]\to K$ such that

Step 1 (padding with zero coefficients). Let $N',L\in\mathbb{N}$ with $N'\le L$, let $c:[N']\to K$, and let $\tilde{c}:[L]\to K$ be the map with $\tilde{c}_{k}=c_{k}$ for $k\in[N']$ and $\tilde{c}_{k}=0$ for those $k\in[L]$ with $k\notin[N']$; this is well defined because $[N']\subseteq[L]$ by claim 4 of Basic Properties of Initial Segments of the Natural Numbers. Then

Fix $x\in K$ and let $E$ be the set of those $L\in\mathbb{N}$ for which the displayed identity holds for every $N'\le L$ and every $c:[N']\to K$.

If $N'\le 1$ then $1\le N'$ by claim 4 of Properties of the Order on the Natural Numbers, hence $N'=1$ by claim 2 of that lemma; then $\tilde{c}=c$ and there is nothing to prove. Thus $1\in E$.

Let $L\in E$ and let $N'\le S(L)$. If $N'=S(L)$ then $\tilde{c}=c$ and the identity is trivial. Otherwise $N'\le L$ by claim 5 of Properties of the Order on the Natural Numbers. The restriction of $\tilde{c}$ to $[L]$ is exactly the extension of $c$ by zeros associated with the pair $N'\le L$, so the restriction part of claim 1 of Properties of Finite Sums, applied to the family $k\mapsto\tilde{c}_{k}x^{k}$, together with $L\in E$ gives

By claim 3 of Basic Properties of Initial Segments of the Natural Numbers we have $S(L)\notin[L]$, hence $S(L)\notin[N']$, so $\tilde{c}_{S(L)}=0$ and therefore $\tilde{c}_{S(L)}x^{S(L)}=0$ by claim 1 of Zero Products and Elementary Identities in a Field. The recursion part of claim 1 of Properties of Finite Sums now gives

so $S(L)\in E$. By Principle of Induction for the Natural Numbers, $E=\mathbb{N}$, which proves Step 1.

Step 2 (claim 1). For the constant map with value $\lambda$, take $N'=1$, $c_{0}=\lambda$ and $c:[1]\to K$ with $c_{1}=0$. For every $x\in K$, claim 1 of Properties of Finite Sums gives $\sum_{k=1}^{1}c_{k}x^{k}=c_{1}x^{1}$, and $x^{1}=x$ by claim 1 of Properties of Natural Number Powers in a Field, so $c_{1}x^{1}=0x=0$ by claim 1 of Zero Products and Elementary Identities in a Field. Hence $\lambda+\sum_{k=1}^{1}c_{k}x^{k}=\lambda+0=\lambda$, and the constant map is a polynomial function.

Now fix $n\in\mathbb{N}$ and take $N'=n$, $c_{0}=0$ and $c:[n]\to K$ with $c_{n}=1$ and $c_{k}=0$ for those $k\in[n]$ with $k\ne n$; here $n\in[n]$ by claim 1 of Basic Properties of Initial Segments of the Natural Numbers. Fix $x\in K$. For $k\in[n]$ with $k\ne n$ we have $c_{k}x^{k}=0x^{k}=0$ by claim 1 of Zero Products and Elementary Identities in a Field, so claim 7 of Properties of Finite Sums, applied to the family $k\mapsto c_{k}x^{k}$, gives $\sum_{k=1}^{n}c_{k}x^{k}=c_{n}x^{n}=1x^{n}=x^{n}$. Hence $0+\sum_{k=1}^{n}c_{k}x^{k}=x^{n}$, and $x\mapsto x^{n}$ is a polynomial function.

Step 3 (claim 2). For the scalar multiple, distributivity and claim 3 of Properties of Finite Sums give, for every $x\in K$,

the last equality by associativity of multiplication. Thus $\lambda p$ is a polynomial function, with coefficients $\bigl(M,\lambda a_{0},k\mapsto\lambda a_{k}\bigr)$.

For the sum, by trichotomy (claim 3 of Properties of the Order on the Natural Numbers) and claim 1 of that lemma, at least one of $M\le N$ and $N\le M$ holds; let $L$ be $N$ in the first case and $M$ in the second, so that $M\le L$ and $N\le L$. Let $\tilde{a},\tilde{b}:[L]\to K$ be the extensions of $a$ and $b$ by zeros as in Step 1. For every $x\in K$, Step 1 gives

which by commutativity and associativity of addition, claim 2 of Properties of Finite Sums and distributivity equals

Thus $p+q$ is a polynomial function.

Step 4 (finite sums of polynomial functions). Let $n\in\mathbb{N}$ and, for each $k\in[n]$, let $g_{k}:K\to K$ be a polynomial function on $K$. Then the map $x\mapsto\sum_{k=1}^{n}g_{k}(x)$ is a polynomial function on $K$.

Let $E$ be the set of those $n\in\mathbb{N}$ for which this holds for every such family. For $n=1$, claim 1 of Properties of Finite Sums gives $\sum_{k=1}^{1}g_{k}(x)=g_{1}(x)$, so $1\in E$. Let $n\in E$ and let $g_{k}$, $k\in[S(n)]$, be polynomial functions. By the restriction and recursion parts of claim 1 of Properties of Finite Sums,

where the first summand on the right is, as a function of $x$, a polynomial function by $n\in E$ applied to the restricted family. By claim 2, already proved, the right-hand side is a polynomial function of $x$, so $S(n)\in E$. By Principle of Induction for the Natural Numbers, $E=\mathbb{N}$.

Step 5 (claim 3). Define $A,B:K\to K$ by $A(x)=\sum_{k=1}^{M}a_{k}x^{k}$ and $B(x)=\sum_{l=1}^{N}b_{l}x^{l}$. Taking $c_{0}=0$ in Polynomial Function on a Field shows that $A$ and $B$ are polynomial functions on $K$. By distributivity,

The constant map with value $a_{0}b_{0}$ is a polynomial function by claim 1, and $a_{0}B$ and $b_{0}A$ are polynomial functions by claim 2; so by claim 2 again it suffices to show that $x\mapsto A(x)B(x)$ is a polynomial function.

Fix $x\in K$. Applying claim 3 of Properties of Finite Sums with the factor $B(x)$, and then again with the factor $a_{k}x^{k}$ for each $k\in[M]$,

using commutativity of multiplication. By commutativity and associativity of multiplication and by Addition of Exponents for Natural Number Powers in a Field,

so that

For fixed $k\in[M]$ and $l\in[N]$ the map $x\mapsto(a_{k}b_{l})x^{k+l}$ is a scalar multiple of the map $x\mapsto x^{k+l}$, hence a polynomial function by claims 1 and 2. By Step 4 the map $x\mapsto\sum_{l=1}^{N}(a_{k}b_{l})x^{k+l}$ is therefore a polynomial function for each $k\in[M]$, and by Step 4 once more so is $x\mapsto A(x)B(x)$. Hence $pq$ is a polynomial function on $K$.