# explicit definition of polynomial rings in arbitrarly many variables

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

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

If $\mathbb{X}=\{X_{1},\ldots,X_{n}\}$ then the elements of $\mathcal{F}(\mathbb{X})$ can be interpreted as monomials

 $X_{1}^{\alpha_{1}}\cdots X_{n}^{\alpha_{n}}.$

Now define

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

The addition in $R[\mathbb{X}]$ is defined as pointwise addition.

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

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

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

Now for any $f\in\mathcal{F}(\mathbb{X})$ define

 $M(h)=\{(f,g)\in\mathcal{F}(\mathbb{X})^{2}\ |\ h=fg\},$

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

 $FG:\mathcal{F}(\mathbb{X})\to R$

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 $R[\mathbb{X}]$ 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

 $E:R\to R[\mathbb{X}]$

which is defined as follows: for any $r\in R$ let $F_{r}:\mathcal{F}(\mathbb{X})\to R$ be the function such that if $f:\mathbb{X}\to\mathbb{N}$ is such that $f(x)=0$ for any $x\in\mathbb{X}$, then put $F_{r}(f)=r$ and for any other function $f\in\mathcal{F}(\mathbb{X})$ 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 $R[\mathbb{X}]$. Anyway we can always interpret $R$ as a subset of $R[\mathbb{X}]$ if put $r=F_{r}$ for $r\in R$.

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

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

 $f_{x}:\mathbb{X}\to\mathbb{N}$

by $f_{x}(x)=1$ and $f_{x}(y)=0$ for any $y\neq x$. Then define

 $F_{x}:\mathcal{F}(\mathbb{X})\to R$

by putting $F_{x}(f_{x})=1$ and $F_{x}(f)=0$ for any $f\neq f_{x}$. 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. $R,\mathbb{X}\subseteq R[\mathbb{X}]$) we have that elements of $R[\mathbb{X}]$ are exactly polynomials in the set of variables $\mathbb{X}$ with coefficients in $R$.

Title explicit definition of polynomial rings in arbitrarly many variables ExplicitDefinitionOfPolynomialRingsInArbitrarlyManyVariables 2013-03-22 19:18:10 2013-03-22 19:18:10 joking (16130) joking (16130) 8 joking (16130) Definition msc 12E05 msc 13P05 msc 11C08