explicit definition of polynomial rings in arbitrarly many variables

Let $R$ be a ring and let $𝕏$ be any set (possibly empty). We wish to give an explicit and formal definition of the polynomial ring $R[𝕏]$.

We start with the set

$$ℱ(𝕏)=\{f:𝕏\to \mathbb{N}|f(x)=0\text{ for almost all }x\}.$$

If $𝕏=\{X_1,\mathrm{\dots },X_n\}$ then the elements of $ℱ(𝕏)$ can be interpreted as monomials

$$X_1^{{\alpha }_1}\mathrm{\cdots }{X}_n^{{\alpha }_n}.$$

Now define

$$R[𝕏]=\{F:ℱ(𝕏)\to R|F(f)=0\text{ for almost all }f\}.$$

The addition in $R[𝕏]$ is defined as pointwise addition.

Now we will define multiplication. First note that we have a multiplication on $ℱ(𝕏)$. For any $f,g:𝕏\to \mathbb{N}$ put

$$(fg)(x)=f(x)+g(x).$$

This is the same as multiplying $x^a\cdot x^b=x^{a+b}$.

Now for any $f\in ℱ(𝕏)$ define

$$M(h)=\{(f,g)\in ℱ{(𝕏)}^2|h=fg\},$$

Now if $F,G\in R[𝕏]$ then we define the multiplication

$$FG:ℱ(𝕏)\to R$$

by putting

$$(FG)(h)=\sum _{(f,g))\in M(h)}F(f)G(g).$$

Note that all of this well-defined, since both $F$ and $G$ vanish almost everywhere.

It can be shown that $R[𝕏]$ with these operations is a ring, even an $R$-algebra. This algebra is commutative if and only if $R$ is. Furthermore we have an algebra homomorphism

$$E:R\to R[𝕏]$$

which is defined as follows: for any $r\in R$ let $F_r:ℱ(𝕏)\to R$ be the function such that if $f:𝕏\to \mathbb{N}$ is such that $f(x)=0$ for any $x\in 𝕏$, then put $F_r(f)=r$ and for any other function $f\in ℱ(𝕏)$ put $F_r(f)=0$. Then

$$E(r)=F_r$$

is our function, which is a monomorphism. Furthermore if $R$ is unital with the identity $1$, then

$$E(1)$$

is the identity in $R[𝕏]$. Anyway we can always interpret $R$ as a subset of $R[𝕏]$ if put $r=F_r$ for $r\in R$.

Note, that if $𝕏=\mathrm{\varnothing }$, then $R[\mathrm{\varnothing }]$ is still defined and $E:R\to R[𝕏]$ is an isomorphism of rings (it is ,,onto”). Actually these two conditions are equivalent.

Also note, that $𝕏$ itself can be interpreted as a subset of $R[𝕏]$. Indeed, for any $x\in 𝕏$ define

$$f_x:𝕏\to \mathbb{N}$$

by $f_x(x)=1$ and $f_x(y)=0$ for any $y\ne x$. Then define

$$F_x:ℱ(𝕏)\to R$$

by putting $F_x(f_x)=1$ and $F_x(f)=0$ for any $f\ne f_x$. It can be easily seen that $F_x=F_y$ if and only if $x=y$. Thus we will use convention $x=F_x$.

With these notations (i.e. $R,𝕏\subseteq R[𝕏]$) we have that elements of $R[𝕏]$ are exactly polynomials in the set of variables $𝕏$ with coefficients in $R$.