Theorem 3, Section 3.5 of Hungerford’s Abstract Algebra

Question

Let $R$ be a ring and denote by $R[x_1 ,…, x_n]$ the set of all functions $f:N^n \to R$ such that $f(u)\neq 0$ for at most a finite number of elements $u$ of $N^n$.

(i) $R[x_1,…,x_n]$ is a ring with addition and multiplication defined by $$(f+g)(u)=f(u)+g(u)\text{ and } (fg)(u)=\sum_{v+w=u}f(v)g(w)$$ where $f,g \in R[x_1,…,x_n]$ and $u,v,w \in N^n$.

Proof: Exercise

Here $N=\Bbb{N}\cup \{0\}$. I am trying to show $R[x_1,…,x_n]$ is indeed a ring. To be specific, $R[x_1,…,x_n]$ satisfy the axioms of multiplication for a ring.

Below I have written proof of multiplication is well defined, closed under multiplication, left and right distributive law. I can’t prove multiplication is associative. Let $f,g,h\in R[x_1,…,x_n]$. Then $$\begin{align}[(fg)h](u) &= \sum_{v+w=u}(fg)(v)h(w)\\ &=\sum_{v+w=u}\left(\sum_{p+q=v}f(p)g(q)\right)h(w) \end{align}$$ How to progress from here?

Answer 1

We have to show number of terms in the summation, $\sum_{v+w=u}$, is finite because then only defined multiplication make sense. Let $u=(u_1,…,u_n)\in N^n$. Let’s consider first component of $u$, that is $u_1$. There are total $u_1+1$ pairs of natural number $(v_1,w_1)\in N^2$ such that $v_1+w_1=u_1$. Those $u_1+1$ pairs are following: $(0,u_1), (1, u_1-1), \cdots , (j,u_1-j), \cdots ,(u_1,0)$. Similarly, for all $1\lt i\leq n$, there are total $u_i+1$ pairs of natural number $(v_i,w_i)\in N^2$ such that $v_i+w_i=u_i$. Since $(u_1,…,u_n)=(v_1+w_1,…,v_n+w_n)$, there are $u_i+1$ choices for $i$th component. Therefore number of terms in the summation, $\sum_{v+w=u}$, is $\prod_{i=1}^n(u_i+1)$.

Now we will show $fg\in R[x_1,…,x_n]$. Let $\{u\in N^n \mid (fg)(u)\neq 0\}=\{u_i\in N^n\mid i\in I\}$, where $I$ is an indexing set. Then $$(fg)(u_i)=\sum_{v_i+w_i=u_i}f(v_i)g(w_i)\neq 0$$ implies $\exists v_i,w_i\in N^n$ such that $v_i+w_i=u_i$ and $f(v_i)g(w_i)\neq 0$. So $f(v_i)\neq 0$ and $g(w_i)\neq 0$, for all $i\in I$. By definition of $f$ and $g$, $|\{v\in N^n\mid f(v)\neq 0\}|$ and $|\{w\in N^n\mid g(w)\neq 0\}|$ are finite. So $\{v_i\in N^n\mid i\in I\}=\{v_{i_1},…,v_{i_k}\}$ and $\{w_i\in N^n\mid i\in I\}=\{w_{j_1},…,w_{j_m}\}$ where $k,m\in \Bbb{N}$. Since $u_i=v_i+w_i=v_{i_p}+w_{j_q}$ ($1\leq p\leq k$ and $1\leq q\leq m$), we have $|\{u_i\in N^n\mid i\in I\}|$ is finite. Thus $fg\in R[x_1,…,x_n]$.

We show multiplication is left distributive. Let $f,g,h\in R[x_1,…,x_n]$. Then $$\begin{align} [f(g+h)](u) &= \sum_{v+w=u}f(v)(g+h)(w)\\ &= \sum_{v+w=u}f(v) (g(w)+h(w)) \\ &= \sum_{v+w=u}f(v)g(w)+\sum_{v+w=u}f(v)h(w)\\ &= (fg)(u)+(fh)(u) \\ &= (fg+fh)(u) \end{align}$$ Thus $f(g+h)=fg+fh$. Proof is similar for right distributive.

Edit: Since comments are temporary, I am adding proof of associativity. Let $f,g,h\in R[x_1,…,x_n]$. Then $$\begin{align}[(fg)h](u) &= \sum_{v+w=u}(fg)(v)h(w)\\ &=\sum_{v+w=u}\left(\sum_{p+q=v}f(p)g(q)\right)h(w) \\ &= \sum_{v+w=u} \sum_{p+q=v}f(p)g(q)h(w) \\ &= \sum_{p+q+w=u}f(p)g(q)h(w) \\ &= \sum_{p+l=u}\sum_{q+w=l} f(p)g(q)h(w) \\ &= \sum_{p+l=u} f(p)\left( \sum_{q+w=l}g(q)h(w)\right) \\ &= \sum_{p+l=u} f(p) (gh)(l) \\ &= [f(gh)](u) \end{align}$$ for all $u\in N^n$. Thus $f(gh)=(fg)h$.